HELP NOW BRAINLIST AND 15 POINTS

Mathematics

2 answers:

Greeley [361]3 years ago

5 0

Using Vertical angles and Supplementary angles see the picture for all the angles.

lidiya [134]3 years ago

3 0

If 8= 49 then 6, 4, and 2 = 49 as well. That would mean 1, 3, 5 and 7 all equal 41. We can figure this out because we know vertical angles are equal and corresponding angles are equal

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Can you help me plz? I'll mark brainilest

Vladimir79 [104]

The answer is amplitude

8 0

3 years ago

For every truck a police officer encounters she writes down T, for every car C, for every van V, and for every vehicle that is n

Kruka [31]

Answer:

5 = T

8 = C

3 = V

3 = O

which means its other because it said it is not a car, truck or van so its other meaning O

O = other

Answer: O

Step-by-step explanation: Theirs really nothing to it just pay attention to the hint :)

4 0

3 years ago

Reduce to lowest terms.

Furkat [3]

Hi friend,
((5+a)/a)/((a^2 -25) /5a)
=(a+5)/((a^2 - 25)/5)
=(a+5)(5)/(a+5)(a-5)
=5/(a-5)
Therefore your answer is B)
Hope this helps you!

3 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

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