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2 years ago

A metric ruler can be used to measure what type of object?? Please help me!!

Mathematics

2 answers:

Xelga [282]2 years ago

8 0

Answer:

A pencil

Step-by-step explanation:

Reil [10]2 years ago

3 0

Answer: a water bottle

Step-by-step explanation:

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Last summer ,at camp okey fun okey,the ratio of the number of boys campers to the girls campers was 8:7. If there were a total o

olya-2409 [2.1K]

<span>With a boys to girls ratio of 8:7, this means 8/15 of the campers are boys and 7/15 of the campers are girls.

We are told that there are 195 total campers.

To find # of boys: 195 (the total # of campers) x 8/15 (fraction of the campers that are boys) = 104 boys

To find # of girls: 195 (total # of campers) x 7/15 (fraction of the campers that are girls) = 91 girls

Note: If you do not know how to multiply by fractions, let me know, I have another trick that takes more time but doesn't require the use of fractions.

</span>

3 0

3 years ago

I need help finding the slope

Studentka2010 [4]

Answer:

its negative 1/3

Step-by-step explanation:

shawtyyyy

8 0

3 years ago

What is 1.53 to the nearest tenth

Marat540 [252]

1.53 rounded to the nearest tenth is 1.5

7 0

3 years ago

Read 2 more answers

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

During a science experiment,the temperature of a solution in beaker 1 was 5 degrees below zero. The temperature of a solution in

GaryK [48]

Answer:
5 degrees

Step-by-step explanation:

We are given that during a science experiment
The temperature of a solution in a beaker 15 degrees below zero
The temperature of solution in a beaker 2 was opposite of the temperature in beaker 1
We have to determine the temperature of beaker 2.
It mean temperature of beaker 1 solution=-5 degrees
According to question
Temperature of beaker 2 solution =5 degrees
Hence, the temperature of solution in a beaker 2=5 degrees

7 0

1 year ago