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3 years ago

HELP MEEEEEE PLESEEE

Mathematics

2 answers:

lara [203]3 years ago

5 0

Answer:

648 cubic centimeters

Step-by-step explanation:

6×9×12=648

viktelen [127]3 years ago

4 0

Answer:

648 cubic centimeters

Step-by-step explanation:

All you need to do is multiply all of the measures that they show you.

6 x 9 x 12 which equals 648

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What is the value of the exponential expression below?<br> 36 1/2

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Answer:

D. 6

Step-by-step explanation:

$36 {}^{ \frac{1}{2} } = \sqrt{36} = 6$

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3 years ago

Please help me and thanks!! <33 :]

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3 years ago

Jason spends 3 1/12 hours biking and 2 1/3 hours at the mall. How much less time does Jason spend at the mall compared to biking

aleksley [76]

We must find the difference between 3 1/12 and 2 1/3. So we can first subtract 3 from 2 and we earn 1 whole number. Next we must subtract 1/12 and 1/3. This is the same as 3/36-12/36, which is -9/36 or -1/4. Since we can't have -1/4 we must subtract 1/4 from 1 and earn 3/4 as our answer.

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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

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