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TiliK225 [7]
2 years ago
12

Which statement is true about the graphed function?

Mathematics
2 answers:
Ksenya-84 [330]2 years ago
7 0

Answer:

d

Step-by-step explanation:

Nady [450]2 years ago
3 0

Answer:

the last option or D

Step-by-step explanation:

f(x) > 0 over the interval (-∞, -4)

i hope this helps! quizlet is also very helpful lol, have a wonderful rest of your day/night, xx, nm <3 :)

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alexgriva [62]
Shane and Isabella have 154 dollars
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3 years ago
GEOMETRY- What is the area of the figure, show your work please!
Kamila [148]
Since this is a 45-45-90 right triangle, the hypotenuse = sqrt2 * leg
24 is the hypotenuse so we need to find the leg.
24 = sqrt2 * leg
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leg = 16.97
A = (1/2)b*h
A = (1/2)*16.97*16.97
A = 144ft^2
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Which of the following expressions is equal to 5 - 2m?
zmey [24]

Answer:

the second one

Step-by-step explanation:

7 - (2m - 5)

3 0
2 years ago
Read 2 more answers
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
Solve for x.
olga nikolaevna [1]

Step-by-step explanation:

4.6 -.75 =

? \3 \frac{7}{20}

so x <3 7/20

3 0
3 years ago
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