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Strike441 [17]
2 years ago
7

Beth will you help Andy with his algebra?

Mathematics
2 answers:
katrin [286]2 years ago
8 0
A,
Beth, will you help Andy with his algebra?
Luda [366]2 years ago
4 0

Answer:

A)

Step-by-step explanation:

mark me brainliest plzzzzzzz

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Find the next three terms of the sequence 80, –160, 320, –640, . . .
posledela
1280, -2560, 5120

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Someone please help me I’ll give out brainliest please dont answer if you don’t know
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it's 1130.97

Step-by-step explanation:

V = A h.

Since the area of a circle = π r 2 , then the formula for the volume of a cylinder is:

V = π r 2 h.

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Simplify the expression -3 devided by -2/5. A. -7 1/2 B. -1 1/5 C. 1 1/5 D. 7 1/2
Llana [10]

Answer:

D. 7 1/2

Step-by-step explanation:

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Hope it helps

7 0
3 years ago
Two clothing stores are having a sale on all their dress shirts. Store A is charging $15 for each dress shirt. Store B is $18 fo
Klio2033 [76]

Answer:

C

Step-by-step explanation:

When the steps are followed. They both pay the same amount

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1. Express <img src="https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bx%282x%2B3%29%20%7D" id="TexFormula1" title="\frac{1}{x(2x+3) }" a
katovenus [111]

1. Let a and b be coefficients such that

\dfrac1{x(2x+3)} = \dfrac ax + \dfrac b{2x+3}

Combining the fractions on the right gives

\dfrac1{x(2x+3)} = \dfrac{a(2x+3) + bx}{x(2x+3)}

\implies 1 = (2a+b)x + 3a

\implies \begin{cases}3a=1 \\ 2a+b=0\end{cases} \implies a=\dfrac13, b = -\dfrac23

so that

\dfrac1{x(2x+3)} = \boxed{\dfrac13 \left(\dfrac1x - \dfrac2{2x+3}\right)}

2. a. The given ODE is separable as

x(2x+3) \dfrac{dy}dx} = y \implies \dfrac{dy}y = \dfrac{dx}{x(2x+3)}

Using the result of part (1), integrating both sides gives

\ln|y| = \dfrac13 \left(\ln|x| - \ln|2x+3|\right) + C

Given that y = 1 when x = 1, we find

\ln|1| = \dfrac13 \left(\ln|1| - \ln|5|\right) + C \implies C = \dfrac13\ln(5)

so the particular solution to the ODE is

\ln|y| = \dfrac13 \left(\ln|x| - \ln|2x+3|\right) + \dfrac13\ln(5)

We can solve this explicitly for y :

\ln|y| = \dfrac13 \left(\ln|x| - \ln|2x+3| + \ln(5)\right)

\ln|y| = \dfrac13 \ln\left|\dfrac{5x}{2x+3}\right|

\ln|y| = \ln\left|\sqrt[3]{\dfrac{5x}{2x+3}}\right|

\boxed{y = \sqrt[3]{\dfrac{5x}{2x+3}}}

2. b. When x = 9, we get

y = \sqrt[3]{\dfrac{45}{21}} = \sqrt[3]{\dfrac{15}7} \approx \boxed{1.29}

8 0
2 years ago
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