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3 years ago

Solve: (w)(3w+5)=0. Multiple answers, if any, should be separated by commas. Enter only integer(s) or fraction(s).

Mathematics

1 answer:

sp2606 [1]3 years ago

7 0

Answer:

$-\frac{5}{3}$

Step-by-step explanation:

Given expression:

(w)(3w + 5) = 0

Expand the parentheses;

3w² + 5w = 0

3w² = -5w

Divide both sides by w;

3w = - 5

w = $-\frac{5}{3}$

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Answer:

$\dfrac{-1}{6}$

Step-by-step explanation:

Given the limit of a function expressed as $\lim_{ x\to \ 0} \dfrac{sin(x)-tan(x)}{x^3}$ , to evaluate the following steps must be carried out.

Step 1: substitute x = 0 into the function

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Step 3: substitute x = 0 into the resulting function

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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

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Step 7: substitute x = 0 into the resulting function in step 6

$= \dfrac{[ -cos(0)-2(sec^4(0)+2sec^2(0)tan^2(0)]}{6}\\\\= \dfrac{-1-2(0)}{6} \\= \dfrac{-1}{6}$

<em>Hence the limit of the function </em> $\lim_{ x\to \ 0} \dfrac{sin(x)-tan(x)}{x^3} \ is \ \dfrac{-1}{6}$ .

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