2 years ago

What is the area of an equilateral triangle with perimeter 24 inches?

Mathematics

1 answer:

Anna11 [10]2 years ago

8 0

Divide the perimeter by 3 to get 8 then use the equation for the area of a triangle (a= BH1/2) and your answer is 32

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Explain your answer please.

andriy [413]

Just do length x width x height

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3 years ago

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timofeeve [1]

Answer:

y-8

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4 0

2 years ago

Y + 12 = -26 thank you if you answer me

tia_tia [17]

Answer:

-38

Step-by-step explanation:

1. Isolate y to one side.

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2 years ago

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Connor earned $85 at his job when he worked for 10 hours. What was his hourly pay rate in dollars per hour?

Rus_ich [418]

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3 years ago

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Use l'Hôpital's Rule to find the following limit. ModifyingBelow lim With x right arrow 0StartFraction 3 sine (x )minus 3 x Ove

steposvetlana [31]

Answer:

$\lim_{x \to 0} \frac{3sinx-3x}{7x^3}=-\frac{1}{14}$

Step-by-step explanation:

The limit is:

$\lim_{x \to 0} \frac{3sinx-3x}{7x^3}=\frac{0}{0}$

so, you have an indeterminate result. By using the l'Hôpital's rule you have:

$\lim_{x \to 0} \frac{a(x)}{b(x)}= \lim_{x \to 0} \frac{a'(x)}{b'(x)}$

by replacing, and applying repeatedly you obtain:

$\lim_{x \to 0} \frac{3sinx-3x}{7x^3}= \lim_{x \to 0}\frac{3cosx-3}{21x^2}= \lim_{x \to 0}\frac{-3sinx}{42x}= \lim_{x \to 0}\frac{-3cosx}{42}\\\\ \lim_{x \to 0} \frac{3sinx-3x}{7x^3}=\frac{-3cos0}{42}=-\frac{1}{14}$

hence, the limit of the function is -1/14

8 0

3 years ago