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Arturiano [62]
4 years ago
12

HELP!!!

Mathematics
1 answer:
san4es73 [151]4 years ago
6 0
Choosing "r" things from a set of "n" can be done in this many ways:
combinations = n! / [r! * (n-r)!]
combinations = 7! / [3! * 4!]

combinations = 7 * 6 * 5 / 6
combinations = 35

Source:http://www.1728.org/combinat.htm

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Can someone pls help me with this?
Misha Larkins [42]

Answer:

the answer is C

Step-by-step explanation:

3 0
3 years ago
37-(-3)squared+30\(-3)x2
vampirchik [111]
Simplify: 
37 - (-3)^2  + 30/(-3)x^2
= 37 +  - 9 +  - 10x^2

Combine like terms:
= 37 +  - 9 +  - 10x^2
= ( - 10x^2) + (37 +   - 90
= - 10X^2 + 28
Answer: - 10x^2  +  28



To Factor: 
37 - (- 3)^2 + 30/- 3x^2

- 10x^2 + 28
- 2(5x^2 - 14)

Answer:  - 2(5x^2 - 14)




I simplify, and factor, because, you didn't really specify what you wanted help on with this equation. - Thanks -





Hope that helps!!!!
6 0
3 years ago
A polynomial p has zeros when x=5, x=-1 and x=-1/4
kiruha [24]

Answer:

p(x) = (5x - 1) (x + 4) (x - 2)

4 0
3 years ago
Read 2 more answers
Teach me how to add this problem
Arada [10]

start with the ones and every time it goes over ten take the first number and add it to the tens for example 1+8+4+8=21 take the one and put it in the one spot below. Then take the 2 and add it to the six in the tens spot so it would be Instead of 6+8+3+2 it would be 8+8+3+2 which would equal 21 then put the 1 u=in the ten spot below add the 2 to the hundreds spot making is 5+5+7=4=14 put the 4 Below and add the 1 the the thousands making it 3+4+5+1=13 put the one on the ten thousands and put the three below the whole problem would be 13,411.

7 0
4 years ago
Pls help with steps 2 and 4!!! (Will give brainliest)
Natalka [10]

The Answer to your problem is:

#2 Answer:

13x + 7y


#4 Answer:

13x + 7y


I believe it's Distribute

6 0
3 years ago
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