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3 years ago

-7(5h + 9) = 17 - 8h please help i need this FAST

Mathematics

2 answers:

Katyanochek1 [597]3 years ago

4 0

Answer:

$h = -2 \frac{26}{27}$

Step-by-step explanation:

Isolate the variable by dividing each side by factors that don't contain the variable.

lisabon 2012 [21]3 years ago

3 0

Answer:

h = - 80/27

Step-by-step explanation:

-7(5h + 9) = 17 - 8h

Distribute the -7 to the terms inside the parentheses

-35h - 63 = 17 - 8h

Add 63 to both sides

-35h - 63 + 63 = 17 - 8h + 63

Simplify

-35h = 80 - 8h

Add 8h to both sides

-35h + 8h = 80- 8h + 8h

Simplify

-27h = 80

Divide both sides by -27

h = - 80/27

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Step-by-step explanation:

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Step 1: substitute x = 0 into the function

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Step 2: Apply L'Hôpital's rule, by differentiating the numerator and denominator of the function

$= \lim_{ x\to \ 0} \dfrac{\frac{d}{dx}[ sin(x)-tan(x)]}{\frac{d}{dx} (x^3)}\\= \lim_{ x\to \ 0} \dfrac{cos(x)-sec^2(x)}{3x^2}\\$

Step 3: substitute x = 0 into the resulting function

$= \dfrac{cos(0)-sec^2(0)}{3(0)^2}\\= \frac{1-1}{0}\\= \frac{0}{0} (ind)$

Step 4: Apply L'Hôpital's rule, by differentiating the numerator and denominator of the resulting function in step 2

$= \lim_{ x\to \ 0} \dfrac{\frac{d}{dx}[ cos(x)-sec^2(x)]}{\frac{d}{dx} (3x^2)}\\= \lim_{ x\to \ 0} \dfrac{-sin(x)-2sec^2(x)tan(x)}{6x}\\$

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Step 6: Apply L'Hôpital's rule, by differentiating the numerator and denominator of the resulting function in step 4

$= \lim_{ x\to \ 0} \dfrac{\frac{d}{dx}[ -sin(x)-2sec^2(x)tan(x)]}{\frac{d}{dx} (6x)}\\= \lim_{ x\to \ 0} \dfrac{[ -cos(x)-2(sec^2(x)sec^2(x)+2sec^2(x)tan(x)tan(x)]}{6}\\\\= \lim_{ x\to \ 0} \dfrac{[ -cos(x)-2(sec^4(x)+2sec^2(x)tan^2(x)]}{6}\\$