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lutik1710 [3]
3 years ago
15

WILL GIVE BRAINLIEST

Mathematics
2 answers:
Natalka [10]3 years ago
6 0

Answer:

A and B

Step-by-step explanation:

Inessa05 [86]3 years ago
4 0
THE ANSWER IS B AND A
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The question is in the picture PLEASE HELP ME!!! idk how to do this
IrinaVladis [17]

Answer:

35

Step-by-step explanation:

Put the numbers in the formula and do the arithmetic.

nCk = n!/(k!(n -k)!)

7C3 = 7!/(3!(7-3)!) = 7·6·5/(3·2·1) = 7·5 = 35

_____

It is convenient to use the largest of the factorials in the denominator to cancel as many factors as you can from the numerator, then cancel factors from the remaining numbers. Here after canceling 4! = 4·3·2·1 from the numerator, we are left with 7·6·5 divided by 3! = 3·2·1 = 6. Obviously, this will cancel the 6 in the numerator product, leaving only 7·5 = 35.

Some graphing and/or scientific calculators will have this function built in.

5 0
3 years ago
What is the factored form of x^12y^18+1
Black_prince [1.1K]

Answer:  The required factored form of the given expression is

(x^4y^6+1)(x^8y^{12}-x^4y^6+1).

Step-by-step explanation:  We are given to find the factored form of the following algebraic expression:

E=x^{12}y^{18}+1.

We will be using the following formula:

a^3+b^3=(a+b)(a^2+ab+b^2).

Now, we have

E\\\\=x^{12}y^{18}+1\\\\=(x^4y^6)^3+1^3\\\\=(x^4y^6+1)\{(x^4y^6)^2-x^4y^6\times1+1^2\}\\\\=(x^4y^6+1)(x^8y^{12}-x^4y^6+1).

Thus, the required factored form of the given expression is

(x^4y^6+1)(x^8y^{12}-x^4y^6+1).

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3 years ago
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vivado [14]

Answer:

trapezoid

Hope help

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