Show one of several blocks of text depending on the value of a question variable.

As well as a simple check for the value of a variable, the condition to display a block of text can be a complex expression in any of the question variables - in this example, depending on the discriminant of the generated quadratic.

A simple ideal gas law question, using number of molecules, that asks the student to calculate one of the four variables. he question can ask for either L atm, or kPa m3, and either K or C for temp. As such the boltzmann constant is in either J/K or L atm/K.

pV = 1/3 Nm<v2>

A simple ideal gas law question, using number of molecules, that asks the student to calculate one of the four variables. he question can ask for either L atm, or kPa m3, and either K or C for temp. As such the boltzmann constant is in either J/K or L atm/K.

A simple ideal gas law question, that asks the student to calculate one of the four variables. he question can ask for either L atm, or kPa m3, and either K or C for temp.

A simple ideal gas law question, using number of molecules, that asks the student to calculate one of the four variables. he question can ask for either L atm, or kPa m3, and either K or C for temp. As such the boltzmann constant is in either J/K or L atm/K.

pV = 1/3 Nm<v2>

A simple ideal gas law question, that asks the student to calculate one of the four variables. he question can ask for either L atm, or kPa m3, and either K or C for temp.

Esta es la pregunta para la semana 4 del curso MA100 en el LSE. Examina el material de los capítulos 7 y 8. A continuación se describe cómo se definió un polinomio en la pregunta. Esto puede ser útil para cualquier persona que necesite editar esta pregunta.

Para las partes a a c, utilizamos un polinomio definido como m * (x ^ 4 - 2a ^ 2 x ^ 2 + a ^ 4 + b), donde las variables "a" y "b" se seleccionan al azar de un conjunto de tamaño reajustable, y la variable $ m $ se elige aleatoriamente del conjunto {+1, -1}. Podemos ver fácilmente que este polinomio tiene puntos estacionarios en -a, 0 y a. Introdujimos la variable "m" para que estos puntos estacionarios no siempre tuvieran la misma clasificación. La variable "b" es siempre positiva, y esto asegura que nuestro polinomio no cruce el eje x. Los primeros y segundos derivados; puntos estacionarios; la evaluación de la segunda derivada en los puntos estacionarios; la clasificación de los puntos estacionarios; y las intersecciones de los ejes se pueden expresar fácilmente en términos de las variables "a", "b" y "m". En efecto,

Solve 4 first order differential equations of two types:$\displaystyle \frac{dy}{dx}=\frac{ax}{y},\;\;\frac{dy}{dx}=\frac{by}{x},\;y(2)=1$ for all 4.

rebelmaths

Find $\displaystyle \int \frac{2ax + b}{ax ^ 2 + bx + c}\;dx$

Find $\displaystyle \int \frac{2ax + b}{ax ^ 2 + bx + c}\;dx$

Find $\displaystyle \int \frac{2ax + b}{ax ^ 2 + bx + c}\;dx$

Find $\displaystyle I=\int \frac{2 a x + b} {a x ^ 2 + b x + c}\;dx$ by substitution or otherwise.

An object moves in a straight line, acceleration given by:

$\displaystyle f(t)=\frac{a}{(1+bt)^n}$. The object starts from rest. Find its maximum speed. 

Multiple response question (3 correct out of 6) re properties of convergent and divergent sequences. Selection of questions from a pool.

Create a truth table for a logical expression of the form $((a \operatorname{op1} b) \operatorname{op2}(c \operatorname{op3} d))\operatorname{op4}(e \operatorname{op5} f) $ where each of $a, \;b,\;c,\;d,\;e,\;f$ can be one the Boolean variables $p,\;q,\;\neg p,\;\neg q$ and each of $\operatorname{op1},\;\operatorname{op2},\;\operatorname{op3},\;\operatorname{op4},\;\operatorname{op5}$ one of $\lor,\;\land,\;\to$.

For example: $((q \lor \neg p) \to (p \land \neg q)) \to (p \lor q)$

Create a truth table for a logical expression of the form $a \operatorname{op} b$ where $a, \;b$ can be the Boolean variables $p,\;q,\;\neg p,\;\neg q$ and $\operatorname{op}$ one of $\lor,\;\land,\;\to$.

For example $\neg q \to \neg p$.

Create a truth table for a logical expression of the form $((a \operatorname{op1} b) \operatorname{op2}(c \operatorname{op3} d))\operatorname{op4}e $ where each of $a, \;b,\;c,\;d,\;e$ can be one the Boolean variables $p,\;q,\;\neg p,\;\neg q$ and each of $\operatorname{op1},\;\operatorname{op2},\;\operatorname{op3},\;\operatorname{op4}$ one of $\lor,\;\land,\;\to$.

For example: $((q \lor \neg p) \to (p \land \neg q)) \lor \neg q$

Create a truth table for a logical expression of the form $a \operatorname{op} b$ where $a, \;b$ can be the Boolean variables $p,\;q,\;\neg p,\;\neg q$ and $\operatorname{op}$ one of $\lor,\;\land,\;\to$.

For example $\neg q \to \neg p$.

Create a truth table for a logical expression of the form $((a \operatorname{op1} b) \operatorname{op2}(c \operatorname{op3} d))\operatorname{op4}e $ where each of $a, \;b,\;c,\;d,\;e$ can be one the Boolean variables $p,\;q,\;\neg p,\;\neg q$ and each of $\operatorname{op1},\;\operatorname{op2},\;\operatorname{op3},\;\operatorname{op4}$ one of $\lor,\;\land,\;\to$.

For example: $((q \lor \neg p) \to (p \land \neg q)) \lor \neg q$

Create a truth table for a logical expression of the form $(a \operatorname{op1} b) \operatorname{op2}(c \operatorname{op3} d)$ where $a, \;b,\;c,\;d$ can be the Boolean variables $p,\;q,\;\neg p,\;\neg q$ and each of $\operatorname{op1},\;\operatorname{op2},\;\operatorname{op3}$ one of $\lor,\;\land,\;\to$.

For example: $(p \lor \neg q) \land(q \to \neg p)$.

Create a truth table for a logical expression of the form $(a \operatorname{op1} b) \operatorname{op2}(c \operatorname{op3} d)$ where $a, \;b,\;c,\;d$ can be the Boolean variables $p,\;q,\;\neg p,\;\neg q$ and each of $\operatorname{op1},\;\operatorname{op2},\;\operatorname{op3}$ one of $\lor,\;\land,\;\to$.

For example: $(p \lor \neg q) \land(q \to \neg p)$.

Create a truth table for a logical expression of the form $((a \operatorname{op1} b) \operatorname{op2}(c \operatorname{op3} d))\operatorname{op4}(e \operatorname{op5} f) $ where each of $a, \;b,\;c,\;d,\;e,\;f$ can be one the Boolean variables $p,\;q,\;\neg p,\;\neg q$ and each of $\operatorname{op1},\;\operatorname{op2},\;\operatorname{op3},\;\operatorname{op4},\;\operatorname{op5}$ one of $\lor,\;\land,\;\to$.

For example: $((q \lor \neg p) \to (p \land \neg q)) \to (p \lor q)$

Create a truth table for a logical expression of the form $((a \operatorname{op1} b) \operatorname{op2}(c \operatorname{op3} d))\operatorname{op4}e $ where each of $a, \;b,\;c,\;d,\;e$ can be one the Boolean variables $p,\;q,\;r,\;\neg p,\;\neg q,\;\neg r$ and each of $\operatorname{op1},\;\operatorname{op2},\;\operatorname{op3},\;\operatorname{op4}$ one of $\lor,\;\land,\;\to$.

For example: $((q \lor \neg r) \to (p \land \neg q)) \land \neg r$

Equations which can be written in the form

\[\dfrac{\mathrm{d}y}{\mathrm{d}x} = f(x),   \dfrac{\mathrm{d}y}{\mathrm{d}x} = f(y),   \dfrac{\mathrm{d}y}{\mathrm{d}x} = f(x)f(y)\]

can all be solved by integration.

In each case it is possible to separate the $x$'s to one side of the equation and the $y$'s to the other

Solving such equations is therefore known as solution by separation of variables

Provided with information on a sample with sample mean and known population variance, use the z test to either accept or reject a given null hypothesis.

These basic questions will help you expand one set of brackets for linear variables

Create a truth table for a logical expression of the form $a \operatorname{op} b$ where $a, \;b$ can be the Boolean variables $p,\;q,\;\neg p,\;\neg q$ and $\operatorname{op}$ one of $\lor,\;\land,\;\to$.

For example $\neg q \to \neg p$.

Shows how to define variables to stop degenerate examples.

This is the question for Lent Term week 7 of the MA100 course at the LSE. It looks at material from chapters 33 and 34.

The following is a description of parts a and b. In particular it describes the varaibles used for those parts.

This question (parts a and b) looks at optimisation problems using the langrangian method. parts a and b of the question we will ask the student to optimise the objective function f(x,y) = y + (a/b)x subject to the constraint function r^2 = (x-centre_x)^2 + (y-centre_y)^2.

The variables centre_x and centre_y take values randomly chosen from {6,7,...,10} and r takes values randomly chosen from {1,2,...,5}.

We have the ordered set of variables (a,b,c) defined to be randomly chosen from one of the following pythagorean triplets: (3,4,5) , (5,12,13) , (8,15,17) , (7,24,25) , (20,21,29). The a and b variables here are the same as those in the objective function. They are defined in this way because the minimum will occur at (centre_x - (a/c)*r , centre_y - (b/c)*r) with value centre_y - (b/c)r + (a/b) * centre_x - (a^2/bc)*r , and the maximum will occur at (centre_x + (a/c)*r , centre_y + (b/c)*r) with value centre_y + (b/c)r + (a/b) *centre_x + (a^2/bc)r. The minimisation problem has lambda = -c/(2br) and the maximation problem has lambda* = c/(2br).

We can see that all possible max/min points and values are nice rational numbers, yet we still have good randomisation in this question. :)

This is the question for week 9 of the MA100 course at the LSE. It looks at material from chapters 17 and 18.

Description of variables for part b:
For part b we want to have four functions such that the derivative of one of them, evaluated at 0, gives 0; but for the rest we do not get 0. We also want two of the ones that do not give 0, to be such that the derivative of their sum, evaluated at 0, gives 0; but when we do this for any other sum of two of our functions, we do not get 0. Ultimately this part of the question will show that even if two functions are not in a vector space (the space of functions with derivate equal to 0 when evaluated at 0), then their sum could nonetheless be in that vector space. We want variables which statisfy:

a,b,c,d,f,g,h,j,k,l,m,n are variables satisfying

Function 1: x^2 + ax + b sin(cx)
Function 2: x^2 + dx + f sin(gx)
Function 3: x^2 + hx + j sin(kx)
Function 4: x^2 + lx + m sin(nx)

u,v,w,r are variables satifying
u=a+bc
v=d+fg
w=h+jk
r=l+mn

The derivatives of each function, evaluated at zero, are:
Function 1: u
Function 2: v
Function 3: w
Function 4: r

So we will define
u as random(-5..5 except(0))
v as -u
w as 0
r as random(-5..5 except(0) except(u) except(-u))

Then the derivative of function 3, evaluated at 0, gives 0. The other functions give non-zero.
Also, the derivative of function 1 + function 2 gives 0. The other combinations of two functions give nonzero.

We now take b,c,f,g,j,k,m,n to be defined as \random(-3..3 except(0)).
We then define a,d,h,l to satisfy
u=a+bc
v=d+fg
w=h+jk
r=l+mn

Description for variables of part e:

Please look at the description of each variable for part e in the variables section, first.
As described, the vectors V3_1 , V3_2 , V3_3 are linearly independent. We will simply write v1 , v2 , v3 here.
In part e we ask the student to determine which of the following sets span, are linearly independent, are both, are neither:

both: v1,v2,v3
span: v1,v1+v2,v1+v2+v3, v1+v2+v3,2*v1+v2+v3
lin ind: v1+v2+v3
neither: v2+v3 , 2*v2 + 2*v3
neither:v1+v3,v1-2*v3,2*v1-v3
neither: v1+v2,v1-v2,v1-2*v2,2*v1-v2