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Helen [10]
3 years ago
7

Help anyone need it asap 8th grade math

Mathematics
2 answers:
jolli1 [7]3 years ago
8 0

Answer:

you answer is 0

Step-by-step explanation:

If you do the math it goes to this

4(3x-5)=6x-2

x=?

so you would do the distributive property for the perenthasese

so..

4(4x3X-4x5) = 4x3= 12x. 4x5=20. 12x-20= -8x

then you would do 6x-2 = 4x

then 8x+4x=0

nekit [7.7K]3 years ago
3 0

Answer:

Step-by-step explanation:

4(3x - 5) = 6x - 2                     Remove the brackets

12x - 20 = 6x - 2                     Subtract 6x from both sides

12x - 6x - 20 = -2                   Add 20 to both sides

6x = -2 + 20

6x = 18                                    Divide both sides by 6

x = 18/6

x = 3

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

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\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

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\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}

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\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}

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

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Therefore:

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