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Nikitich [7]
3 years ago
10

In each figure below, find m<1 and m<2 if a is parallel to b. You don't have to show work.

Mathematics
1 answer:
Gekata [30.6K]3 years ago
5 0

Answer:

m <5 = 71 degrees.

m <8 = 109 degrees.

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Find the probability that a randomly
Lerok [7]

Fine the area of the circle:

Area of a circle = pi x r ^2

Area of the circle = 3.14 x 4^2 = 50.24 square cm

Find the area of the triangle:

Area of triangle = 1/2 x base x height

Area =1/2 x 8 x 4 = 16 square cm

Find the area of the white part by subtracting the area of the triangle from the circle:

50.24 - 16 = 34.24 square cm.

The probability of landing in white is the area of white/ area of circle:

34.24/50.24 = 0.682

Multiply by 100 to get percent:

0.682 x 100 = 68.2%

4 0
2 years ago
Solve 12|x|−7=8. Does anyone know how to do this?<br>​
Travka [436]

Answer:

8+8=16/12

x=16/12, 4/3

4 0
3 years ago
The book rack sold 24 paperbacks on Saturday, 17 of them mysteries. What percent of the books were mysteries?
igomit [66]

Answer:

About 71% of the books were mystery books

Step-by-step explanation:

To find the percentage of something you simply divide the amounts. 17 of the 24 books were mystery so to find the percentage of mystery books you do 17 divided by 24

17/24= 0.70833333333

and round it to 71

8 0
2 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
YO PLEASE HELP ME ASAP!!! Which explicit function defines this arithmetic sequence?
BartSMP [9]

Answer:

351

Step-by-step explanation:

d

6 0
3 years ago
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