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Y_Kistochka [10]
3 years ago
13

How can you tell when a pattern shows counting on by 10

Mathematics
2 answers:
Sunny_sXe [5.5K]3 years ago
8 0
If you see a zero in the end of a number it might be counting on by ten, the hint is most likely the zero. Let's say you had a question like: 10_30. If you see the zero, you can make sense out of it and try the tens times table. So add ten to 10 and bam! You found your answer
Viefleur [7K]3 years ago
6 0
If you see a zero that means that you are counting by 10s For example: 10,20,30,40,50,60,70,80
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24 red marbles is40% of ? Marbles
Anvisha [2.4K]

Answer: The answer is 60

Step-by-step explanation: 40% = .4 because % means hundredths. So 40% = 40/100 = .4

Let x be the total number of marbles.

Then, .4x = 24

 Divide both sides by .4

         x = 60 and 24 is 40% of 60 marbles

5 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
I NEED HELP!!! 21-22
Nesterboy [21]

21. P=118

21. Q=142

22. P=120

22. R=60

22. S=60

3 0
3 years ago
Pls help me out i’ll give brainliest to best answer
Vedmedyk [2.9K]

Answer:

5. ABC and XYZ

Step-by-step explanation:

matching angle values

5 0
2 years ago
The fit line equation is Y = -2.61x +152.51, where x is variable 1 and Y is variable 2. According to the equation, what is the e
egoroff_w [7]

The expected value of y when x is equals to 45 in the equation is 35.06

<h3>How to find variable from an equation?</h3>

The equation is given as follows;

y = -2.61x + 152.51

where

  • x = variable 1
  • y = variable 2

Therefore,

when

x = 45

y = -2.61x + 152.51

y = -2.61(45) + 152.51

y = - 117.45 + 152.51

y = 35.06

learn more on equation here: brainly.com/question/14279419

#SPJ1

5 0
2 years ago
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