All categories

3 years ago

How many times larger is the value of 150,000 than 150?

Mathematics

2 answers:

stiks02 [169]3 years ago

4 0

1,000 times larger! You're welcome :) All you do it divide the two. Hope this helped!

V125BC [204]3 years ago

3 0

1000. Because 150a = 150,000. So 1000x150=150,000 so a=1000

You might be interested in

Colin rides his bike 10 and a half miles every morning and 12 and 1/4 miles every afternoon on Monday he wrote his usual distanc

jeka94

12 1/4- 1 1/2 >>> Convert fractions to decimals:
12.25- 1.5 = 10.75 >>> Convert decimals back to fractions:
10 3/4 >>> Add miles from the morning: 10 + 10 3/4 = 20 3/4 miles

3 0

2 years ago

Maria bought seven boxes. A week later half of all her boxes were destroyed in a fire. There are only 22 boxes left. With how ma

Sloan [31]

Answer:

22*2

=44

Step-by-step explanation:

7 0

2 years ago

Which expression is equivalent to y x 48?

Nostrana [21]

Answer: (y*40)+(y*8)

Step-by-step explanation:

7 0

2 years ago

First make a substitution and then use integration by parts to evaluate the integral. (Use C for the constant of integration.) x

e-lub [12.9K]

Answer:

$(\frac{x^{2}-25}{2})ln(5+x)-\frac{x^{2}}{4}+\frac{5x}{2}+C$

Step-by-step explanation:

Ok, so we start by setting the integral up. The integral we need to solve is:

$\int x ln(5+x)dx$

so according to the instructions of the problem, we need to start by using some substitution. The substitution will be done as follows:

U=5+x

du=dx

x=U-5

so when substituting the integral will look like this:

$\int (U-5) ln(U)dU$

now we can go ahead and integrate by parts, remember the integration by parts formula looks like this:

$\int (pq')=pq-\int qp'$

so we must define p, q, p' and q':

p=ln U

$p'=\frac{1}{U}dU$

$q=\frac{U^{2}}{2}-5U$

q'=U-5

and now we plug these into the formula:

$\int (U-5)lnUdU=(\frac{U^{2}}{2}-5U)lnU-\int \frac{\frac{U^{2}}{2}-5U}{U}dU$

Which simplifies to:

$\int (U-5)lnUdU=(\frac{U^{2}}{2}-5U)lnU-\int (\frac{U}{2}-5)dU$

Which solves to:

$\int (U-5)lnUdU=(\frac{U^{2}}{2}-5U)lnU-\frac{U^{2}}{4}+5U+C$

so we can substitute U back, so we get:

$\int xln(x+5)dU=(\frac{(x+5)^{2}}{2}-5(x+5))ln(x+5)-\frac{(x+5)^{2}}{4}+5(x+5)+C$

and now we can simplify:

$\int xln(x+5)dU=(\frac{x^{2}}{2}+5x+\frac{25}{2}-25-5x)ln(5+x)-\frac{x^{2}+10x+25}{4}+25+5x+C$

$\int xln(x+5)dU=(\frac{x^{2}-25}{2})ln(5+x)-\frac{x^{2}}{4}-\frac{5x}{2}-\frac{25}{4}+25+5x+C$

$\int xln(x+5)dU=(\frac{x^{2}-25}{2})ln(5+x)-\frac{x^{2}}{4}+\frac{5x}{2}+C$

notice how all the constants were combined into one big constant C.

7 0

3 years ago

Solve this problem on paper using all four steps. A girl scout troop sold cookies. If the girls sold 5 more boxes the second wee

zhenek [66]

Assume the girls sold X boxes in the first week. They sold in the second week X+5 and 2X+10 in the third week.

The sum X + X + 5 + 2X + 10 = 431
4X = 416. Therefore X = 104
The answers are:
a0 = 104
a1 = 109
a2 = 218

Hope that helps you :)

6 0

3 years ago