2 years ago

Algebra PLS HELP!!!!!!!!!!!!!

1 answer:

4 0

Answer:

C - 53 Degrees

Step-by-step explanation:

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Mars2501 [29]

By the binomial theorem,

$\displaystyle \left(x^2-\frac1x\right)^6 = \sum_{k=0}^6 \binom 6k (x^2)^{6-k} \left(-\frac1x\right)^k = \sum_{k=0}^6 \binom 6k (-1)^k x^{12-3k}$

I assume you meant to say "independent", not "indecent", meaning we're looking for the constant term in the expansion. This happens for k such that

12 - 3k = 0 ===> 3k = 12 ===> k = 4

which corresponds to the constant coefficient

$\dbinom 64 (-1)^4 = \dfrac{6!}{4!(6-4)!} = \boxed{15}$

3 0

2 years ago

bearhunter [10]

Answer:

1/2 mile

Step-by-step explanation:

$\frac{2}{3} (\frac{3}{4} )=\frac{6}{12} =\frac{1}{2}$

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3 years ago

lapo4ka [179]

Answer:

155

Step-by-step explanation:

6 0

2 years ago

Vitek1552 [10]

The mode of the values listed is 9.

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2 years ago

NARA [144]

$9 \times 2 + 2 \times {5}^{2} \\ = 18 + 2 \times 25 \\ = 18 + 50 \\ = 68$

6 0

2 years ago