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Law Incorporation [45]
3 years ago
6

PLEASE HELP, I WILL MARK BRAINLIEST!

Mathematics
1 answer:
Rainbow [258]3 years ago
4 0

Answer:

5

Step-by-step explanation:

There are 5 different segments.

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Answer the equations:<br> 1. 4x+16<br> 2. 8 + 4x+8<br> 3. x+8+x<br> 4. 4+2x+4<br> 5. x+1+2x+8
andrew-mc [135]

Answer:

Answers below

Step-by-step explanation:

1)  4 (x+4)

2) 4x + 16

3) 2x+8

4) 2x+8

5) 3x+9

3 0
3 years ago
Read 2 more answers
3) (7,5), (9,-4)<br> A) (-1, 4.5)<br> G) (6, 2.5)<br> B) (11.-13)<br> D) (8,0.5)
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Step-by-step explanation:  474747-4747474-

6 0
3 years ago
3.6 terminating or repeating fraction
devlian [24]

Answer:

11/3

Step-by-step explanation:

7 0
4 years ago
Complete similarity statement, then state what the scale factor is.
geniusboy [140]

Step-by-step explanation:

XVW

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7 0
3 years ago
CALC- limits<br> please show your method
gladu [14]
A. Factor the numerator as a difference of squares:

\displaystyle\lim_{x\to9}\frac{x-9}{\sqrt x-3}=\lim_{x\to9}\frac{(\sqrt x-3)(\sqrt x+3)}{\sqrt x-3}=\lim_{x\to9}(\sqrt x+3)=6

c. As x\to\infty, the contribution of the terms of degree less than 2 becomes negligible, which means we can write

\displaystyle\lim_{x\to\infty}\frac{4x^2-4x-8}{x^2-9}=\lim_{x\to\infty}\frac{4x^2}{x^2}=\lim_{x\to\infty}4=4

e. Let's first rewrite the root terms with rational exponents:

\displaystyle\lim_{x\to1}\frac{\sqrt[3]x-x}{\sqrt x-x}=\lim_{x\to1}\frac{x^{1/3}-x}{x^{1/2}-x}

Next we rationalize the numerator and denominator. We do so by recalling

(a-b)(a+b)=a^2-b^2
(a-b)(a^2+ab+b^2)=a^3-b^3

In particular,

(x^{1/3}-x)(x^{2/3}+x^{4/3}+x^2)=x-x^3
(x^{1/2}-x)(x^{1/2}+x)=x-x^2

so we have

\displaystyle\lim_{x\to1}\frac{x^{1/3}-x}{x^{1/2}-x}\cdot\frac{x^{2/3}+x^{4/3}+x^2}{x^{2/3}+x^{4/3}+x^2}\cdot\frac{x^{1/2}+x}{x^{1/2}+x}=\lim_{x\to1}\frac{x-x^3}{x-x^2}\cdot\frac{x^{1/2}+x}{x^{2/3}+x^{4/3}+x^2}

For x\neq0 and x\neq1, we can simplify the first term:

\dfrac{x-x^3}{x-x^2}=\dfrac{x(1-x^2)}{x(1-x)}=\dfrac{x(1-x)(1+x)}{x(1-x)}=1+x

So our limit becomes

\displaystyle\lim_{x\to1}\frac{(1+x)(x^{1/2}+x)}{x^{2/3}+x^{4/3}+x^2}=\frac{(1+1)(1+1)}{1+1+1}=\frac43
3 0
3 years ago
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