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3 years ago

5

the circumference of a circle is blank. a)the distance around the circle b)the length of a radius c)2pie d)the length of the dia

meter

1 answer:

Misha Larkins [42]3 years ago

8 0

Answer:

a. distance around the circle

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I NEED FAST HELP!!!!!!!!!!!!!!!!!!!

Sindrei [870]

Answer:

The student is expected to spend <em>15.4 hours </em>doing homework

Step-by-step explanation:

The scattered plot shows there is a close correlation between the variables. A line of best fit will go through the 'center' of the points. Since we are not required to find an exact line, we'll draw it in red color as shown below

To know the equation of that line, we must take two clear points of it from the graph. We'll pick (28,4) and (4,25)

The equation of a line, given two points (a,b) and (c,d) is

$\displaystyle y-b=\frac{d-b}{c-a}(x-a)$

Using the selected points

$\displaystyle y-4=\frac{25-4}{4-28}(x-28)$

Simplifying and computing results, the equation is

$\displaystyle y=-\frac{7}{8}x+\frac{57}{2}$

Using that equation, we can predict how many hours the students will spend doing homework if they spend 15 hours watching TV

$\displaystyle y=-\frac{7}{8}(15)+\frac{57}{2}$ =15.4 hours

So the student is expected to spend 15.4 hours doing homework

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3 years ago

When evaluating, what will be the exponent of 10 in the quotient?<br> 3<br> 4<br> 8<br> 16

qaws [65]

In this case it would be 8

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3 years ago

Read 2 more answers

What is 6.25% into 261

natta225 [31]

13.5 hope this helps

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3 years ago

Calculate the area of the circular field if its radius is 110m and express this area in hectares?

Contact [7]

Hi There! :)

<span>Calculate the area of the circular field if its radius is 110m and express this area in hectares?

</span>A≈38013.27<span>m²
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5 0

3 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 <em>f</em> 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