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2 years ago

What is the solution to the linear equation?

Mathematics

2 answers:

Burka [1]2 years ago

8 0

$p=1$

To better understand the problem see the attached figure

we have

$\frac{2}{5}+p=\frac{4}{5}+\frac{3}{5}p$

Multiply by $5$ both sides

$5(\frac{2}{5}+p)=5(\frac{4}{5}+\frac{3}{5}p)$

$2+5p=4+3p$

Group terms that contain the same variable p and move the constant to the other side of the equation

$5p-3p=4-2$

$2p=2$

$p=1$

MrRissso [65]2 years ago

4 0

Answer:

p=1

Step-by-step explanation:

We are given that an equation

$\frac{2}{5}+p=\frac{4}{5}+\frac{3}{5}p$

We have to find the solution of given linear equation .

Subtraction property of equality

$p-\frac{3}{5}p+\frac{2}{5}=\frac{4}[5}$

$\frac{5p-3p}{5}+\frac{2}{5}=\frac{4}{5}$

Combine like terms

$\frac{2}{5}p+\frac{2}{5}=\frac{4}{5}$

$\frac{2}{5}p=\frac{4}{5}-\frac{2}{5}$

Subtraction property of equality

$\frac{2}{5}p=\frac{2}{5}$

When a fraction term in multiply goes form one side to other side then the term will be reciprocal.

$p=\frac{2}{5}\times \frac{5}{2}$

$p=1$

Hence, p=1

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$\displaystyle \int \text{f}(x)\:\text{d}x=\text{F}(x)+\text{C} \iff \text{f}(x)=\dfrac{\text{d}}{\text{d}x}(\text{F}(x))$

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$\boxed{\textsf{For }\sqrt{a^2-x^2} \textsf{ use the substitution }x=a \sin \theta}$

$\textsf{Let }x=3 \sin \theta$

$\begin{aligned}\implies \sqrt{3^2-x^2} & =\sqrt{3^2-(3 \sin \theta)^2}\\ & = \sqrt{9-9 \sin^2 \theta}\\ & = \sqrt{9(1-\sin^2 \theta)}\\ & = \sqrt{9 \cos^2 \theta}\\ & = 3 \cos \theta\end{aligned}$

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$\implies \text{d}x=3 \cos \theta\:\:\text{d}\theta$

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Take out the constant:

$\implies \displaystyle \dfrac{1}{9} \int \dfrac{1}{\cos^2 \theta}\:\:\text{d}\theta$

$\textsf{Use the trigonometric identity}: \quad\sec^2 \theta=\dfrac{1}{\cos^2 \theta}$

$\implies \displaystyle \dfrac{1}{9} \int \sec^2 \theta\:\:\text{d}\theta$

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