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Olin [163]
3 years ago
15

What is thesimlified form of sqrt(48n^9)

Mathematics
1 answer:
garik1379 [7]3 years ago
4 0
4n^4 √3n I believe would be the simplified form.
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Giovanna has a score of
12345 [234]

Answer:'

-9.7 minus, in 13

Step-by-step explanation:

7 0
2 years ago
The areas of two similar triangles are 72dm2 and 50dm2. The sum of their perimeters is 226dm. What is the perimeter of each of t
astraxan [27]

Answer:

Hence, the perimeter of the triangles are:

P=123.2727 dm

P'=102.7272 dm

Step-by-step explanation:

In two similar triangles:

The ratio of the areas of two triangle is equal to the square of their perimeters.

Let A and A' represents the area of two triangles and P and P' represents their perimeter.

Then they are related as:

\dfrac{A}{A'}=\dfrac{P^2}{P'^2}

We are given:

A=72 dm^2  , A'=50 dm^2

and P+P'=226 dm.-----------(1)

i.e. \dfrac{72}{50}=\dfrac{P^2}{P'^2}\\\\\dfrac{36}{25}=\dfrac{P^2}{P'^2}

on taking square root on both the side we get:

\dfrac{P}{P'}=\dfrac{6}{5}\\\\P=\dfrac{6}{5}P'

Now putting the value of P in equation (1) we obtain:

\dfrac{6}{5}P'+P'=226\\\\\dfrac{6P'+5\times P'}{5}=226\\\\\dfrac{6P'+5P'}{5}=226\\\\11P'=226\times 5\\\\11P'=1130\\\\P'=\dfrac{1130}{11}=102.7272

Hence,

P=226-102.7272=123.2727

Hence, the perimeter of the triangles are:

P=123.2727 dm

P'=102.7272 dm

6 0
3 years ago
___ cosb =1/2 sin(a+b)+sin(a-b)?
vodomira [7]

Answer:

That would be sina.

Step-by-step explanation:

sin(a+b) = sinacosb + cosasinb

sin(a-b) = sinacosb -  cosasinb

Adding we get  sin(a+b) + sin(a-b) = 2sinaccosb

so sinacosb = 1/2sin(a+b) + sin(a-b)

8 0
3 years ago
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Need some help with this math question, can someone help me out pls
Sergeeva-Olga [200]

Okay I'm not too sure on this but I believe it is the third one.

I hope this helps!!

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3 years ago
You plan to go bowling at grand station. It cost $3 to rent shoes and $4 per game. You want to spend exactly $35 to use all your
raketka [301]
You can play 5 games to spend exactly $35
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3 years ago
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