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brilliants [131]
2 years ago
13

What is the answer does this problem

Mathematics
1 answer:
Lapatulllka [165]2 years ago
3 0
The first letter or word is not there
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Someone please help it’s for home work
kifflom [539]
1 6 dis 9/3 please please come out with the best way
8 0
3 years ago
I need help, i do not understand, this is pythagorean theorem
ser-zykov [4K]

Answer:

11.2

Step-by-step explanation:

First draw an imaginary straight line to separate the two shapes.

You will have a side of 10 in the traingle

And a side of 13 - 8 =5

Then

Using pythagoras theorem

Sqrt(10^2+5^2)=11.180

5 0
3 years ago
PLEASE HELP I GIVE THANKS
drek231 [11]
A, C, And D

Could be these answers...
7 0
3 years ago
In how many ways can a
Tomtit [17]

Answer:

The answer is 364. There are 364 ways of choosing a recorder, a facilitator and a questioner froma club containing 14 members.

This is a Combination problem.  

Combination is a branch of mathematics that deals with the problem relating to the number of iterations which allows one to select a sample of elements which we can term "<em>r</em>" from a collection or a group of distinct objects which we can name "<em>n</em>". The rules here are that replacements are not allowed and sample elements may be chosen in any order.

Step-by-step explanation:

Step I

The formula is given as

C (n,r) = \frac{n}{r} = \frac{n!}{(r!(n-r)!)}

n (objects) = 14

r (sample) = 3

Step 2 - Insert Figures

C (14, 3) = (\frac{14}{3}) = \frac{14!}{(3!(14-3)!)}

= \frac{87178291200}{(6 X 39916800)}

= \frac{87178291200}{239500800}

= 364

Step 3

The total number of ways a recorder, a facilitator and a questioner can be chosen in a club containing 14 members therefore is 364.

Cheers!

6 0
3 years ago
PLEASE HELP FAST!!!
Llana [10]

Answer:

4500 miles

Step-by-step explanation:

750 rounded to the nearest hundred is 800. 2,488 rounded to the nearest hundred is 2,500. 155 rounded to the nearest hundred is 200. 1,015 rounded to the nearest hundred is 1,000. So 800+2,500+200+1,000= 4,500

8 0
3 years ago
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