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3 years ago

I’ve looked at this for a while, can someone clear up the answers for me? I can’t figure it out.

Mathematics

1 answer:

navik [9.2K]3 years ago

4 0

What answers? I don't see any

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How do you solve number 27 and 28

Elanso [62]

27) if a = 1/a

multiply both sides by a

a * a = 1/a * a

a^ 2 = 1

Answer is D) 1

28)

6 x 75 = 450

7 x 80 = 560

8 x 85 = 680

9 x 90 = 810

Total all students in class = 6 + 7 + 8 + 9 = 30 students

Total score of whole class:

450 + 560 + 680 + 810 = 2500

Average = 2500 /30 = 83.3 = 83 1/3

Answer is C) 83 1/3

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3 years ago

Please help I need this ASAP.

Serjik [45]

Answer:

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Step-by-step explanation:

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3 years ago

The price of an item has been reduced by 65% the original price was $95

EastWind [94]

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3 years ago

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

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3 years ago

A^3 b^2/a^-1 b^-3 please help!!!!!!!!!!!!!!!!!!!!!!

SashulF [63]

Answer:

Hope the picture will help you.............

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2 years ago