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Alenkinab [10]
3 years ago
7

What is the solution set for 4x−3=2x+5 , given the replacement set {2, 3, 4, 5} ?

Mathematics
2 answers:
antiseptic1488 [7]3 years ago
8 0
The answer is C. x=4
Illusion [34]3 years ago
5 0

Answer:

C.x=4

Step-by-step explanation:

We are given that an equation

4x-3=2x+5

The replacement set ={2,3,4,5}

We have to find the solution of given equation.

4x-3=2x+5

4x-2x-3=5

Using subtraction property of equality

2x-3=5

Combine like terms

2x=5+3

Subtraction property of equality

2x=8

Addition property of integers.

x=\frac{8}{2}=4

Division property of equality

Hcene, the solution of given equation is  given by

x=4

Therefore, option C is true.

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

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

Using <u>integration by parts</u>:

\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}

\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}

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