Type in {@ta@}.
[[input:ans1]]
This question just tests answer boxes of multiple sizes.
[[input:ans1]] [[validation:ans1]]
[[input:ans2]] [[validation:ans2]]
[[input:ans3]] [[validation:ans3]]
[[input:ans4]] [[validation:ans4]]
[[input:ans5]] [[validation:ans5]]
[[input:ans7]] [[validation:ans7]]
[[input:ans10]] [[validation:ans10]]
[[input:ans15]] [[validation:ans15]]
[[input:ans20]] [[validation:ans20]]
This just takes account of the first answer box!
Differentiate {@p@} with respect to \(x\).
Solve {@p@}.
sangwinc
Type in {@M@}
Show your working in this box! \(x\).
Type in \(x\)
Find \[ \int {@p@} d{@v@}\] [[input:ans1]] [[validation:ans1]]
We can either do this question by inspection (i.e. spot the answer) or in a more formal manner by using the substitution \[ u = ({@v@}-{@a@}).\] Then, since \(\frac{d}{d{@v@}}u=1\) we have \[ \int {@p@} d{@v@} = \int u^{@n@} du = \frac{u^{@n+1@}}{@n+1@}+c = {@ta@}+c.\]
[[feedback:prt1]]
Correct answer, well done.
Your answer is partially correct.
Incorrect answer.
A rectangle has length {@sg@}cm greater than its width. If it has an area of \({@abs(ar)@}cm^2\), find the dimensions of the rectangle.
1. Write down an equation which relates the side lengths to the area of the rectangle. [[input:ans1]] [[validation:ans1]] [[feedback:eq]]
2. Solve your equation. Enter your answer as a set of numbers. [[input:ans2]] [[validation:ans2]] [[feedback:sol]]
3. Hence, find the length of the shorter side. [[input:ans3]] cm [[validation:ans3]] [[feedback:short]]
If \(x\)cm is the width then \((x+{@sg@})\) is the length. Now the area is \({@abs(ar)@}cm^2\) and so \[ {@x*(x+sg)=-ar@}.\] \[ {@x^2+sg*x+ar@}=0\] \[ {@(x+rp)*(x+rn)=0@} \] So that \(x={@-rp@}\) or \(x={@-rn@}\). Since lengths are positive quantities \(x>0\) and we discard the negative root. Hence the length of the shorter side is \(x={@-rn@}\)cm.
You have correctly solved the equation you have entered in part 1. Please try both parts again!
Calculate \[ {@A@}.{@B@}\]
To multiply matrices \(A\) and \(B\) we need to remember that the \((i,j)\)th entry is the scalar product of the \(i\)th row of \(A\) with the \(j\)th column of \(B\).
\[ {@A@}.{@B@} = {@C@} = {@D@}.\]
Remember, you don't multiply matrices by multiplying the corresponding entries! A quite different process is needed.
Please multiply the matrices. It looks like you have added them instead!
Consider the real function
\[ f(x) :=\left\{ \begin{array}{ll} {@f0@} & \mbox{for } x\leq {@x0@}, \\ p(x) & \mbox{for } {@x0@} < x < {@x1@}, \\ {@f1@} & \mbox{for } {@x1@}\leq x.\end{array} \right. \]
{@plot(pg,[x,(x0-3),(x1+3)],[y,-3,3])@}
Find the cubic polynomial \(p(x)\) which makes \(f(x)\) continuously differentiable.
\(p(x)=\)[[input:ans1]]
First we need to find the information which \(p(x)\) needs to satisfy. This is
\[ p({@x0@})={@subst(x=x0,f0)@}.\]
\[ p({@x1@})={@subst(x=x1,f1)@}.\]
\[ p'({@x0@})={@subst(x=x0,diff(f0,x))@}.\]
\[ p'({@x1@})={@subst(x=x1,diff(f1,x))@}.\]
If we define \(p(x)\) to be the cubic
\[ p(x) = a_3 x^3 + a_2 x^2 + a_1 x + a_0,\]
We then set up the matrix equation
\[ {@CS@} {@CV@} = {@CT@} \]
Solving this gives the coefficients, from which we get the polynomial.
{@ta@}.
Your answer does not satisfy \({@p(x1)=subst(x=x1,f1)@}\).
Your answer does not satisfy \(p'({@x0@})={@subst(x=x0,diff(f0,x))@}\).
Your answer does not satisfy \(p'({@x1@})={@subst(x=x1,diff(f1,x))@}\).
Your answer is not a cubic!
Your answer is plotted below, although part of your graph might appear out of range of the plot!
{@plot([pg,anspt],[x,(x0-3),(x1+3)],[y,-3,3])@}
Given a complex number \(\displaystyle z={@q@}\) determine
\(|z^{@n@}|=\)[[input:ans1]] [[validation:ans1]] [[feedback:prt1]]
and \(\arg(z^{@n@})=\)[[input:ans2]] [[validation:ans2]] [[feedback:prt2]]
It makes sense that the index laws should still apply. This is called De Moivre's theorem.
\[ {@q^n@} ={@a^n@} e^{@b*n*%i*%pi@}.\]
Recall that
\[ e^{i\theta} = \cos(\theta)+i\sin(\theta).\]
Working with the principle argument \(0\leq \theta \leq 2\pi\) gives us
\[ {@q^n@} = {@a^n@} e^{@b*n*%i*%pi@} = {@a^n@} e^{@ev(b*n,simp)*%i*%pi@} = {@a^n@} e^{@p*%i*%pi@}.\]
1. Give an example of an odd function by typing an expression which represents it. \(f_1(x)=\) [[input:ans1]]. [[validation:ans1]] [[feedback:odd]]
2. Give an example of an even function. \(f_2(x)=\) [[input:ans2]]. [[validation:ans2]] [[feedback:even]]
3. Give an example of a function which is odd and even. \(f_3(x)=\) [[input:ans3]]. [[validation:ans3]] [[feedback:oddeven]]
4. Is the answer to 3. unique? [[input:ans4]] (Or are there many different possibilities.) [[validation:ans4]] [[feedback:unique]]
Your answer is not an even function. Look, \[ f(x)-f(-x)={@sa@} \neq 0.\]
Your answer is not an odd function. Look, \[ f(x)+f(-x)={@sa@} \neq 0.\]
Your answer is not an odd function. Look, \[ f(x)+f(-x)={@sa1@} \neq 0.\]
Your answer is not an even function. Look, \[ f(x)-f(-x)={@sa2@} \neq 0.\]
Solve {@first(ta)@}, by factoring and working line by line. Leave your answer in the form \({@v@}=\cdots \mbox{ or } {@v@}=\cdots\) in fully simplified form.
\[ {@stack_disp_arg(ta, true)@} \]
The question asked you to solve the equation by factoring the equation. The factored form should appear as one line in your working.
All generalizations are false: [[input:ans1]] [[validation:ans1]]
Who knows!
What is gravity?