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3 years ago

The formula for the volume of a cone is v= 1/3 xr^2 h. The radius, r, of the cone may be expressed as

Mathematics

1 answer:

Alexxandr [17]3 years ago

5 0

Answer:

r=√(3v)/(xh)

Step-by-step explanation:

the volume of a cone is given as

v= 1/3 xr^2 h. Then we need the radius, r, of the cone

v= 1/3 xr^2 h

Let us cross multiply we have

3v= xr^2h

Make r^2 subject of the formula we have

r^2= 3v/xh

Find the square root of bother sides we have

r=√(3v)/(xh)

the radius, r, of the cone can be expressed as r=√(3v)/(xh)

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What are the zeros of the function f(x) = x2 + 8x + 4, expressed in simplest radical form? x = –4 ± 2StartRoot 3 EndRoot x = –4

SCORPION-xisa [38]

Answer:

Step-by-step explanation:

Discriminant Δ = √(8²-4×(1)×(4)) = √(48) = √(4×12) = √(4)×√(12) = 2√(12)

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3 years ago

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Explanation:
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Hope this helps!! :)

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3 years ago

Jay and Blair went fishing. Together they catch 27 fish. Jay catch 2 times as many fish as Blair. How many fish did Jay and Blai

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3 years ago

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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

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