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3 years ago

If segment AB segment BC and m<3 =36, find m<4

Mathematics

1 answer:

shepuryov [24]3 years ago

5 0

What are the measures of line segments AB and BC?

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Please help me! in need of help due tomorrow!

julsineya [31]

Answer:

Step-by-step explanation:

(1). A = π r²

A = 4 π

Area of shaded region is $\frac{4\pi *260}{360}$ = $\frac{26}{9} \pi$ or $\frac{26\pi }{9}$

(2). C = 4 π

In this case, the length of the bigger arc and the area of shaded region happen to be the same.

The length of the arc ADB is $\frac{26}{9} \pi$ or $\frac{26\pi }{9}$

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1 year ago

If a is one-factor x, what is the other factor

NARA [144]

B-y is the another factor

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3 years ago

Which equation represents a linear function? Equation 1: y^3 = 2x +1 Equation 2: y= 3x +1 Equation 3: y= 5x^2 - 1 Equation 4: y=

Yuri [45]

the second equation is linear

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3 years ago

9.88,10.19,11.15,11.70,12.01,12.46,13.22,13.29,14.00,14.38,14.50,14.50,15.14,15.16,15.16,15.16,15.54,15.96,16.20,16.22,16.44,17.

Arada [10]

You can use this website called Quartile Calculator to answer this question. You just have to type in the numbers. <span />

6 0

3 years ago

For the function below, find a formula for the upper sum obtained by dividing the interval [a comma b ][a,b] into n equal subin

Vlad [161]

Answer:

See below

Step-by-step explanation:

We start by dividing the interval [0,4] into n sub-intervals of length 4/n

$[0,\displaystyle\frac{4}{n}],[\displaystyle\frac{4}{n},\displaystyle\frac{2*4}{n}],[\displaystyle\frac{2*4}{n},\displaystyle\frac{3*4}{n}],...,[\displaystyle\frac{(n-1)*4}{n},4]$

Since f is increasing in the interval [0,4], the upper sum is obtained by evaluating f at the right end of each sub-interval multiplied by 4/n.

Geometrically, these are the areas of the rectangles whose height is f evaluated at the right end of the interval and base 4/n (see picture)

$\displaystyle\frac{4}{n}f(\displaystyle\frac{1*4}{n})+\displaystyle\frac{4}{n}f(\displaystyle\frac{2*4}{n})+...+\displaystyle\frac{4}{n}f(\displaystyle\frac{n*4}{n})=\\\\=\displaystyle\frac{4}{n}((\displaystyle\frac{1*4}{n})^2+3+(\displaystyle\frac{2*4}{n})^2+3+...+(\displaystyle\frac{n*4}{n})^2+3)=\\\\\displaystyle\frac{4}{n}((1^2+2^2+...+n^2)\displaystyle\frac{4^2}{n^2}+3n)=\\\\\displaystyle\frac{4^3}{n^3}(1^2+2^2+...+n^2)+12$

but

$1^2+2^2+...+n^2=\displaystyle\frac{n(n+1)(2n+1)}{6}$

so the upper sum equals

$\displaystyle\frac{4^3}{n^3}(1^2+2^2+...+n^2)+12=\displaystyle\frac{4^3}{n^3}\displaystyle\frac{n(n+1)(2n+1)}{6}+12=\\\\\displaystyle\frac{4^3}{6}(2+\displaystyle\frac{3}{n}+\displaystyle\frac{1}{n^2})+12$

When $n\rightarrow \infty$ both $\displaystyle\frac{3}{n}$ and $\displaystyle\frac{1}{n^2}$ tend to zero and the upper sum tends to

$\displaystyle\frac{4^3}{3}+12=\displaystyle\frac{100}{3}$

8 0

4 years ago