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You have 5 minutes to complete this quiz.

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Evaluate \\(f(\\var{a3})\\)

\n

\\(f(\\var{a3})\\) = [[0]]

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Evaluating a function

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\\(f(x)=\\var{a1}x^{\\var{a2}}+\\var{b1}x-\\var{c1}\\)

\n

\\(x=\\var{a3}\\)

\n

\\(f(\\var{a3})=\\var{a1}*(\\var{a3})^{\\var{a2}}+\\var{b1}*(\\var{a3})-\\var{c1}\\)

\n

\\(f(\\var{a3})=\\simplify{{a1}*{a3}^{{a2}}}+\\simplify{{b1}*{a3}}-\\var{c1}\\)

\n

\\(f(\\var{a3})=\\simplify{{a1}*{a3}^{{a2}}+{b1}*{a3}-{c1}}\\)

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Given the function:

\n

\\(f(x)=\\var{a1}x^{\\var{a2}}+\\var{b1}x-\\var{c1}\\)

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\\(\\displaystyle{\\lim_{x \\to \\var{a}}}\\frac{x-\\var{a}}{x^2-\\simplify{{a}+{b}}x+\\simplify{{a}*{b}}}\\)

\n

We can factorise the denominator

\n

\\(\\displaystyle{\\lim_{x \\to \\var{a}}}\\frac{x-\\var{a}}{(x-\\var{a})(x-\\var{b})}\\)

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Cancel out the common factor

\n

\\(\\displaystyle{\\lim_{x \\to \\var{a}}}\\frac{1}{x-\\var{b}}\\)

\n

Insert the value \\(\\var{a}\\) in for \\(x\\) to evaluate the limit

\n

\\(-\\frac{1}{\\simplify{{b}-{a}}}\\)

\n

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Input your answer as a fraction.

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Evaluate the following limit

\n

\\(\\displaystyle{\\lim_{x \\to \\var{a}}}\\frac{x-\\var{a}}{x^2-\\simplify{{a}+{b}}x+\\simplify{{a}*{b}}}\\)

\n
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\\(\\var{k}=\\var{a}e^{\\var{m}x+{\\var{c}}}\\)

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\\(\\frac{\\var{k}}{\\var{a}}=e^{\\var{m}x+\\var{c}}\\)

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\\(ln\\left(\\frac{\\var{k}}{\\var{a}}\\right)=\\var{m}x+\\var{c}\\)

\n

\\(ln\\left(\\frac{\\var{k}}{\\var{a}}\\right)-\\var{c}=\\var{m}x\\)

\n

\\(\\frac{ln\\left(\\frac{\\var{k}}{\\var{a}}\\right)-\\var{c}}{\\var{m}}=x\\)

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Solve an exponential equation

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Given the equation \\(f(x)=\\var{a}e^{\\var{m}x+\\var{c}}\\)

\n

Determine the value for \\(x\\) that satisfies the relation \\(f(x)=\\var{k}\\)

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Input your answer correct to three decimal places.

\n

\\(x = \\) [[0]]

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Calculate the value of \\(x\\) that satisfies the equation when  \\(y=\\var{d}\\).

\n

Input your answer correct to three decimal places.

\n

\\(x = \\) [[0]]

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Given the following logarithmic equation:

\n

\\(y=\\var{a}log(\\var{b}x+\\var{c}))\\)

\n

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Solve a logarithmic equation

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\\(\\var{a}log(\\var{b}x+\\var{c})=\\var{d}\\)

\n

Divide across by \\(\\var{a}\\)

\n

\\(log(\\var{b}x+\\var{c})=\\var{d}/\\var{a}=\\simplify{{d}/{a}}\\)

\n

\\(\\var{b}x+\\var{c}=10^{\\simplify{{d}/{a}}}\\)

\n

\\(\\var{b}x+\\var{c}=\\simplify{10^{{d}/{a}}}\\)

\n

\\(\\var{b}x=\\simplify{10^{{d}/{a}}}-\\var{c}\\)

\n

\\(\\var{b}x=\\simplify{10^{{d}/{a}}-{c}}\\)

\n

\\(x=\\simplify{(10^{{d}/{a}}-{c})/{b}}\\)

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Manipulation of an exponential function

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\\(x =\\) [[0]]

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\\(y=\\var{k}(1-\\var{c}e^{\\var{m}x+\\var{d}})\\)

\n

Working from the outside in, we divide across by \\(\\var{k}\\)   

\n

\\(\\frac{y}{\\var{k}}=1-\\var{c}e^{\\var{m}x+\\var{d}}\\)

\n

We can bring the \\(x\\) variable to the left hand side and move the \\(y\\) variable to the right hand side

\n

\\(\\var{c}e^{\\var{m}x+\\var{d}}=1-\\frac{y}{\\var{k}}\\)

\n

Again working from the outside in we divide across by \\(\\var{c}\\)

\n

\\(e^{\\var{m}x+\\var{d}}=\\frac{1-\\frac{y}{\\var{k}}}{\\var{c}}\\)

\n

Taking the natural log of both sides eliminates the \\(e\\) from the left hand side. 

\n

\\(\\var{m}x+\\var{d}=ln\\left(\\frac{1-\\frac{y}{\\var{k}}}{\\var{c}}\\right)\\)

\n

Subtract \\(\\var{d}\\) from both sides

\n

\\(\\var{m}x=ln\\left(\\frac{1-\\frac{y}{\\var{k}}}{\\var{c}}\\right)-\\var{d}\\)

\n

and finally divide by \\(\\var{m}\\) to get

\n

\\(x=\\frac{ln\\left(\\frac{1-\\frac{y}{\\var{k}}}{\\var{c}}\\right)-\\var{d}}{\\var{m}}\\)

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Rearrange the following expression to make \\(x\\) the subject:

\n

               \\(y=\\var{k}(1-\\var{c}e^{\\var{m}x+\\var{d}})\\)

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Manipulation of algebraic fractions

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Express your answer as a fraction:

\n

               \\(V =\\) [[0]]

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When one fraction equals another fraction we can clear both fractions by cross-multiplying:

\n

\\((\\var{a}V+1)*(\\var{d}R+7)=(\\var{b}R+3)*(\\var{c}V+5)\\)

\n

\\(\\simplify{{a}*{d}}VR+\\simplify{7*{a}}V+\\var{d}R+7=\\simplify{{b}*{c}}VR+\\simplify{5*{b}}R+\\simplify{3*{c}}V+15\\)

\n

Gathering all the terms involving \\(V\\) to the left hand side and moving all other terms to the right hand side gives

\n

\\(\\simplify{{a}*{d}-{b}*{c}}VR+\\simplify{7*{a}-3*{c}}V=\\simplify{5*{b}-{d}}R+8\\)

\n

Factoring \\(V\\) out on the left hand side 

\n

\\(V(\\simplify{{a}*{d}-{b}*{c}}R+\\simplify{7*{a}-3*{c}})=\\simplify{5*{b}-{d}}R+8\\)

\n

Thus

\n

\\(V=\\frac{\\simplify{5*{b}-{d}}R+8}{\\simplify{{a}*{d}-{b}*{c}}R+\\simplify{7*{a}-3*{c}}}\\)

", "statement": "

Rearrange the following expression to make V the subject:

\n

     \\(\\frac{\\var{a}V+1}{\\var{b}R+3}=\\frac{\\var{c}V+5}{\\var{d}R+7}\\)

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There are two values that satisfy the quadratic equation:

\n

\\(\\var{a1}x^2+\\simplify{{{a1}*{b1}*{c1}}}=\\simplify{{a1}*{b1}+{a1}{c1}}x\\)

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Type in the greater of the two values that satisfies the equation.

\n

Input your answer correct to three decimal places.  \\(x = \\) [[0]]

\n

Type in the lesser of the two values that satisfies the equation. 

\n

Input your answer correct to three decimal places.  \\(x = \\) [[1]]

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Solving quadratic equations using a formula,

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The formula for solving a quadratic equation of the form  \\(ax^2+bx+c=0\\)  is given by

\n

\\(x=\\frac{-b\\pm \\sqrt{b^2-4ac}}{2a}\\)

\n

In this example  \\(a=\\var{a1},\\,\\,\\,b=\\simplify{+-{a1}*({b1}+{c1})}\\)  and  \\(c=\\simplify{{a1}*{b1}*{c1}}\\)

\n

\\(x=\\frac{\\var{b}\\pm \\sqrt{(-\\var{b})^2-4*\\var{a1}*\\var{c}}}{2*\\var{a1}}\\)

\n

\\(x=\\frac{\\var{b}\\pm \\sqrt{\\simplify{{b}^2-4*{a1}*{c}}}}{\\simplify{2*{a1}}}\\)

\n

\\(x=\\simplify{{b}+({b}^2-4*{a1}*{c})^0.5}/\\simplify{2*{a1}}=\\simplify{({b}+({b}^2-4*{a1}*{c})^0.5)/(2*{a1})}\\)        or        \\(x=\\simplify{{b}-({b}^2-4*{a1}*{c})^0.5}/\\simplify{2*{a1}}=\\simplify{({b}-({b}^2-4*{a1}*{c})^0.5)/(2*{a1})}\\)

\n

\n

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Type in the greater of the two values that satisfies the equation. Input your answer correct to three decimal places.

\n

\\(x\\) = [[0]]

\n

Type in the lesser of the two values that satisfies the equation. Input your answer correct to three decimal places.

\n

\\(x\\) = [[1]]

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There are two values that satisfy the quadratic function below when  \\(y=\\var{c1}\\):

\n

\\(y=\\var{a1}x^2+\\var{b1}x\\)

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Solving quadratic equations using a formula,

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The formula for solving a quadratic equation of the form  \\(ax^2+bx+c=0\\)  is given by

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\\(x=\\frac{-b\\pm \\sqrt{b^2-4ac}}{2a}\\)

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In this example  \\(a=\\var{a1},\\,\\,\\,b=\\var{b1}\\)  and  \\(c=\\var{c1}\\)

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\\(x=\\frac{-\\var{b1}\\pm \\sqrt{\\var{b1}^2-4\\times\\var{a1}\\times\\var{c1}}}{2\\times\\var{a1}}\\)

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\\(x=\\frac{-\\var{b1}\\pm \\sqrt{\\simplify{{b1}^2-4*{a1}*{c1}}}}{\\simplify{2*{a1}}}\\)

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\\(x=\\simplify{(-{b1}+ ({b1}^2-4*{a1}*{c1})^0.5)/(2*{a1})}\\)   or   \\(x=\\simplify{(-{b1}- ({b1}^2-4*{a1}*{c1})^0.5)/(2*{a1})}\\)

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The following equation can be converted into a quadratic equation:

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\\(\\var{a1}x+\\frac{\\simplify{{a1}*{b1}*{c1}}}{x}=\\simplify{{a1}*({b1}+{c1})}\\)

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Solving quadratic equations using a formula

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\\(\\var{a1}x+\\frac{\\simplify{{a1}*{b1}*{c1}}}{x}=\\simplify{{a1}*({b1}+{c1})}\\)

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We clear the fraction in the equation by multiplying across by \\(x\\)

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\\(\\var{a1}x^2+\\simplify{{a1}*{b1}*{c1}}=\\simplify{{a1}*({b1}+{c1})}x\\)

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Bringing all the terms to the left hand side and putting them in order of their powers of \\(x\\) gives

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\\(\\var{a1}x^2-\\simplify{{a1}*({b1}+{c1})}x+\\simplify{{a1}*{b1}*{c1}}=0\\)

\n

The formula for solving a quadratic equation of the form  \\(ax^2+bx+c=0\\)  is given by

\n

\\(x=\\frac{-b\\pm \\sqrt{b^2-4ac}}{2a}\\)

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In this example  \\(a=\\var{a1},\\,\\,\\,b=\\simplify{+-{a1}*({b1}+{c1})}\\)  and  \\(c=\\simplify{{a1}*{b1}*{c1}}\\)

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\\(x=\\frac{\\var{b}\\pm \\sqrt{(-\\var{b})^2-4*\\var{a1}*\\var{c}}}{2*\\var{a1}}\\)

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\\(x=\\frac{\\var{b}\\pm \\sqrt{\\simplify{{b}^2-4*{a1}*{c}}}}{\\simplify{2*{a1}}}\\)

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\\(x=\\simplify{{b}+({b}^2-4*{a1}*{c})^0.5}/\\simplify{2*{a1}}=\\simplify{({b}+({b}^2-4*{a1}*{c})^0.5)/(2*{a1})}\\)        or        \\(x=\\simplify{{b}-({b}^2-4*{a1}*{c})^0.5}/\\simplify{2*{a1}}=\\simplify{({b}-({b}^2-4*{a1}*{c})^0.5)/(2*{a1})}\\)

\n

\n

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Type in the greater of the two values that satisfies the equation. 

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\\(x = \\) [[0]]

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Type in the lesser of the two values that satisfies the equation. 

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\\(x = \\) [[1]]

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This quiz contains questions on functions, limits, logs, exponential functions, simultaneous equations and quadratic equations.

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