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3 years ago

14

Tower A is 90 feet tall. Tower B is 180 feet tall. Tower C is 5 times as tall as both Towers A and B combined. How tall is Tower

C?

1 answer:

noname [10]3 years ago

6 0

Answer:

1350

Step-by-step explanation:

I added tower A and B together (90+180) Which gives me 270. Then I did;

270 * 5= 1350 ft. Hope its correct!

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Someone help me with this algebra question please 100% correct only

STALIN [3.7K]

Let x represent the width. Since the length is 3 more than 4 times the width, that would be equal to 3+4x. The perimeter is the sum of all sides, so that would be x+x+3+4x+3+4=86. Combine like terms, which would result in 10x+6=86. Subtract 6 from both sides giving you 10x=80. Divide by 10 giving you x=8. That means that the width is 8. The length is 3+4x. Substitute 8 for x giving you 3+4(8). Multiply, giving you 3+32, then add, which would give you 35. 35 is the length.

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3 years ago

Use the method of undetermined coefficients to find the general solution to the de y′′−3y′ 2y=ex e2x e−x

djverab [1.8K]

I'll assume the ODE is

$y'' - 3y' + 2y = e^x + e^{2x} + e^{-x}$

Solve the homogeneous ODE,

$y'' - 3y' + 2y = 0$

The characteristic equation

$r^2 - 3r + 2 = (r - 1) (r - 2) = 0$

has roots at $r=1$ and $r=2$ . Then the characteristic solution is

$y = C_1 e^x + C_2 e^{2x}$

For nonhomogeneous ODE (1),

$y'' - 3y' + 2y = e^x$

consider the ansatz particular solution

$y = axe^x \implies y' = a(x+1) e^x \implies y'' = a(x+2) e^x$

Substituting this into (1) gives

$a(x+2) e^x - 3 a (x+1) e^x + 2ax e^x = e^x \implies a = -1$

For the nonhomogeneous ODE (2),

$y'' - 3y' + 2y = e^{2x}$

take the ansatz

$y = bxe^{2x} \implies y' = b(2x+1) e^{2x} \implies y'' = b(4x+4) e^{2x}$

Substitute (2) into the ODE to get

$b(4x+4) e^{2x} - 3b(2x+1)e^{2x} + 2bxe^{2x} = e^{2x} \implies b=1$

Lastly, for the nonhomogeneous ODE (3)

$y'' - 3y' + 2y = e^{-x}$

take the ansatz

$y = ce^{-x} \implies y' = -ce^{-x} \implies y'' = ce^{-x}$

and solve for $c$ .

$ce^{-x} + 3ce^{-x} + 2ce^{-x} = e^{-x} \implies c = \dfrac16$

Then the general solution to the ODE is

$\boxed{y = C_1 e^x + C_2 e^{2x} - xe^x + xe^{2x} + \dfrac16 e^{-x}}$

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1 year ago

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ad-work [718]

Answer: Choice D) $x = a\left(1-\frac{y}{b}\right)$

================================================

Work Shown:

$\frac{x}{a}+\frac{y}{b} = 1\\\\\frac{x}{a} = 1-\frac{y}{b}\\\\x = a\left(1-\frac{y}{b}\right)\\\\$

Explanation:

First I subtracted y/b from both sides. Then I multiplied both sides by 'a' to fully isolate x. You can optionally distribute the 'a' through to each term inside the parenthesis, but your teacher has chosen not to do this.

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2 years ago

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vekshin1

You have the polygon MNOPQR which can be expressed as two rectangles pasted one next to each other.

To see the two rectangles in the picture, you can draw a line parallel to segment MR througn point N.

From the original picture you can state the dimensions of both rectangles.

Call S, the point where the line that you drew intercepts the segment RQ.

Then one rectangle is MNSR and the other rectangle is OPQS.

The measures of the sides of the rectangle MNSR are:

- the length of MN = length of SR = base

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So its area is base * height, which you can all A1.

The measured of the rectangle OPQS are:

- segment OP = segment SQ = segment QR - segment SR = base

- segment PQ = segment OS = height

So its area is base * height, which you can call A2.

Then the area of the polygon MNOPQRS is A1 + A2. One of them is 9 u^2 and the other is what the answer is asking for, and that you have calculated above.

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Answer:

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Step-by-step explanation:

My explanation is in the picture. All I did was replace the known variable with the proper letter!

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