Any help would be appreciated

Mathematics

1 answer:

ale4655 [162]3 years ago

7 0

It looks like 2.25 is the decimal. Since there are four sections between each number, those sections would be divided into quarters. The first quarter is 2 1/4, which is the same as 2.25.

Hope this helps! :)

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If f(x)=4(3x-5), find f^-1(x)

Margaret [11]

Answer:

f^-1(x) = (x+20) / 12

Step-by-step explanation:

f(x) = 4(3x-5)

Let y be the image of f.

y = 4(3x-5)

y = 12x-20

y+20 = 12x

x = (y+20) / 12

f^-1(y) = (y+20) / 12, so

f^-1(x) = (x+20) / 12

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3 years ago

2x2 =4 this is 4 or 3 ??????????<br>

Zolol [24]

Answer:

<h2>4 is the right answer</h2>

<h3>Originally Answered: why does 2*2=4 and 2+2=4? The truth is that a number multiplied by itself does not have to be equal to the same number added to itself. We can prove 2x2=4 easily. We have 2 groups of 2s.</h3>

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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

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