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borishaifa [10]
3 years ago
5

These parallelograms are similar. Compare the first to the second and give the ratio of their perimeters. Write the ratio in sim

plest terms.
P1= 8
P2= 6
A. 64:36
B. 16:9
C. 4:3
D. 8:6
Mathematics
1 answer:
nlexa [21]3 years ago
4 0
Its c. hope it helps. :)
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Please help<br> thank you again
guajiro [1.7K]

<em><u>Answer:</u></em>

True

<em><u>Step-by-step explanation:</u></em>

So, again, from the most right column, we can add 5, -6, and -2 to get -3. We now know that every row or column must be equal to -3.

The question is asking if the number square below is -3. We found out from above that is -3 so it's true.

3 0
3 years ago
Please explain how this is incorrect.
MatroZZZ [7]

Step-by-step explanation:

The spores always x3 after a month so going a month back they ÷3 instead so 900÷3 is 300 spores on the first month

5 0
2 years ago
To approximate the number of bees in a hive, multiply the number of bees that leave the hive in one minute by 3 and divide by 0.
AlekseyPX

Answer:

There are about 5,357 bees in the hive

Step-by-step explanation:

Let

x ----->  the number of bees that leave the hive in one minute

y -----> the approximate number of bees in a hive

we know that

The formula to calculate the approximate number of bees in a hive is equal to

y=\frac{3x}{0.014}

For x=25

substitute

y=\frac{3(25)}{0.014}

y=5,357\ bees

therefore

There are about 5,357 bees in the hive

6 0
3 years ago
Read 2 more answers
Write and solve a real world word problem in which there are three tables for every 8 chairs
mylen [45]
Fsdfsdfsdfsdfsdfsdfsdfsdf
5 0
3 years ago
Find the indefinite integral. (Note: Solve by the simplest method—not all require integration by parts. Use C for the constant o
umka2103 [35]

Answer:

∫▒〖arctan⁡(x).1 dx=arctan⁡(x).x〗-1/2  ln⁡(1+x^2 )+C

Step-by-step explanation:

∫▒〖1st .2nd dx=1st∫▒〖2nd dx〗-∫▒〖(derivative of 1st) dx∫▒〖2nd dx〗〗〗

Let 1st=arctan⁡(x)

And 2nd=1

∫▒〖arctan⁡(x).1 dx=arctan⁡(x) ∫▒〖1 dx〗-∫▒〖(derivative of arctan(x))dx∫▒〖1 dx〗〗〗

As we know that  

derivative of arctan(x)=1/(1+x^2 )

∫▒〖1 dx〗=x

So  

∫▒〖arctan⁡(x).1 dx=arctan⁡(x).x〗-∫▒〖(1/(1+x^2 ))dx.x〗…………Eq1

Let’s solve ∫▒(1/(1+x^2 ))dx by substitution now  

Let 1+x^2=u

du=2xdx

Multiply and divide ∫▒〖(1/(1+x^2 ))dx.x〗 by 2 we get

1/2 ∫▒〖(2/(1+x^2 ))dx.x〗=1/2 ∫▒(2xdx/u)  

1/2 ∫▒(2xdx/u) =1/2 ∫▒(du/u)  

1/2 ∫▒(2xdx/u) =1/2  ln⁡(u)+C

1/2 ∫▒(2xdx/u) =1/2  ln⁡(1+x^2 )+C

Putting values in Eq1 we get

∫▒〖arctan⁡(x).1 dx=arctan⁡(x).x〗-1/2  ln⁡(1+x^2 )+C  (required soultion)

3 0
3 years ago
Read 2 more answers
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