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notsponge [240]
3 years ago
9

A cooling tower for a nuclear reactor is to be constructed in the shape of a hyperboloid of one sheet. The diameter at the base

is 300 m and the minimum diameter, 500 m above the base, is 200 m. Find an equation describing the shape of the tower in the coordinates where the origin is at the center of the narrowest part of the tower. In particular, use coordinates where the origin is 500 m above the ground.
Mathematics
2 answers:
zaharov [31]3 years ago
5 0

Okay so the equation you need to use is (x2/a2)+(y2/b2)-(z2/c2)=1

So an equation describing the shape of the tower in the coordinates where the origin is at the center of the narrowest part of the tower would be:

<span>x^2 / 100^2 + y^2 / 100^2 – z^2 / 2 * 100^2 = 1</span>

I am hoping that this answer has satisfied your query and it will be able to help you in your endeavor, and if you would like, feel free to ask another question.

SpyIntel [72]3 years ago
4 0
Hello,

as  the bases are circles,
equation is :

\dfrac{x^2}{a^2} + \dfrac{y^2}{a^2}- \dfrac{z^2}{c^2}=1\\\\&#10;if\ x=100,y=0,z=0\ then \  \dfrac{100^2}{a^2}=1\\\\&#10;a=100\\\\&#10;if\ x=150,y=0,z=-500\  then \\&#10; \dfrac{150^2}{100^2} + \dfrac{0^2}{100^2}- \dfrac{500^2}{c^2}=1\\\\&#10;c=100*\sqrt{2} \\\\&#10;\boxed{\dfrac{x^2}{100^2} + \dfrac{y^2}{100^2}- \dfrac{z^2}{2*100^2}=1}&#10;&#10;

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Given that aₙ = (-2) (5)ⁿ⁻1 for a geometric sequence, determine the common ratio:
patriot [66]

Answer:

5

Step-by-step explanation:

The nth term of the geometric sequence is

a_n=(-2)(5)^{n-1}

In general the nth term of a geometric sequence is

a_n=a_1(r)^{n-1}

By comparism;

r=5 is the common ratio of the sequence.

Or

a_1=(-2)(5)^{1-1}=-2

a_2=(-2)(5)^{2-1}=-10

Common ratio is

r=\frac{-10}{-2}=5

4 0
3 years ago
Can someone help me i will give brainliest to whom ever right 28=x+13
frozen [14]
The answer is that x equals 15
6 0
2 years ago
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I need help and a right answer.
Juli2301 [7.4K]

Answer:

Step-by-step explanation:

P1(3,3)

P2(1, -5)

M= (-5-3)/(1-3)=-8/-2=4

Y=4x-9

6 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
Find the equation of the line using the point-slope formula. Write the final equation using slope-intercept form. (1,2) with a s
Lilit [14]

Answer:

y-2=\displaystyle -\frac{3}{4}(x-1)

OR

y=\displaystyle -\frac{3}{4}x+\frac{11}{4}

Step-by-step explanation:

Hi there!

Point-slope form: y-y_1=m(x-x_1) where <em>m</em> is the slope of the line and (x_1,y_1) is a given point

Given that the slope is -3/4, we can plug it into y-y_1=m(x-x_1) as <em>m</em>:

y-y_1=\displaystyle -\frac{3}{4}(x-x_1)

We can also plug in the given point (1,2):

y-2=\displaystyle -\frac{3}{4}(x-1)

Slope-intercept form: y=mx+b where <em>m</em> is the slope and <em>b</em> is the y-intercept (the value of y when the line crosses the y-axis)

To write the equation in slope-intercept form, isolate <em>y</em>:

y-2=\displaystyle -\frac{3}{4}(x-1)\\\\y=\displaystyle -\frac{3}{4}(x-1)+2\\\\y=\displaystyle -\frac{3}{4}x+\frac{3}{4}+2\\\\y=\displaystyle -\frac{3}{4}x+\frac{11}{4}

I hope this helps!

8 0
2 years ago
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