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2 years ago

5X-50+2X+80-X=180..............

2 answers:

7 0

If you are solving for X, the answer will be 25. To solve this, isolate the variable by dividing each side by factors that don’t contain the same variable
Hope this helps :)

JulsSmile [24]2 years ago

3 0

Hey there!

5x - 50 + 2x + 80 - x = 180

COMBINE the LIKE TERMS

= 6x + 30 = 180

SUBTRACT 30 to BOTH SIDES

6x + 30 - 30 = 180 - 30

CANCEL out: -30 - 30 because that gives you 0

KEEP: 180 - 30 because it helps solve for the x-value

6x = 180 - 30

180 - 30 = 150

NEW EQUATION: 6x = 150

DIVIDE 6 to BOTH SIDES

6x/6 P 150/6

CANCEL out: 150/6 because that gives you the result of the x-value

x = 150/6

SIMPLIFY IT!

x = 25

Therefore, your answer is: x = 25

Good luck on your assignment and enjoy your day!

~Amphitrite1040:)

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3 years ago

Calculate: ㅤ <img src="https://tex.z-dn.net/?f=%5Clim_%7Bx%20%5Crightarrow%20%2B%5Cinfty%7Dx%28%5Csqrt%7Bx%5E%7B2%7D-1%7D-x%29"

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$\displaystyle \large \boxed{ \lim_{x \rightarrow +\infty} {x\left(\sqrt{x^2-1}-x\right)}=-\dfrac{1}{2}}$

Step-by-step explanation:

Hello, please consider the following.

$\sqrt{(x^2-1)}-x\\\\=\sqrt{x^2(1-\dfrac{1}{x^2})}-x\\\\=x\left( \sqrt{1-\frac{1}{x^2}}-1\right)$

For x close to 0, we can write

$\sqrt{1+x}=1+\dfrac{1}{2}x-\dfrac{1}{8}x^2+o(x^2)\\\\\ \text{x tends to } +\infty \text{ means }\dfrac{1}{x} \text{ tends to 0}\\\\\text{So, when }\dfrac{1}{x}\text{ is close to 0, we can write.}\\\\\sqrt{1-\dfrac{1}{x^2}}=1-\dfrac{1}{2}\dfrac{1}{x^2}-\dfrac{1}{8}\dfrac{1}{x^4}+o(\dfrac{1}{x^4})$

So,

$x\left( \sqrt{1-\frac{1}{x^2}}-1\right)\\\\=x(1-\dfrac{1}{2}\dfrac{1}{x^2}+o(\dfrac{1}{x^2})-1)\\\\=-\dfrac{1}{2x}+o(\dfrac{1}{x})$

It means that

$\displaystyle \lim_{x \rightarrow +\infty} {x\left(\sqrt{x^2-1}-x\right)}\\\\=\lim_{x \rightarrow +\infty} {-\dfrac{x}{2x}}=-\dfrac{1}{2}$

Thank you

8 0

4 years ago