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3 years ago

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Given+the+formula+:+A=Pe+rt++Find+the+amount+of+money+made+if+you+invested+$65,000+at+7.3%25+interest%3F&

1 answer:

anastassius [24]3 years ago

6 0

65,000e^(.073x25)
$403,181.68

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Isabella constructed a cylindrical vase that has a volume of 63 cubic centimeters.Jasonbconstructed a cone that had the same hei

Korvikt [17]

<span>cylindrical vase
r=radius </span>cylinder=radius cone
h= height cylinder=height cone
Volume cylindrical vase (V1)= 63 cm³=pi*r² *h

cone
Volume cone (V2)=pi*r² *h
if radius cylinder=radius cone and height cylinder=height cone
then V1=V2
therefore
V2= 63 cm³

the answer is The volume of cone is 63 cm³

5 0

3 years ago

[Help Asap, Will Mark Brainliest] Answer is in the image,

Vera_Pavlovna [14]

Z=118 as vertically opposite angles are the same

x= (8x-50)

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3 years ago

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Mekhanik [1.2K]

Answer: what is it?

Step-by-step explanation:

6 0

3 years ago

Put 9, -7, and -8 in order from least to greatest. You can use the number line to help. 9 -10 -8 -7 -6 -8 -4 -2 0 2 4 6 8 10

boyakko [2]

-8, -7, 9

Step-by-step explanation:

8 0

2 years ago

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

3 years ago