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3 years ago

What is 8 6/8 ÷ 2 2/5 = ? (show ur work)

Mathematics

2 answers:

grigory [225]3 years ago

8 0

Answer:

Work is down below

in the picture below

Rama09 [41]3 years ago

3 0

Answer:

175/48

Step-by-step explanation:

Convert 8 6/8 and 2 2/5 to fractions
8 6/8 = 70/8 ; 2 2/5 = 12/5

Now, we have : 70/8 : 12/5
= 70/8 x 5/12
= 70x5/8x12

=350/96
Simplify the fraction, we get 175/48

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Find the area of a regular pentagon with an apothem of 11 units. *

docker41 [41]

Answer:

<h2>440 square units</h2>

Step-by-step explanation:

We can find half of a side of the pentagon with the expression

$tan(36\°)=\frac{s_{half} }{11}$

Because, as a regular polygon, all its sectors have the same central angle, and the apothem divides equally each sector in two equal parts.

$s_{half} \approx 8$

Therefore, half of a side is 8 units long, which means each side measures 16 units.

Now, the area of a penthagon is defined by

$A=\frac{p \times a}{2}$

Where $p$ is the perimeter and $a$ is the apothem. Where the perimeter is the sum of all sides, which is 80 units.

$A= \frac{80 \times 11}{2} =440 \ u^{2}$

Therefore, the right answer is the second choice.

3 0

3 years ago

Is 7 1/5 a rational number

kolbaska11 [484]

Yes because when you turn into a decimal it's 7.2

7 0

3 years ago

Read 2 more answers

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

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