// Numbas version: exam_results_page_options {"name": "Maths and Stats Week 3 Knowledge Check", "metadata": {"description": "

This is a formative assessment to see how well you have understood the content from the first 3 weeks of the course. It will not count towards your final mark. If you get less than 40%, then you should take immediate steps to improve your knowledge, such as attending Study Clinics.

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Random integer.

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Random number with 7 decimal places.

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Number of significant figures to round.

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Round $\\var{d1}$ to 1 and {sf} significant figures.

\n

i) $\\var{d1}$ rounded to 1 significant figure is:  [[0]]

\n

ii) $\\var{d1}$ rounded to {sf} significant figures is:  [[1]]

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Round $\\var{e1}$ to 1 and {sf} significant figures.

\n

iii) $\\var{e1}$ rounded to 1 significant figure is:  [[0]]

\n

iv) $\\var{e1}$ rounded to {sf} significant figures is:  [[1]]

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Round numbers to a given number of significant figures.

"}, "preamble": {"css": "", "js": ""}, "advice": "

The first thing to do when we are rounding numbers is to identify the last digit we are keeping.

\n

When you're asked to round your answer to a number of significant figures, you need to decide whether to keep the last digit same (rounding down) or increase it by 1 (rounding up). If the following digit is less than 5 we round down and we round up when the next digit is 5 or more.

\n

To write it down in steps:

\n
    \n
      \n
        \n
          \n
        1. Identify the last digit we need to keep.
        2. \n
        3. Look at the following digit.
        4. \n
        5. If it's 5 or more, increase the previous digit by one.
        6. \n
        7. If it's 4 or less, keep the previous digit the same.
        8. \n
        9. Fill any spaces to the right of the digit with zeros if needed.
        10. \n
        \n
      \n
    \n
\n

It is important to keep in mind that if the digit we are increasing is 9, it becomes zero and we increase the previous digit instead. If this digit is 9 as well, we move along to the left side until we find a digit less than 9.

\n

When asked to round to significant figures, it may seem tricky at first to identify the last digit we want to keep. If we round to one significant figure, we look at the first non-zero digit from left to right. If we round to $n$ significant figures, we want to keep all digits up to the $n$th number of places next to the last zero number. Once we identify this digit, the rounding method is the same as previously. It may be easier to understand with an example:

\n

$0.02554$ rounded to 2 significant figures is $0.026$ as digits $25$ are the 2 significant figures we keep.

\n

Similarly, $52124$ rounded to 3 significant figures is $52100$ since digits $521$ are the 3 significant figures we keep.

\n

a)

\n

i)

\n

We round $\\var{d1}$ to 1 significant figure. The first non-zero digit is $\\var{ddig[5]}$. The following digit is $\\var{ddig[4]}$ so we round updown to get $\\var{dpformat(siground(d1, 1), 0)}$.

\n

ii)

\n

We round $\\var{d1}$ to {sf} significant figures. The first non-zero digit is $\\var{ddig[5]}$. The second following digit is $\\var{ddig[4]}$, the third following digit is $\\var{ddig[3]}$ and the fourth following digit is $\\var{ddig[2]}$. The digit following the last digit we are keeping is $\\var{ddig[3]}$$\\var{ddig[2]}$$\\var{ddig[1]}$, so we round to get $\\var{sigformat(d1, sf)}$. These are our {sf} significant figures. 

\n

\n

b)

\n

i)

\n

We round $\\var{e1}$ to 1 significant figure. The first non-zero digit is $\\var{edig[4]}$, followed by $\\var{edig[3]}$. This is lower than 5 so we round downmore than 5 so we round up to get $\\var{sigformat(e1,1)}$.

\n

ii)

\n

We round $\\var{e1}$ to {sf} significant figures. The first non-zero digit is $\\var{edig[4]}$. The second following digit is $\\var{edig[3]}$, the third following digit is $\\var{edig[2]}$ and the fourth following digit is $\\var{edig[1]}$. The digit following the last digit we are keeping is $\\var{edig[2]}$$\\var{edig[1]}$$\\var{edig[0]}$, so we round to get $\\var{sigformat(e1, sf)}$. These are our {sf} significant figures. 

\n

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Given some numbers in standard index form, convert to decimal form.

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When given a number in the form $A \\times 10^n$, we can think of $n$ as a number telling us how many places to move the decimal point.

\n

When $n$ is positive, we move the decimal point to the right side, for example:

\n

\\[ 1.5 \\times 10^3 = 1500.0 \\text{ .} \\]

\n

When $n$ is negative, we move the decimal point to the left side, for example:

\n

\\[ 1.5 \\times 10^{-3} = 0.0015 \\text{ .} \\]

\n

When $n = 0$, we do not move the decimal point:

\n

\\[ 1.5 \\times 10^0 = 1.5 \\text{ .}\\]

\n

 

\n

a)

\n

In $\\var{A[0]} \\times 10^\\var{ran}$, $n = \\var{ran}$ and so we move the decimal point {ran} places to the right.

\n

\\[\\var{A[0]}  ⇒  \\var{precround((A[0] * 10^ran), 0)}\\]

\n

\n

b)

\n

In $\\var{A[1]} \\times 10^\\var{-ran + 4}$, $n = \\var{-ran +4}$ and so we move the decimal point {ran -4} places to the left.

\n

\\[\\var{A[1]}  ⇒  \\var{A[1]*10^(-ran+4)}\\]

", "statement": "

Write the following in decimal form.

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$\\var{A[0]} \\times 10^\\var{ran} =$  [[0]]

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$\\var{A[1]} \\times 10^\\var{-ran + 4} =$  [[0]]

\n

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Working with numbers that are very large or very small can be tricky.

\n

Standard form allows us to simplify these numbers, using powers of 10.

\n
The standard index form can be defined as
\n
\\[A \\times 10^n,\\]
\n
where $1 ≤ A < 10$ and $n$ is an integer, e.g. $2.26 \\times 10^5$ is a standard form of a number 226000.
\n
\n

 

\n

Write the following in standard index form (for example, for $2.01\\times 10^5$ we would write 2.01*10^5 in the gap).

\n
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$\\var{A[2]*10^2} = $  [[0]]

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$\\var{A3dp*10^(-1)} = $  [[0]]

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$\\var{precround(A5dp*10^7,0)} =$  [[0]]

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$\\var{small5} = $  [[0]]

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Convert a variety of numbers from decimal to standard index form.

"}, "preamble": {"css": "", "js": ""}, "advice": "

Converting from decimal to a standard form, we are looking for $A \\times 10^n$.

\n

We need make the first number ($A$) between 1 and 10, so we put the decimal place after the first non-zero digit.

\n

 

\n

a)

\n

In $\\var{A[2]*10^2}$, the first non-zero digit is $\\var{siground(A[2] - 0.5, 1)}$ so we get $A = \\var{A[2]}$.

\n

If we moved the decimal place in $\\var{A[2]}$ so it matches our original number $\\var{A[2]*10^2}$, we would go 2 places to the right, so $n = 2$.

\n

 

\n

b)

\n

In $\\var{A3dp*10^(-1)}$, the first non-zero digit is $\\var{siground(A3dp - 0.5, 1)}$ so we get $A = \\var{A3dp}$.

\n

If we moved the decimal place in $\\var{A3dp}$ so it matches our original number $\\var{A3dp*10^(-1)}$, we would go 1 place to the left, so $n = -1$.

\n

 

\n

c)

\n

In $\\var{precround(A5dp*10^7,0)}$ the first non-zero digit is $\\var{siground(A5dp - 0.5, 1)}$ so we get $A = \\var{A5dp}$.

\n

If we moved the decimal place in $\\var{A5dp}$ so it matches our original number $\\var{precround(A5dp*10^7,0)}$, we would go 7 places to the right, so $n = 7$.

\n

 

\n

d)

\n

In $\\var{small5}$ the first non-zero digit is {siground({{small5}*10^5} - 0.5, 1)} so we get $A = \\var{small5*10^5}$.

\n

If we moved the decimal place in $\\var{small5*10^5}$ so it matches our original number $\\var{small5}$, we would go 5 places to the left, so $n = -5$.

\n

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Multiply two numbers in standard form, then divide two numbers in standard form.

", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "

Simplify, leaving the result in standard index form.

", "advice": "

We use the rules for multiplying and dividing powers.

\n

To multiply powers we add, for example $10^2 \\times 10^6 = 10^{(2+6)} = 10^{8}$.

\n

To divide powers we subtract, for example $10^2 \\div 10^6 = 10^{(2-6)} = 10^{-4}$.

\n

 

\n

a)

\n

\\[ \\begin{align} (\\var{int} \\times 10^\\var{ran - 3} ) \\times (\\var{int - 11} \\times 10^\\var{ran})  &= \\var{int} \\times (\\var{int-11}) \\times 10^{(\\var{ran - 3} + \\var{ran})} \\\\&= (\\var{int*(int-11)}) \\times 10^{\\var{ran - 3 + ran}} \\end{align} \\]

\n

We want to write this number in standard form so we move decimal place one place to the left to get

\n

\\[(\\var{int*(int-11)/10}) \\times 10^{(\\var{ran - 3 + ran} + 1)} = (\\var{int*(int-11)/10}) \\times 10^{\\var{ran - 2 + ran}} \\text{.}\\]

\n

 

\n

b)

\n

Similarly,

\n

\\[ \\begin{align} (\\var{int*int} \\times 10^\\var{ran -4} ) \\div (\\var{int} \\times 10^\\var{ran - 2})  &= \\var{int*int} \\div \\var{int} \\times 10^{(\\var{ran - 4} - \\var{ran -2})} \\\\&= \\var{int} \\times 10^{-2} \\text{.}\\end{align} \\]

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$(\\var{int} \\times 10^\\var{ran - 3} ) \\times (\\var{int - 11} \\times 10^\\var{ran})  =$  [[0]]

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$(\\var{int*int} \\times 10^\\var{ran -4} ) \\div (\\var{int} \\times 10^\\var{ran - 2})  =$  [[0]]

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{number} moles of a solute are dissolved to create {litres} L of solution. The molar concentration is [[0]] mol/L (to one decimal place).

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Molar concentration is the number of moles divided by the volume of solution in litres.

\n

In particular, if a {litres} L solution is made by mixing {number} moles of a solute then the molar concentration is calculated as follows:

\n

\n\n\n\n\n\n\n\n\n\n\n\n\n\n
$\\dfrac{\\text{number of moles}}{\\text{litres}}$$=$$\\dfrac{\\var{number}}{\\var{litres}}$ mol/L
$=$$\\var{onedec}$ mol/L  (1 dec. pl)
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This question determines the students ability to understand the relationship between moles and concentration.

", "advice": "

The formula mass of NaOH is 40 g/mol. That is, 40 g of NaOH dissolved into 1000ml (1 litre) of solvent will give a 1 Molar solution (1 M). We need only {vol} mL ({decimal(vol/1000)} L) at {molar} M so 40 x {decimal(vol/1000)} L x {molar} M = {correct} g.

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How many grams of NaOH would you need to make {vol} mL of a {molar} M solution, given the molecular weight of NaOH is 40 g/mol?

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Never seen before.

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Practice calculating number of moles of a substance given the concentration and volume of a solution.

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How many moles of glucose are there in $\\var{a}$L of a $\\var{0.25 * b}$M (mol/L) solution?

\n

[[0]] moles

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How many moles of glucose are there in $\\var{25 * c}$mL of a $\\var{0.25 * d}$M (mol/L) solution?

\n

[[0]] moles

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To answer these questions, we use the formula

\n

$\\text{volume of liquid (in litres)} \\times \\text{concentration (in mol/L)} = \\text{number of moles of substance}$.

\n

a)

\n

How many moles of glucose are there in $\\var{a}$L of a $\\var{0.25 * b}$M (mol/L) solution?

\n

Solution:

\n

Putting our numbers into the formula, we find that there are

\n

$\\var{a} \\times \\var{0.25 * b} = \\var{a * 0.25 * b}$ moles

\n

of glucose in $\\var{a}$L of a $\\var{0.25 * b}$M (mol/L) solution.

\n

b)

\n

How many moles of glucose are there in $\\var{25 * c}$mL of a $\\var{0.25 * d}$M (mol/L) solution?

\n

Solution:

\n

Our formula uses the volume of liquid in litres so first we have to convert $\\var{25 * c}$mL to a volume in litres. There are 1000ml in 1L so $\\var{25 * c}$mL is equal to 

\n

$\\dfrac{\\var{25 * c}}{1000} = \\var{25 * c / 1000}$L.

\n

Putting our numbers into the formula, we find that there are

\n

$\\begin{align}\\var{25 * c / 1000} \\times \\var{0.25 * d} & = \\var{(25 * c / 1000) * 0.25 * d} \\text{ moles} \\\\ & = \\var{precround(((25 * c / 1000) * 0.25 * d),2 )} \\text{ moles to 2 d.p.}\\end{align}$

\n

\n

of glucose in $\\var{25 * c}$mL of a $\\var{0.25 * d}$M (mol/L) solution.

", "functions": {}, "statement": "

Answer the following questions. Please enter your answers as decimals, not as fractions. Give your answers to 2 decimal places.

\n

If you would like to see how to do this question, click on 'Reveal answers' at the bottom of the page.

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Solve the simultaneous equations:

\n

$x+2y=\\var{x+2y}$

\n

$x^2+y^2=\\var{x^2+y^2}$

", "advice": "

Rearranging the first equation gives $x=\\var{x+2y}-2y$

\n

Substituting this expression for $x$ into our second equation gives $(\\var{x+2y}-2y)^2+y^2=\\var{x^2+y^2}$

\n

Expanding brackets gives $\\simplify[]{{(x+2y)^2}-{4*(x+2y)}}$ $y+4y^2+y^2=\\var{x^2+y^2}$

\n

Rearranging gives $\\simplify[]{5y^2-{4*(x+2y)}y-{x^2+y^2-(x+2y)^2}}=0$

\n

Solving by factorising or using the quadratic formula gives $y=\\frac{\\var{4a} \\pm \\sqrt{\\var{16a^2+20(b-a^2)}}}{10}$

\n

Hence $y = \\var{y1}$ or $\\var{y2}$

\n

and the corresponding values of $x$ are $\\var{x1}$ and $\\var{x2}$ respectively (found by substituting each value of $y$ into $x+2y=\\var{x+2y}$)

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Lower value of $y=$[[0]]

\n

Corresponding value of $x=$[[1]]

\n

\n

Greater value of $y=$[[2]]

\n

Corresponding value of $x=$[[3]]

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Quadratic equations of the form

\n

\\[x^2+bx+c=0\\]

\n

can be factorised to create an equation of the form

\n

\\[(x+m)(x+n)=0\\text{.}\\]

\n

When we expand a factorised quadratic expression we obtain

\n

\\[(x+m)(x+n)=x^2+(m+n)x+(m \\times n)\\text{.}\\]

\n

To factorise an equation of the form $x^2+bx+c$, we need to find two numbers which add together to make $b$, and multiply together to make $c$.

\n

a)

\n

\\[\\simplify{x^2+{v1+v2}x+{v1*v2}=0}\\]

\n

We need to find two values that add together to make $\\var{v1+v2}$ and multiply together to make $\\var{v1*v2}$.

\n

\\[\\begin{align}
\\var{v1} \\times \\var{v2}&=\\var{v1*v2}\\\\
\\var{v1}+\\var{v2}&=\\var{v1+v2}\\\\
\\end{align} \\]

\n

So the factorised form of the equation is

\n

\\[\\simplify{(x+{v1})(x+{v2})}=0\\text{.}\\]

\n

\n

b)

\n

We can begin factorising by finding factors of $\\var{v3*v4}$ that add together to give $\\var{v3+v4}$.

\n

\\[\\begin{align}
\\var{v3} \\times \\var{v4}&=\\var{v3*v4}\\\\
\\var{v3}+\\var{v4}&=\\var{v3+v4}\\\\
\\end{align} \\]

\n

So the factorised form of the equation is

\n

\\[\\simplify{(x+{v3})(x+{v4})}=0\\text{.}\\]

\n

c)

\n

When factorising the quadratic expression

\n

\\[\\simplify{x^2+{v5*v6}=0}\\]

\n

we need to find two values that add together to make $0$ and multiply together to make $\\var{v5*v6}$.

\n

\\begin{align}
\\var{v5} \\times \\var{v6}& = \\var{v5*v6}\\\\
\\simplify[]{ {v5} + {v6}} &= 0 \\\\
\\end{align}

\n

So the factorised form of the equation is

\n

\\[\\simplify{(x+{v5})(x+{v6})}=0\\text{.}\\]

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Factorise the following quadratic equations.

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$\\simplify{x^2+{v1+v2}x+{v1*v2}=0}$

\n

[[0]] $=0$

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$\\simplify{x^2+{v3+v4}x+{v3*v4}}=0$

\n

[[0]] $=0$

\n

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$\\simplify{x^2+{v5*v6}}=0$

\n

[[0]] $=0$

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Factorise three quadratic equations of the form $x^2+bx+c$.

\n

The first has two negative roots, the second has one negative and one positive, and the third is the difference of two squares.

"}, "ungrouped_variables": [], "variable_groups": [{"variables": ["v1", "v2", "v3", "v4", "v5", "v6"], "name": "Part A "}]}, {"name": "Simultaneous Equations", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Mark Patterson", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/5064/"}], "tags": [], "metadata": {"description": "

Exercises in solving simultaneous equations with 2 variables.

", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "

Simultaneous Equations

\n

\n

Solve the following simultaneous equations.

", "advice": "

To solve simultaneous equations, first rewrite the equations in matrix form of $Ab = C$, where the matrix $A$ will contain the coefficients of the equation, $b$ will contain the unknown values and $C$ will contain the constants.

\n

We rearrange $Ab = C$ to get $A^{-1}C = b$.

\n

So, we need to find the inverse of $A$ and multiple it with $C$. This will give us $b$, i.e. the values for $x$ and $y$.

\n

\n

For the first question:

\n

$A = \\simplify{{maA}}$  $b = \\begin{pmatrix}x\\\\y\\end{pmatrix}$ $C = \\begin{pmatrix}\\var{a1}\\\\ \\var{a2}\\end{pmatrix}$

\n

To find the inverse of $A$, switch the elements on the leading diagonal, change the signs on the non-leading diagonal, and divide all elements by the determinant.

\n

The determinant of $A$ is $|A| = (\\var{maA[0][0]}\\times\\var{maA[1][1]}) - (\\var{maA[0][1]}\\times\\var{maA[1][0]}) = \\var{detA}$

\n

The inverse of $A$ is $A^{-1} = \\dfrac{1}{\\var{detA}}\\times\\begin{pmatrix}
\\var{maAA[0][0]} & \\var{maAA[0][1]} \\\\
\\var{maAA[1][0]} & \\var{maAA[1][1]}
\\end{pmatrix} =
\\begin{pmatrix}
\\frac{\\var{maAA[0][0]}}{\\var{detA}} & \\frac{\\var{maAA[0][1]}}{\\var{detA}} \\\\
\\frac{\\var{maAA[1][0]}}{\\var{detA}} & \\frac{\\var{maAA[1][1]}}{\\var{detA}}
\\end{pmatrix}$

\n

\n

Rearranging for $A^{-1}C = b$:

\n

$A^{-1}\\times C = 
\\begin{pmatrix}
\\frac{\\var{maAA[0][0]}}{\\var{detA}} & \\frac{\\var{maAA[0][1]}}{\\var{detA}} \\\\
\\frac{\\var{maAA[1][0]}}{\\var{detA}} & \\frac{\\var{maAA[1][1]}}{\\var{detA}}
\\end{pmatrix}
\\times\\begin{pmatrix}\\var{a1}\\\\ \\var{a2}\\end{pmatrix} =
\\begin{pmatrix}
\\frac{\\var{maAA[0][0]}}{\\var{detA}}\\times\\var{a1} + \\frac{\\var{maAA[0][1]}}{\\var{detA}}\\times\\var{a2} \\\\
\\frac{\\var{maAA[1][0]}}{\\var{detA}}\\times\\var{a1} + \\frac{\\var{maAA[1][1]}}{\\var{detA}}\\times\\var{a2}
\\end{pmatrix} =
\\begin{pmatrix}\\var{ax}\\\\ \\var{ay}\\end{pmatrix}$

\n

Therefore $x = \\var{ax}$ and $y = \\var{ay}$.

\n

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$\\simplify{{maA[0][0]}}x + \\simplify{{maA[0][1]}}y = \\simplify{{a1}}$

\n

$\\simplify{{maA[1][0]}}x +\\simplify{{maA[1][1]}}y = \\simplify{{a2}}$

\n

\n

$x =$ [[0]]

\n

$y =$ [[1]]

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$\\simplify{{maB[0][0]}}x + \\simplify{{maB[0][1]}}y = \\simplify{{b1}}$

\n

$\\simplify{{maB[1][0]}}x +\\simplify{{maB[1][1]}}y = \\simplify{{b2}}$

\n

\n

$x =$ [[0]]

\n

$y =$ [[1]]

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rearranging the Michelas-Menten equation to make the substrate the subject.

\n

rebelmaths

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Rearrange the following equation to make S the subject.

\n

\n

$ V=\\frac{\\var{a}S}{S+\\var{b}}$

\n

\n

to write a fraction you type (numerator)/(denominator)

\n

S=[[0]]

\n

", "extendBaseMarkingAlgorithm": true}], "rulesets": {}, "tags": [], "advice": "

start by multiplying both sides by the denominator

\n

for example if you have $V=\\frac{5S}{S+12}$ then multiply both sides by $(S+12)$

\n

this gives:  $V(S+12)=\\frac{5S}{S+12} (S+12) $

\n

the (S+12) term on the right hand side cancels out to give: $V(S+12)=5S$

\n

now expand out the brackets:  $VS+12V=5S$

\n

then collect the like terms, you want to get all the terms with S in them onto one side, so subtract VS from both sides:

\n

$VS-VS+12V=5S-VS$

\n

this becomes $12V=5S-VS$

\n

now you can factorise the right hand side: $12V=S(5-V)$

\n

then divide both sides by (5-V) to leave S on its own: $\\frac{12V}{5-V}=S$

\n

"}, {"name": "Laws of Indices", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Christian Lawson-Perfect", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/7/"}, {"name": "Chris Graham", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/369/"}, {"name": "Elliott Fletcher", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/1591/"}], "tags": ["indices", "laws of indices", "powers", "taxonomy"], "metadata": {"description": "

This question aims to test understanding and ability to use the laws of indices.

", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "

Using the laws of indices, simplify each expression down to its simplest form. Recall that $a^{0} = 1$ for any number $a$.

", "advice": "

a)

\n

Here we are using the rule of indices: $a^m \\times a^n = a^{m+n}$.

\n

Using this rule, 

\n

\\[
\\begin{align}
a^\\var{x} \\times a^\\var{y}\\ &= a^\\simplify[all, !collectNumbers]{{x}+{y}}\\\\
&= a^\\var{x+y}.
\\end{align}
\\]

\n

b)

\n

We are asked to find $\\var{c}a^\\var{p} \\times \\var{d}a^\\var{q}$.

\n

Notice there is a constant in front of each of the terms.

\n

To do this, write the product out explicitly, as

\n

\\[\\var{c}a^\\var{p} \\times \\var{d}a^\\var{q} = \\var{c} \\times \\var{d} \\times a^\\var{p} \\times a^\\var{q}.\\]

\n

We know that $\\var{c} \\times \\var{d} = \\var{c*d}$, and using the rule of indices: $a^\\var{p} \\times a^\\var{q} = a^\\var{p+q}$.

\n

Therefore:

\n

\\begin{align}
\\var{c}a^\\var{p} \\times \\var{d}a^\\var{q}&= \\var{c*d} \\times a^\\var{p+q} \\\\
&= \\simplify{{c*d}*a^{p+q}}.
\\end{align}

\n

c)

\n

Here we are using: $a^m \\div a^n = a^{m-n}$.

\n

We are asked to simplify the expression, $\\displaystyle\\simplify{{b}*a^{x}/({g}*a^{y})}$.

\n

To do this, we just have to use the previously mentioned rule of indices. We write this out explicity as

\n

\\[\\simplify{{b}*a^{x}/({g}*a^{y})} = \\simplify{{b}/{g}} \\times \\simplify{a^{x}/(a^{y})}.\\]

\n

Using rules of indices,

\n

\\begin{align}                                                                                                                                                                                                                                                                                           \\frac{a^\\var{x}}{a^\\var{y}} &= a^\\var{x} \\div a^\\var{y}\\\\
&= a^\\simplify[all, !collectNumbers]{{x}-{y}}\\\\
&= a^\\var{x-y}.
\\end{align}

\n

Therefore,

\n

\\begin{align}
\\frac{\\var{b}a^\\var{x}}{\\var{g}a^\\var{y}} &= \\simplify{{b}/{g}} \\times \\simplify{a^{{x}-{y}}}\\\\
&= \\simplify{{b}/{g}*a^{x-y}}.
\\end{align}

\n

Alternatively, 

\n

Using the rule of indices: $a^{-m}  = \\displaystyle\\frac{1}{a^{m}}$, we can rewrite the question as:

\n

\\begin{align}
\\frac{\\var{b}a^\\var{x}}{\\var{g}a^\\var{y}} &= \\simplify{{b}/{g}} \\times \\frac{a^\\var{x}}{a^\\var{y}}\\\\
&= \\simplify{{b}/{g}} \\times a^\\var{x} \\times a^{-\\var{y}}.
\\end{align}

\n

And then using the rule: $a^m \\times a^n = a^{m+n}$, this becomes:

\n

\\begin{align}
\\simplify{{b}/{g}} \\times a^\\var{x} \\times a^{-\\var{y}} &= \\simplify{{b}/{g}} \\times a^\\simplify[all,!collectNumbers]{{x}+(-{y})}\\\\
&= \\simplify{{b}/{g}*a^{x-y}}.
\\end{align}

\n

d)

\n

The question asks us to simplify $(\\simplify{{c}*a^{p}})^{\\var{q}}$.

\n

To do this we use the rules:

\n

\\[(a^{m})^{n} = a^{mn},\\]

\n

\\[(ab)^m = a^mb^m.\\]

\n

We can then expand the equation as

\n

\\[(\\simplify{{c}*a^{p}})^{\\var{q}}= \\var{c}^{\\var{q}} \\times (a^{\\var{p}})^{\\var{q}}.\\]

\n

Then using the rule of indices mentioned previously,

\n

\\[
\\begin{align}
(\\simplify{{c}*a^{p}})^{\\var{q}}&= \\simplify{{c}^{q}} \\times a^\\var{p*q}\\\\
&= \\simplify{{c}^{q}*a^{p*q}}.
\\end{align}
\\]

\n

e)

\n

The question asks us to simplify $\\sqrt[\\var{d}]{\\var{x}^\\var{d}a}$.

\n

To do this we use the rules:

\n

\\[a^\\frac{1}{m} = \\sqrt[m]{a},\\]

\n

\\[(ab)^m = a^mb^m.\\]

\n

We can expand the expression as follows:

\n

\\[
\\begin{align}
\\sqrt[\\var{d}]{a} &= (\\simplify{a})^\\frac{1}{\\var{d}}\\\\
&= a^\\frac{1}{\\var{d}}.
\\end{align}
\\]

\n

f)

\n

The question requires us to simplify $\\sqrt[\\var{c}]{a^\\var{q}}$.

\n

Here, we use the rule of indices: $a^\\frac{n}{m} = \\sqrt[m]{a^n}$, allowing us to expand the expression as follows:

\n

\\[
\\begin{align}
\\sqrt[\\var{c}]{\\simplify{a^{q}}} &= \\simplify[fractionnumbers,all]{(a^{q})^{{1}/{{c}}}}\\\\
&= \\simplify[fractionnumbers,all]{a^{{q}/{c}}}.
\\end{align}
\\]

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Used in part c

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Used in parts b,d and f

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Used in parts a,c and e

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Used in parts b and e

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Used in parts b and d

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Used in parts b,d and f

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Used in part c

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\n

Used in parts a,c and f

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Write $a^{\\var{x}} \\times a^{\\var{y}}$ as a single power of $a$.

\n

\n

$a^{\\var{x}} \\times a^{\\var{y}} =$ [[0]].

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Use the rule: $a^m \\times a^n = a^{m+n}$.

"}], "gaps": [{"type": "jme", "useCustomName": false, "customName": "", "marks": "2", "showCorrectAnswer": true, "showFeedbackIcon": true, "scripts": {}, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "adaptiveMarkingPenalty": 0, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "answer": "a^{x+y}", "showPreview": true, "checkingType": "absdiff", "checkingAccuracy": 0.001, "failureRate": 1, "vsetRangePoints": 5, "vsetRange": [0, 1], "checkVariableNames": false, "mustmatchpattern": {"pattern": "a^?`?", "partialCredit": 0, "message": "You haven't simplified: your answer is not in the form $a^?$.", "nameToCompare": ""}, "valuegenerators": [{"name": "a", "value": ""}]}], "sortAnswers": false}, {"type": "gapfill", "useCustomName": false, "customName": "", "marks": 0, "showCorrectAnswer": true, "showFeedbackIcon": true, "scripts": {}, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "adaptiveMarkingPenalty": 0, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "prompt": "

Write $\\var{c}a^\\var{p} \\times \\var{d}a^\\var{q}$ as an integer multiplied by a single power of $a$.

\n

$\\var{c}a^\\var{p} \\times \\var{d}a^\\var{q} =$ [[0]].

\n

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Write $\\displaystyle\\simplify{{b}*a^{x}/({g}*a^{y})}$ as a number multiplied by a single power of $a$.

\n

$\\displaystyle\\simplify{{b}*a^{x}/({g}*a^{y})} =$ [[0]].

\n

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You could use one of the following rules:

\n

$a^m \\div a^n = a^{m-n}$.

\n

$a^{-m} = \\displaystyle\\frac{1}{a^m}$.

"}], "gaps": [{"type": "jme", "useCustomName": false, "customName": "", "marks": "2", "showCorrectAnswer": true, "showFeedbackIcon": true, "scripts": {}, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "adaptiveMarkingPenalty": 0, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "answer": "{b}/{g}a^{x-y}", "showPreview": true, "checkingType": "absdiff", "checkingAccuracy": 0.001, "failureRate": 1, "vsetRangePoints": 5, "vsetRange": [0, 1], "checkVariableNames": false, "mustmatchpattern": {"pattern": "$n/$n`? * a^?`?", "partialCredit": 0, "message": "You haven't simplified: your answer is not in the form $\\frac{?}{?} \\cdot a^?$.", "nameToCompare": ""}, "valuegenerators": [{"name": "a", "value": ""}]}], "sortAnswers": false}, {"type": "gapfill", "useCustomName": false, "customName": "", "marks": 0, "showCorrectAnswer": true, "showFeedbackIcon": true, "scripts": {}, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "adaptiveMarkingPenalty": 0, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "prompt": "

Write $(\\simplify{{c}*a^{p}})^{\\var{q}}$ as an integer multiplied by a single power of $a$.

\n

$(\\simplify{{c}*a^{p}})^{\\var{q}} =$ [[0]].

\n

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Use the rules:

\n

$(ab)^m = a^mb^m$.

\n

$(a^m)^n = a^{mn}$.

"}], "gaps": [{"type": "jme", "useCustomName": false, "customName": "", "marks": "2", "showCorrectAnswer": true, "showFeedbackIcon": true, "scripts": {}, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "adaptiveMarkingPenalty": 0, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "answer": "{c^{q}}*a^{p*q}", "showPreview": true, "checkingType": "absdiff", "checkingAccuracy": 0.001, "failureRate": 1, "vsetRangePoints": 5, "vsetRange": [0, 1], "checkVariableNames": false, "mustmatchpattern": {"pattern": "$n`?*a^?`?", "partialCredit": 0, "message": "You must write your answer as an integer multiplied by a power of $a$.", "nameToCompare": ""}, "valuegenerators": [{"name": "a", "value": ""}]}], "sortAnswers": false}, {"type": "gapfill", "useCustomName": false, "customName": "", "marks": 0, "showCorrectAnswer": true, "showFeedbackIcon": true, "scripts": {}, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "adaptiveMarkingPenalty": 0, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "prompt": "

Write $\\sqrt[\\var{d}]{a}$ as a single power of $a$. 

\n

$\\sqrt[\\var{d}]{a} =$ [[0]].

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Use the rule: $a^\\frac{1}{m} = \\sqrt[m]{a}$.

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You must input your answer as a single power of a.

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Write $\\sqrt[\\var{q}]{a^\\var{c}}$ as a single power of $a$.

\n

$\\sqrt[\\var{q}]{a^\\var{c}} =$ [[0]].

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Use the rule: $a^\\frac{n}{m} = \\sqrt[m]{a^n}$.

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You must input your answer as a single power of a.

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Finding the roots by factorisation.

\n

Finding a factorisation of a quadratic $q(x)=a(x-r)(x-s)$ where $a$ is the coefficient of $x^2$ gives the roots $x=r$, $x=s$ immediately.

\n

If you cannot find a factorisation then there are several other methods you can use.

\n

Using the formula for the roots.

\n

You can find the roots by using the formula for finding the roots of a general quadratic equation $ax^2+bx+c=0$

\n

The two roots are:

\n

\\[ x = \\frac{-b +\\sqrt{b^2-4ac}}{2a}\\mbox{ and } x = \\frac{-b -\\sqrt{b^2-4ac}}{2a}\\]
there are three main types of solutions depending upon the value of the discriminant $\\Delta=b^2-4ac$

\n

1. $\\Delta \\gt 0$. The roots are real and distinct

\n

2. $\\Delta=0$. The roots are real and equal. Their value is $\\displaystyle \\frac{-b}{2a}$

\n

3. $\\Delta \\lt 0$. There are no real roots. The root are complex and form a complex conjugate pair.

\n

 

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Solve for $x$:

\n

\\[\\simplify[std]{{a*b} * x ^ 2 + ( {-b*c-a * d}) * x + {c * d}}=0\\]

\n

$x=$ [[0]] or [[1]].

\n

You can get more information on solving a quadratic by clicking on Show steps. You will lose 1 mark if you do so.

\n

Enter the roots as fractions or integers, not as decimals.

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Input numbers as fractions or integers not as a decimals.

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Find the roots of the following quadratic equation.

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Solve for $x$: $\\displaystyle ax ^ 2 + bx + c=0$.

\n

Entering the correct roots in any order is marked as correct. However, entering one correct and the other incorrect gives feedback stating that both are incorrect.

"}, "advice": "\n\t

Direct Factorisation

\n\t

If you can spot a direct factorisation then this is the quickest way to do this question.

\n\t

For this example we have the factorisation

\n\t

\\[\\simplify{{a*b} * x ^ 2 + ( {-b*c-a * d}) * x + {c * d} = ({a} * x + { -c}) * ({b} * x + { -d})}\\]

\n\t

Hence we find the roots:
\\[\\begin{eqnarray} x&=& \\simplify{{n1-n4}/{2*a*b}}\\\\ x&=& \\simplify{{n1+n4}/{2*a*b}} \\end{eqnarray} \\]

\n\t

Other Methods.

\n\t

There are several methods of finding the roots – here are the main methods.

\n\t

Finding the roots of a quadratic using the standard formula.

\n\t

We can use the following formula for finding the roots of a general quadratic equation $ax^2+bx+c=0$

\n\t

The two roots are

\n\t

\\[ x = \\frac{-b +\\sqrt{b^2-4ac}}{2a}\\mbox{ and } x = \\frac{-b -\\sqrt{b^2-4ac}}{2a}\\]
there are three main types of solutions depending upon the value of the discriminant $\\Delta=b^2-4ac$

\n\t

1. $\\Delta \\gt 0$. The roots are real and distinct

\n\t

2. $\\Delta=0$. The roots are real and equal. Their common value is $\\displaystyle -\\frac{b}{2a}$

\n\t

3. $\\Delta \\lt 0$. There are no real roots. The root are complex and form a complex conjugate pair.

\n\t

For this question the discriminant of $\\simplify{{a*b}x^2+{-b*c-a*d}x+{c*d}}$ is $\\Delta = \\simplify[std]{{-n1}^2-4*{a*b*c*d}}=\\var{disc}$

\n\t

{rdis}.

\n\t

So the {rep} roots are:

\n\t

\\[\\begin{eqnarray} x = \\frac{\\var{n1} - \\sqrt{\\var{disc}}}{\\var{n3}} &=& \\frac{\\var{n1} - \\var{n4} }{\\var{n3}} &=& \\simplify{{n1 - n4}/ {n3}}\\\\ x = \\frac{\\var{n1} + \\sqrt{\\var{disc}}}{\\var{n3}} &=& \\frac{\\var{n1} + \\var{n4} }{\\var{n3}} &=& \\simplify{{n1 + n4}/ {n3}} \\end{eqnarray}\\]

\n\t

Completing the square.

\n\t

First we complete the square for the quadratic expression $\\simplify{{a*b}x^2+{-n1}x+{c*d}}$
\\[\\begin{eqnarray} \\simplify{{a*b}x^2+{-n1}x+{c*d}}&=&\\var{n5}\\left(\\simplify{x^2+({-n1}/{a*b})x+ {c*d}/{a*b}}\\right)\\\\ &=&\\var{n5}\\left(\\left(\\simplify{x+({-n1}/{2*a*b})}\\right)^2+ \\simplify{{c*d}/{a*b}-({-n1}/({2*a*b}))^2}\\right)\\\\ &=&\\var{n5}\\left(\\left(\\simplify{x+({-n1}/{2*a*b})}\\right)^2 -\\simplify{ {n2^2}/{4*(a*b)^2}}\\right) \\end{eqnarray} \\]
So to solve $\\simplify{{a*b}x^2+{-n1}x+{c*d}}=0$ we have to solve:
\\[\\begin{eqnarray} \\left(\\simplify{x+({-n1}/{2*a*b})}\\right)^2&\\phantom{{}}& -\\simplify{ {n2^2}/{4*(a*b)^2}}=0\\Rightarrow\\\\ \\left(\\simplify{x+({-n1}/{2*a*b})}\\right)^2&\\phantom{{}}&=\\simplify{ {n2^2}/{4*(a*b)^2}=({abs(n2)}/{2*a*b})^2} \\end{eqnarray}\\]
So we get the two {rep} solutions:
\\[\\begin{eqnarray} \\simplify{x+({-n1}/{2*a*b})}&=&\\simplify{-{abs(n2)}/{2*a*b}} \\Rightarrow &x& = \\simplify{({-abs(n2)+n1}/{2*a*b})}\\\\ \\simplify{x+({-n1}/{2*a*b})}&=&\\simplify{({abs(n2)}/{2*a*b})} \\Rightarrow &x& = \\simplify{({n1+abs(n2)}/{2*a*b})} \\end{eqnarray}\\]

\n\t \n\t \n\t \n\t \n\t \n\t \n\t \n\t \n\t \n\t \n\t \n\t"}]}], "navigation": {"allowregen": true, "reverse": true, "browse": true, "allowsteps": true, "showfrontpage": true, "showresultspage": "oncompletion", "navigatemode": "sequence", "onleave": {"action": "none", "message": ""}, "preventleave": false, "startpassword": ""}, "timing": {"allowPause": true, "timeout": {"action": "none", "message": ""}, "timedwarning": {"action": "none", "message": ""}}, "feedback": {"showactualmark": true, "showtotalmark": true, "showanswerstate": true, "allowrevealanswer": true, "advicethreshold": 0, "intro": "", "reviewshowscore": true, "reviewshowfeedback": true, "reviewshowexpectedanswer": true, "reviewshowadvice": true, "feedbackmessages": []}, "contributors": [{"name": "Paul Mackay", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/2697/"}], "extensions": [], "custom_part_types": [], "resources": []}