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1 year ago

What is the relationship between the number of sides of the base of a prism and the number of lateral faces?

Mathematics

2 answers:

bezimeni [28]1 year ago

5 0

Answer: The number of sides of the base of a prism and the number of lateral faces is equal.

Step-by-step explanation:

Arte-miy333 [17]1 year ago

4 0

Answer:

Step-by-step explanation:

edge exam

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Kenji is saving money to buy a new bicycle that costs $125. So far he has saved his weekly allowance of $5 for the past 8 weeks.

Alla [95]

Answer:

$85

Step-by-step explanation:

$125 - $5 - $35 = $85

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2 years ago

Jessica made $95 for 5 hours of work.

poizon [28]

Answer:

$247

Step-by-step explanation:

95/5= 19 to find rate per unit which means,

Jessica makes $19 dollars a hour.

19 x 13= $247

So Jessica made $247 in 13 hours of work.

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2 years ago

Read 2 more answers

LOTS OF POINTS, WILL MARK BRAINLIEST! pls show working!!!!!

sp2606 [1]

Answer: ΔTAU ≈ ΔUAV ≈ ΔTUV

Step-by-step explanation:

I'm not really sure what "work" you really need; this is a problem that can be solved easily by simply looking at the triangles and seeing which sides have the same ratio of distances for each side.

Best of luck with your assignment. :) Feel free to give me Brainliest if you feel this helped. Have a good day.

8 0

1 year ago

If the legs of an isosceles right triangle have a length of ft, the hypotenuse will have a length of?

lesya [120]

Isoceleese means then that 2 sides are equal
right triangle so the legs are equal

a^2+b^2=c^2
the legs aer a and b
in isocoleese, a=b
a^2+a^2=c^2
2a^2=c^2
if isocoleese=a then c=?
solve for c
sqrt both sides
a√2=c

if the legs are x feet, then the hyppotonuse is x√2 feet

6 0

2 years ago

Consider the following. (A computer algebra system is recommended.) y'' + 3y' = 2t4 + t2e−3t + sin 3t (a) Determine a suitable f

drek231 [11]

First look for the fundamental solutions by solving the homogeneous version of the ODE:

$y''+3y'=0$

The characteristic equation is

$r^2+3r=r(r+3)=0$

with roots $r=0$ and $r=-3$ , giving the two solutions $C_1$ and $C_2e^{-3t}$ .

For the non-homogeneous version, you can exploit the superposition principle and consider one term from the right side at a time.

$y''+3y'=2t^4$

Assume the ansatz solution,

${y_p}=at^5+bt^4+ct^3+dt^2+et$

$\implies {y_p}'=5at^4+4bt^3+3ct^2+2dt+e$

$\implies {y_p}''=20at^3+12bt^2+6ct+2d$

(You could include a constant term f here, but it would get absorbed by the first solution $C_1$ anyway.)

Substitute these into the ODE:

$(20at^3+12bt^2+6ct+2d)+3(5at^4+4bt^3+3ct^2+2dt+e)=2t^4$

$15at^4+(20a+12b)t^3+(12b+9c)t^2+(6c+6d)t+(2d+e)=2t^4$

$\implies\begin{cases}15a=2\\20a+12b=0\\12b+9c=0\\6c+6d=0\\2d+e=0\end{cases}\implies a=\dfrac2{15},b=-\dfrac29,c=\dfrac8{27},d=-\dfrac8{27},e=\dfrac{16}{81}$

$y''+3y'=t^2e^{-3t}$

$e^{-3t}$ is already accounted for, so assume an ansatz of the form

$y_p=(at^3+bt^2+ct)e^{-3t}$

$\implies {y_p}'=(-3at^3+(3a-3b)t^2+(2b-3c)t+c)e^{-3t}$

$\implies {y_p}''=(9at^3+(9b-18a)t^2+(9c-12b+6a)t+2b-6c)e^{-3t}$

Substitute into the ODE:

$(9at^3+(9b-18a)t^2+(9c-12b+6a)t+2b-6c)e^{-3t}+3(-3at^3+(3a-3b)t^2+(2b-3c)t+c)e^{-3t}=t^2e^{-3t}$

$9at^3+(9b-18a)t^2+(9c-12b+6a)t+2b-6c-9at^3+(9a-9b)t^2+(6b-9c)t+3c=t^2$

$-9at^2+(6a-6b)t+2b-3c=t^2$

$\implies\begin{cases}-9a=1\\6a-6b=0\\2b-3c=0\end{cases}\implies a=-\dfrac19,b=-\dfrac19,c=-\dfrac2{27}$

$y''+3y'=\sin(3t)$

Assume an ansatz solution

$y_p=a\sin(3t)+b\cos(3t)$

$\implies {y_p}'=3a\cos(3t)-3b\sin(3t)$

$\implies {y_p}''=-9a\sin(3t)-9b\cos(3t)$

Substitute into the ODE:

$(-9a\sin(3t)-9b\cos(3t))+3(3a\cos(3t)-3b\sin(3t))=\sin(3t)$

$(-9a-9b)\sin(3t)+(9a-9b)\cos(3t)=\sin(3t)$

$\implies\begin{cases}-9a-9b=1\\9a-9b=0\end{cases}\implies a=-\dfrac1{18},b=-\dfrac1{18}$

So, the general solution of the original ODE is

$y(t)=\dfrac{54t^5 - 90t^4 + 120t^3 - 120t^2 + 80t}{405}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-\dfrac{3t^3+3t^2+2t}{27}e^{-3t}-\dfrac{\sin(3t)+\cos(3t)}{18}$

3 0

3 years ago