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3 years ago

What value can you multiply by 3/4 to get a product of 1

Mathematics

1 answer:

Rus_ich [418]3 years ago

4 0

Multiply by the reciprocal, 4/3

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The expression f(x) = 12(1.035)* models the monthly growth of membership in the new drama club at a school. According to the fun

olga nikolaevna [1]

Answer:

The monthly growth rate is 3.5%.

Step-by-step explanation:

The exponential growth function is given as follows:

$y=a(1+r)^{x}$

Here,

y = final value

a = initial value

r = growth rate

x = time taken

The provided expression for the monthly growth of membership in the new drama club at a school is:

$f(x) = 12\cdot(1.035)^{x}$

Comparing this function with the exponential growth function:

$a(1+r)^{x}=12(1.035)^{x}\\\\a(1+r)^{x}=12(1+0.035)^{x}$

Then value of r is 0.035 or 3.5%.

Thus, the monthly growth rate is 3.5%.

7 0

3 years ago

Find the slope of the line (Image attached) A) 5 B) -5 C) 1/5 D) -1/5

Inessa [10]

The slope is 1/5 as [rise/run]

7 0

3 years ago

Read 2 more answers

Find x 6787+x=99798999? dom join my remind ......28gc97d

wolverine [178]

Answer:

x=99792212

Step-by-step explanation:

get the x alone by subtracting 6787 from both sides

6787-6787+x=99798999-6787

x=99792212

Hope this helps!

5 0

2 years ago

Read 2 more answers

Can someone check whether its correct or no? this is supposed to be the steps in integration by parts

Gwar [14]

Answer:

$\displaystyle - \int \dfrac{\sin(2x)}{e^{2x}}\: \text{d}x=\dfrac{\sin(2x)}{4e^{2x}}+\dfrac{\cos(2x)}{4e^{2x}}+\text{C}$

Step-by-step explanation:

$\boxed{\begin{minipage}{5 cm}\underline{Integration by parts} \\\\$\displaystyle \int u \dfrac{\text{d}v}{\text{d}x}\:\text{d}x=uv-\int v\: \dfrac{\text{d}u}{\text{d}x}\:\text{d}x$ \\ \end{minipage}}$

Given integral:

$\displaystyle -\int \dfrac{\sin(2x)}{e^{2x}}\:\text{d}x$

$\textsf{Rewrite }\dfrac{1}{e^{2x}} \textsf{ as }e^{-2x} \textsf{ and bring the negative inside the integral}:$

$\implies \displaystyle \int -e^{-2x}\sin(2x)\:\text{d}x$

Using integration by parts:

$\textsf{Let }\:u=\sin (2x) \implies \dfrac{\text{d}u}{\text{d}x}=2 \cos (2x)$

$\textsf{Let }\:\dfrac{\text{d}v}{\text{d}x}=-e^{-2x} \implies v=\dfrac{1}{2}e^{-2x}$

Therefore:

$\begin{aligned}\implies \displaystyle -\int e^{-2x}\sin(2x)\:\text{d}x & =\dfrac{1}{2}e^{-2x}\sin (2x)- \int \dfrac{1}{2}e^{-2x} \cdot 2 \cos (2x)\:\text{d}x\\\\& =\dfrac{1}{2}e^{-2x}\sin (2x)- \int e^{-2x} \cos (2x)\:\text{d}x\end{aligned}$

$\displaystyle \textsf{For }\:-\int e^{-2x} \cos (2x)\:\text{d}x \quad \textsf{integrate by parts}:$

$\textsf{Let }\:u=\cos(2x) \implies \dfrac{\text{d}u}{\text{d}x}=-2 \sin(2x)$

$\textsf{Let }\:\dfrac{\text{d}v}{\text{d}x}=-e^{-2x} \implies v=\dfrac{1}{2}e^{-2x}$

$\begin{aligned}\implies \displaystyle -\int e^{-2x}\cos(2x)\:\text{d}x & =\dfrac{1}{2}e^{-2x}\cos(2x)- \int \dfrac{1}{2}e^{-2x} \cdot -2 \sin(2x)\:\text{d}x\\\\& =\dfrac{1}{2}e^{-2x}\cos(2x)+ \int e^{-2x} \sin(2x)\:\text{d}x\end{aligned}$

Therefore:

$\implies \displaystyle -\int e^{-2x}\sin(2x)\:\text{d}x =\dfrac{1}{2}e^{-2x}\sin (2x) +\dfrac{1}{2}e^{-2x}\cos(2x)+ \int e^{-2x} \sin(2x)\:\text{d}x$