HELP PLS is 8.4 or 8.40 bigger?

Mathematics

1 answer:

Sati [7]3 years ago

5 0

They are the same number.

8.4 is 8 units and 4 tenths.

8.40 is 8 units, 4 tenths and zero hundreths.

So, you've actually added nothing! Indeed, you can add as many zeroes as you like at the end of the decimal expansion:

$8.4 = 8.40 = 8.400 = 8.40000 = \ldots$

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An organization supports 125 villages financially. This figure is only 11% as compared to the organization's worldwide support.

Fittoniya [83]

If organisation supports 125 villages financially and it is 11% of world wide then the total villages in world wide that needs help is 1136.36 villages.

Given An organization supports 125 villages financially. This figure is only 11% as compared to the organization's worldwide support.

If 125 villages are 11% of worldwide villages who need help financially then we have to find 100% value according to organisation figure which can be calculated as under:

if 125 is 11% then 1% will be 125/11

then 100% will be 125/11*100

=12500/11

=1136.36

=1136 Approx

Hence 1136 villages are totally present worldwide which need help financially.

Learn more about percentage at brainly.com/question/24304697

#SPJ10

3 0

2 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

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