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3 years ago

Value 23.7 is 1/10 of

Mathematics

1 answer:

inna [77]3 years ago

6 0

Is 19:17 because u have to value the number of the first like this 78:807y :997

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What is the answer to 9 m+ 9

klemol [59]

Answer: 18m

Step-by-step explanation:

9m + 9

Simplify:

9+9=18

Hence, the answer to 9m + 9 is 18m.

3 0

3 years ago

Please help me solve this equation and show the work show me how you got the answer

Nesterboy [21]

Answer: k=3

Step-by-step explanation:

To solve for k, you want to isolate k.

$\frac{2}{5} (5k+35)-8=12$ [add both sides by 8]

$\frac{2}{5} (5k+35)=20$ [multiply both sides by 5/2]

$5k+35=50$ [subtract both sides by 35]

$5k=15$ [divide both sides by 5]

$k=3$

Now, we know that k=3.

7 0

2 years ago

Which solution is correct? Be sure to check for extraneous solutions.

denis23 [38]

That answer is d)x=-4

4 0

2 years ago

Whats the answer for 12 (2w-3)=6w

puteri [66]

12(2w-3)=6w
Distribute
24w-36=6w
Rearrange
18w=36
Divide both sides by 18
W=36/18
Final answer:
W=2
Hope I could help with the work, but give makeitnasty credit for the answer, he got it first :)

8 0

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

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

When $n\rightarrow \infty$ both $\displaystyle\frac{3}{n}$ and $\displaystyle\frac{1}{n^2}$ tend to zero and the upper sum tends to

$\displaystyle\frac{4^3}{3}+12=\displaystyle\frac{100}{3}$

8 0

3 years ago