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3 years ago

Solve the proportion 5/7=x/70 A. 10 B. 35 C. 98 D. 50

Mathematics

2 answers:

myrzilka [38]3 years ago

7 0

Answer:

D. 50

Step-by-step explanation:

Multiply the top and bottom number by 10. The bottom number comes out to be 70 and the top number comes out to be 50.

natta225 [31]3 years ago

3 0

Answer: D. 50

Step-by-step explanation:

So hopefully this work that i Showed you, helped you.

* Mark me the Brainliest:) !!!

⇒70* 5 = 350

⇒ 350/7 = 50 as your answer.

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If I read 2/3 of a book that has 222 pages how many pages do I have left to read

joja [24]

Multiply 2/3 by 222 then subtract the number from 222 and you should get your answer

3 0

3 years ago

Read 2 more answers

You are designing a rectangular garden. the garden will be enclosed by fencing on three sides and by a house on the fourth side.

RUDIKE [14]

Suppose the dimensions of the rectangle is x by y and let the side enclosed by a house be one of the sides measuring x, then the sides that is to be enclosed are two sides measuring y and one side measuring x.

Thus, the length of fencing needed is given by

P = x + 2y

The area of the rectangle is given by xy,

i.e.

$xy = 288 \\ \\ \\ \\ \Rightarrow y= \frac{288}{x}$

Substituting for y into the equation for the length of fencing needed, we have

$P=x+2\left( \frac{288}{x} \right)=x+ \frac{576}{x}$

For the amount of fencing to be minimum, then

$\frac{dP}{dx} =0 \\ \\ \Rightarrow1- \frac{576}{x^2} =0 \\ \\ \Rightarrow \frac{576}{x^2} =1 \\ \\ \Rightarrow x^2=576 \\ \\ \Rightarrow x=\sqrt{576}=24$

Now, recall that

$y= \frac{288}{x} = \frac{288}{24} =12$

Thus, the length of fencing needed is given by

P = x + 2y = 24 + 2(12) = 24 + 24 = 48.

Therefore, 48 feets of fencing is needed to enclose the garden.

8 0

3 years ago

Someone help with this, please. It's for my test ;(

Gnom [1K]

Answer:

y=4

Step-by-step explanation:

90 - 66 = 24

therefore, 8y-8 = 24

8y = 32

divide both sides by 8

y = 4

8 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

Translate the phrase to an algebraic expression. State what each variable represents.

mezya [45]

X+ 4? idk I don't really understand the question...

4 0

3 years ago