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3 years ago

My bank account started the day with $550 in it. During the day, I made 2

Mathematics

2 answers:

umka21 [38]3 years ago

8 0

Answer: 740$

Step-by-step explanation:

125x2= 250, 250-60= 190, 550+190= 740

Elanso [62]3 years ago

4 0

Answer:

$740

Step-by-step explanation:

550 + 2(125) -60 =

550 + 250 - 60

800 - 60 = 740

$740

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Step-by-step explanation:

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3 years ago

Solve 18n - 7p-15n = 5p for n.<br> A n = 1<br> B. n=2p<br> C. n = 1/2<br> D. n=4p

andrey2020 [161]

Answer:

<h3>The answer is option D</h3>

Step-by-step explanation:

18n - 7p-15n = 5p

To solve for n first make n the subject

Group like terms

That's

18n - 15n = 5p + 7p

Simplify

3n = 12p

Divide both sides by 3 to make n stand alone

That's

3n/3 = 12p/3

We have the final answer as

Hope this helps you

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3 years ago

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tino4ka555 [31]

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3 years ago

Find the value of x given the figure below.

kogti [31]

Here’s the step-by-step 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

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

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

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

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3 years ago