$course$/Default for Scratch/input_samples Algebraic input Type in {@ta@}.

[[input:ans1]]

[[validation:ans1]]

]]> 1.0000000 0.1000000 0 ta:a*b [[feedback:prt1]] 1 0 0 Correct answer, well done. Your answer is partially correct. Incorrect answer. dot 1 i cos-1 [ ans1 algebraic ta 15 1 2 0 solve 1 0 1 1 1 prt1 1.0000000 1 0 AlgEquiv ans1 ta 0 = 1.0000000 -1 prt1-1-T = 0.0000000 -1 prt1-1-F Algebraic input (answer box sizes test) 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]]

]]> 1.0000000 0.1000000 0 [[feedback:prt1]] 1 0 0 Correct answer, well done. Your answer is partially correct. Incorrect answer. dot 1 i cos-1 [ ans1 algebraic 2 1 1 0 2 0 1 0 0 1 1 ans10 algebraic 2 10 1 0 2222222222 0 1 0 0 1 1 ans15 algebraic 2 15 1 0 222222222222222 0 1 0 0 1 1 ans2 algebraic 2 2 1 0 22 0 1 0 0 1 1 ans20 algebraic 2 20 1 0 12345123451234512345 0 1 0 0 1 1 ans3 algebraic 2 3 1 0 222 0 1 0 0 1 1 ans4 algebraic 2 4 1 0 2222 0 1 0 0 1 1 ans5 algebraic 2 5 1 0 22222 0 1 0 0 1 1 ans7 algebraic 2 5 1 0 2222222 0 1 0 0 1 1 prt1 1.0000000 1 0 AlgEquiv ans1 2 0 = 1.0000000 -1 prt1-1-T This just takes account of the first answer box!

]]> = 0.0000000 -1 prt1-1-F This just takes account of the first answer box!

]]> Checkbox Differentiate {@p@} with respect to $x$.

[[input:ans1]]

[[validation:ans1]]

]]> 1.0000000 0.1000000 0 [[feedback:prt1]] 1 0 0 Correct answer, well done. Your answer is partially correct. Incorrect answer. dot 1 i cos-1 [ ans1 checkbox ta 15 1 0 0 1 0 0 1 2 prt1 1.0000000 1 0 Diff first(ans1) diff(p,x) x 0 = 1.0000000 -1 prt1-1-T = 0.0000000 -1 prt1-1-F Dropdown (shuffle) Differentiate {@p@} with respect to $x$.

[[input:ans1]]

[[validation:ans1]]

]]> 1.0000000 0.1000000 0 [[feedback:prt1]] {@ta@} 1 0 0 Correct answer, well done. Your answer is partially correct. Incorrect answer. dot 1 i cos-1 [ ans1 dropdown ta 15 1 0 0 1 0 0 1 2 prt1 1.0000000 1 0 Diff first(ans1) diff(p,x) x 0 = 1.0000000 -1 prt1-1-T = 0.0000000 -1 prt1-1-F Equiv input test Solve {@p@}.

[[input:ans1]]

[[validation:ans1]]

]]> sangwinc

]]> 1.0000000 0.1000000 0 v:x p:3*v+7=4 ta:[p,x=(4-7)/3,x=-1] [[feedback:prt1]] 1 0 0 Correct answer, well done. Your answer is partially correct. Incorrect answer. dot 1 i cos-1 [ ans1 equiv ta 15 1 5 firstline 0 1 0 1 1 1 prt1 1.0000000 1 0 AlgEquiv last(ans1) last(ta) 0 = 1.0000000 -1 prt1-1-T = 0.0000000 -1 prt1-1-F Matrix Type in {@M@}

[[input:ans1]]

[[validation:ans1]]

]]> 1.0000000 0.1000000 0 M:matrix([1,2],[3,4]) [[feedback:prt1]] 1 0 0 Correct answer, well done. Your answer is partially correct. Incorrect answer. dot 1 i cos-1 [ ans1 matrix M 15 1 0 0 1 0 1 1 1 prt1 1.0000000 1 0 AlgEquiv ans1 M 0 = 1.0000000 -1 prt1-1-T = 0.0000000 -1 prt1-1-F Notes Show your working in this box! $x$.

[[input:ans1]]

[[validation:ans1]]

]]> 1.0000000 0.1000000 0 [[feedback:prt1]] 1 0 0 Correct answer, well done. Your answer is partially correct. Incorrect answer. dot 1 i cos-1 [ ans1 notes true 15 1 0 0 1 0 0 1 1 prt1 1.0000000 1 0 AlgEquiv ans1 true 0 = 1.0000000 -1 prt1-1-T = 0.0000000 -1 prt1-1-F Radio Differentiate {@p@} with respect to $x$.

[[input:ans1]]

[[validation:ans1]]

]]> 1.0000000 0.1000000 0 [[feedback:prt1]] 1 0 0 Correct answer, well done. Your answer is partially correct. Incorrect answer. dot 1 i cos-1 [ ans1 radio ta 15 1 0 0 1 0 0 1 1 prt1 1.0000000 1 0 Diff first(ans1) diff(p,x) x 0 = 1.0000000 -1 prt1-1-T = 0.0000000 -1 prt1-1-F Single char Type in $x$

[[input:ans1]]

[[validation:ans1]]

]]> 1.0000000 0.1000000 0 [[feedback:prt1]] 1 0 0 Correct answer, well done. Your answer is partially correct. Incorrect answer. dot 1 i cos-1 [ ans1 singlechar x 15 1 0 0 1 0 0 1 1 prt1 1.0000000 1 0 AlgEquiv ans1 x 0 = 1.0000000 -1 prt1-1-T = 0.0000000 -1 prt1-1-F $course$/Default for Scratch test_1_integration 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 \[ \int {@p@} d{@v@} = \int u^{@n@} du = \frac{u^{@n+1@}}{@n+1@}+c = {@ta@}+c.\]

]]> 1.0000000 0.1000000 0 n:rand(5)+3; a:rand(5)+3; v:rand([x,t]); p:(v-a)^n; ta:(v-a)^(n+1)/(n+1); [[feedback:prt1]]

]]> $\int {@p@} d{@v@} = {@ta@}$ 1 0 0 Correct answer, well done.

]]> Your answer is partially correct.

]]> Incorrect answer.

]]> dot 1 i cos-1 [ ans1 algebraic ta+c 20 1 0 0 int 1 1 1 1 1 prt1 1.0000000 1 0 Int ans1 ta v 0 = 1.0000000 -1 1-0-T = 0.0000000 -1 1-0-F 1 1001758021 1 ans1 ta+c prt1 1.0000000 0.0000000 1-0-T 2 ans1 ta prt1 0.0000000 0.1000000 1-0-F 3 ans1 n*(v-a)^(n-1) prt1 0.0000000 0.1000000 1-0-F 4 ans1 (v-a)^(n+1) prt1 0.0000000 0.1000000 1-0-F test_2_rectangle 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.

]]> 1.0000000 0.1000000 0 rn:-1*(rand(4)+2); rp:9+rand(6); ar:rn*rp; sg:rn+rp; ta1:x*(x+sg)=-ar; ta2:x*(x-sg)=-ar; tas1:setify(map(rhs,solve(ta1,x))); tas2:setify(map(rhs,solve(ta2,x))); {@ta1@}, {@tas1@}. 1 0 0 Correct answer, well done.

]]> Your answer is partially correct.

]]> Incorrect answer.

]]> dot 1 i cos-1 [ ans1 algebraic ta1 15 1 1 0 1 1 1 1 1 ans2 algebraic tas1 15 1 1 0 1 1 1 1 1 ans3 algebraic rp 5 1 1 0 1 1 1 1 1 eq 1.0000000 1 0 SubstEquiv ans1 ta1 1 = 1.0000000 -1 eq-0-T = 0.0000000 1 eq-0-F 1 SubstEquiv ans1 ta2 1 = 1.0000000 -1 eq-1-T = 0.0000000 -1 eq-1-F short 1.0000000 1 0 AlgEquiv ans3 -rn 0 = 1.0000000 -1 short-0-T = 0.0000000 -1 short-0-F sol 1.0000000 1 v1:first(listofvars(ans1)); ftm:setify(map(rhs,solve(ans1,v1))); 0 SubstEquiv ans1 ta1 1 = 1.0000000 3 sol-0-T = 0.0000000 1 sol-0-F 1 SubstEquiv ans1 ta2 1 = 1.0000000 4 sol-1-T = 0.0000000 2 sol-1-F 2 AlgEquiv ans2 ftm 0 = 1.0000000 -1 sol-2-T You have correctly solved the equation you have entered in part 1. Please try both parts again!

]]> = 0.0000000 -1 sol-2-F 3 AlgEquiv ans2 tas1 0 = 1.0000000 -1 sol-3-T = 0.0000000 -1 sol-3-F 4 AlgEquiv ans2 tas2 0 = 1.0000000 -1 sol-5-T - 0.0000000 -1 sol-5-F 92 1621764605 79313047 1740562616 1 ans1 ta1 ans2 ev({-rp,-rn},simp) ans3 ev(-rn,simp) eq 1.0000000 0.0000000 eq-0-T short 1.0000000 0.0000000 short-0-T sol 1.0000000 0.0000000 sol-3-T 2 ans1 ta2 ans2 ev({rp,rn},simp) ans3 ev(-rn,simp) eq 1.0000000 0.0000000 eq-1-T short 1.0000000 0.0000000 short-0-T sol 1.0000000 0.0000000 sol-5-T 3 ans1 x+sg=-ar ans2 {-ar-sg} ans3 eq 0.0000000 0.1000000 eq-1-F short NULL sol 1.0000000 0.0000000 sol-2-T test_3_matrix Calculate \[ {@A@}.{@B@}\]

[[input:ans1]] [[validation:ans1]]

]]> 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 $B$.

\[ {@A@}.{@B@} = {@C@} = {@D@}.\]

]]> 1.0000000 0.1000000 0 [[feedback:prt1]]

]]> ${@A@}.{@B@}={@TA@}$ 0 0 0 Correct answer, well done.

]]> Your answer is partially correct.

]]> Incorrect answer.

]]> dot 1 i cos-1 [ ans1 matrix TA 3 1 0 0 1 1 1 1 1 prt1 1.0000000 1 0 AlgEquiv ans1 TA 1 = 1.0000000 -1 1-0-T = 0.0000000 1 1-0-F 1 AlgEquiv ans1 TB 1 = 0.0000000 -1 1-1-T Remember, you don't multiply matrices by multiplying the corresponding entries! A quite different process is needed.

]]> = 0.0000000 2 1-1-F 2 AlgEquiv ans1 A+B 1 = 0.0000000 -1 1-3-T Please multiply the matrices. It looks like you have added them instead!

]]> = 0.0000000 -1 1-3-F 86 219862533 1167893775 1 ans1 TA prt1 1.0000000 0.0000000 1-0-T 2 ans1 TB prt1 0.0000000 0.1000000 1-1-T 4 ans1 1 prt1 NULL 5 ans1 A prt1 0.0000000 0.1000000 1-3-F test_5_cubic_spline 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]]

[[validation: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@}.

]]> 1.0000000 0.0000000 0 [[feedback:prt1]]

]]> {@ta@} 1 0 0 Correct answer, well done.

]]> Your answer is partially correct.

]]> Incorrect answer.

]]> dot 1 i cos-1 [ ans1 algebraic ta 15 1 0 0 1 0 0 1 1 prt1 1.0000000 1 x0 and x 0 AlgEquiv subst(x=x0,ans1) subst(x=x0,f0) 0 + 0.2500000 1 prt1-1-T = 0.0000000 1 prt1-1-F Your answer does not satisfy ${@p(x0)=subst(x=x0,f0)@}$.

]]> 1 AlgEquiv subst(x=x1,ans1) subst(x=x1,f1) 0 + 0.2500000 2 prt1-2-T + 0.0000000 2 prt1-2-F Your answer does not satisfy ${@p(x1)=subst(x=x1,f1)@}$.

]]> 2 AlgEquiv subst(x=x0,diff(ans1,x)) subst(x=x0,diff(f0,x)) 0 + 0.2500000 3 prt1-3-T + 0.0000000 3 prt1-3-F Your answer does not satisfy $p'({@x0@})={@subst(x=x0,diff(f0,x))@}$.

]]> 3 AlgEquiv subst(x=x1,diff(ans1,x)) subst(x=x1,diff(f1,x)) 0 + 0.2500000 4 prt1-4-T + 0.0000000 4 prt1-4-F Your answer does not satisfy $p'({@x1@})={@subst(x=x1,diff(f1,x))@}$.

]]> 4 AlgEquiv degree(ans1,x) 3 0 + 0.0000000 5 prt1-5-T = 0.0000000 5 prt1-5-F Your answer is not a cubic!

]]> 5 AlgEquiv ans1 ta 0 = 1.0000000 -1 prt1-6-T + 0.0000000 -1 prt1-6-F 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])@}

]]> 1 ans1 ta prt1 1.0000000 0.0000000 prt1-6-T 2 ans1 x^2 prt1 0.0000000 0.0000000 prt1-6-F 3 ans1 (2*%pi/9+2/27)*x^3-x^2/9+(-2*%pi/3-4/9)*x-4*%pi/9+20/27 prt1 0.7500000 0.0000000 prt1-6-F text_4_complex-De Moivre's 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

\[ {@q^n@} = {@a^n@} e^{@b*n*%i*%pi@} = {@a^n@} e^{@ev(b*n,simp)*%i*%pi@} = {@a^n@} e^{@p*%i*%pi@}.\]

]]> 1.0000000 0.1000000 0 a : ev(2+rand(15),simp); b : ev((-1)^rand(2)*((1+rand(10)))/(2+rand(15)),simp); n : ev(3+rand(20),simp); q : a*%e^(b*%i*%pi); p : ev(mod(b*n,2),simp); {@q^n = a^n*(cos(p*%i*%pi)+%i*sin(p*%i*%pi))@} 0 0 0 Correct answer, well done.

]]> Your answer is partially correct.

]]> Incorrect answer.

]]> dot 1 i cos-1 [ ans1 algebraic a^n 15 1 0 0 1 0 0 1 1 ans2 algebraic p*%pi 15 1 0 0 1 0 0 1 1 prt1 1.0000000 1 0 AlgEquiv ans1 a^n 0 = 1.0000000 0.0000000 -1 prt1-1-T = 0.0000000 0.0000000 -1 prt1-1-F prt2 1.0000000 1 0 AlgEquiv [cos(ans2),sin(ans2)] [cos(b*%pi*n),sin(b*%pi*n)] 1 = 1.0000000 -1 prt2-1-T = 0.0000000 -1 prt2-1-F 1 ans1 a^n ans2 prt1 1.0000000 0.0000000 prt1-1-T prt2 NULL 2 ans1 ans2 b*n*%pi prt1 NULL prt2 1.0000000 0.0000000 prt2-1-T 3 ans1 ans2 p*%pi prt1 NULL prt2 1.0000000 0.0000000 prt2-1-T text_6_odd_even 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]]

]]> 1.0000000 0.3333333 0 1 0 0 Correct answer, well done.

]]> Your answer is partially correct.

]]> Incorrect answer.

]]> dot 1 i cos-1 [ ans1 algebraic x^3 15 1 0 0 1 1 1 1 1 ans2 algebraic x^4 15 1 0 0 1 1 1 1 1 ans3 algebraic 0 15 1 0 0 1 1 1 1 1 ans4 boolean true 15 1 0 0 1 1 1 1 1 even 1.0000000 1 sa:ans2-subst(x=-x,ans2); 0 AlgEquiv sa 0 0 = 1.0000000 -1 even-0-T = 0.0000000 -1 even-0-F Your answer is not an even function. Look, \[ f(x)-f(-x)={@sa@} \neq 0.\]

]]> odd 1.0000000 1 sa:subst(x=-x,ans1)+ans1; 0 AlgEquiv sa 0 0 = 1.0000000 -1 odd-0-T = 0.0000000 -1 odd-0-F Your answer is not an odd function. Look, \[ f(x)+f(-x)={@sa@} \neq 0.\]

]]> oddeven 2.0000000 1 sa1:subst(x=-x,ans3)+ans3; sa2:ans3-subst(x=-x,ans3); 0 AlgEquiv sa1 0 0 = 0.5000000 1 ODD = 0.0000000 1 oddeven-0-F Your answer is not an odd function. Look, \[ f(x)+f(-x)={@sa1@} \neq 0.\]

]]> 1 AlgEquiv sa2 0 0 + 0.5000000 -1 EVEN + 0.0000000 -1 oddeven-1-F Your answer is not an even function. Look, \[ f(x)-f(-x)={@sa2@} \neq 0.\]

]]> unique 1.0000000 1 0 AlgEquiv ans4 true 0 = 1.0000000 -1 unique-0-T = 0.0000000 -1 unique-0-F 1 ans1 x^3 ans2 cos(x) ans3 0 ans4 true even 1.0000000 0.0000000 even-0-T odd 1.0000000 0.0000000 odd-0-T oddeven 1.0000000 0.0000000 EVEN unique 1.0000000 0.0000000 unique-0-T 2 ans1 x^2 ans2 x^3 ans3 x^3 ans4 false even 0.0000000 0.3333333 even-0-F odd 0.0000000 0.3333333 odd-0-F oddeven 0.5000000 0.3333333 oddeven-1-F unique 0.0000000 0.3333333 unique-0-F text_7_solve_quadratic

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.

[[input:ans1]] [[validation:ans1]]

]]> \[ {@stack_disp_arg(ta, true)@} \]

]]> 1.0000000 0.1000000 0 n1:2 n2:n1+3 v:x p:expand((v-n1)*(v-n2)) simp:false ta:[p=0,(v-n1)*(v-n2)=0,v-n1=0 nounor v-n2=0,v=n1 nounor v=n2] [[feedback:prt1]] {@ta@} 1 0 0 Correct answer, well done. Your answer is partially correct. Incorrect answer. none 1 i cos-1 [ ans1 equiv ta 15 1 5 firstline 0 1 0 0 1 1 firstline prt1 1.0000000 0 foundfac1:ev(sublist(ans1,lambda([ex], equationp(ex) and is(rhs(ex)=0))),simp); foundfac2:ev(any_listp(lambda([ex], second(ATFacForm(lhs(ex),lhs(ex),x))), foundfac1),simp); 0 EquivFirst ans1 ta x 0 = 1.0000000 1 prt1-1-T = 0.0000000 -1 prt1-1-F 1 EqualComAss last(ans1) last(ta) 1 + 0.0000000 2 prt1-2-T = 0.0000000 2 prt1-2-F

]]> 2 AlgEquiv foundfac2 true 1 + 0.0000000 -1 prt1-3-T = 0.0000000 -1 prt1-3-F The question asked you to solve the equation by factoring the equation. The factored form should appear as one line in your working.

]]> 1 ans1 ta prt1 1.0000000 0.0000000 prt1-3-T 2 ans1 [p=0, (x-n1)*(x-n1) = 0, x=n1 nounor n2] prt1 NULL 3 ans1 [p=0,x-n1 = 0 nounor x-n2 = 0,x = n1 nounor x = n2] prt1 0.0000000 0.1000000 prt1-3-F 4 ans1 [p=0,(x-(n1+n2)/2)^2-ev((n1+n2)^2/4-n1*n2,simp)=0,(x-(n1+n2)/2)^2=ev((n1+n2)^2/4-n1*n2,simp),(x-(n1+n2)/2)=+-ev(sqrt((n1+n2)^2/4-n1*n2),simp),x=ev((n1+n2)/2+sqrt((n1+n2)^2/4-n1*n2),simp) nounor x=ev((n1+n2)/2-sqrt((n1+n2)^2/4-n1*n2),simp)] prt1 0.0000000 0.1000000 prt1-3-F $course$/Default for Scratch/input_samples Textarea test [[input:ans1]] [[validation:ans1]]]]> 1.0000000 0.1000000 0 ta:[x=1,x=2] vendor/bin/phpunit --group qtype_stack 1 0 0 Correct answer, well done. Your answer is partially correct. Incorrect answer. dot 1 i cos-1 [ ans1 textarea ta 15 1 1 0 1 0 0 1 1 prt1 1.0000000 1 0 AlgEquiv ans1 ta 0 = 1.0000000 -1 prt1-1-T = 0.0000000 -1 prt1-1-F True/false All generalizations are false: [[input:ans1]] [[validation:ans1]]

]]> 1.0000000 0.1000000 0 [[feedback:prt1]] 1 0 0 Correct answer, well done. Your answer is partially correct. Incorrect answer. dot 1 i cos-1 [ ans1 boolean false 15 1 0 0 1 0 0 1 1 prt1 1.0000000 1 0 AlgEquiv ans1 false 0 = 0.5000000 -1 prt1-1-T Who knows!

]]> = 0.5000000 -1 prt1-1-F Who knows!

]]> Units What is gravity?

[[input:ans1]]

[[validation:ans1]]

]]> 1.0000000 0.1000000 0 ta:9.81*m*s^-2 [[feedback:prt1]] 1 0 0 Correct answer, well done. Your answer is partially correct. Incorrect answer. dot 1 i cos-1 [ ans1 units ta 15 1 0 0 1 0 0 1 1 prt1 1.0000000 1 0 Units ans1 ta 3 0 = 1.0000000 -1 prt1-1-T = 0.0000000 -1 prt1-1-F