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SVETLANKA909090 [29]
3 years ago
9

Factor the GCF: 30a3b4 + 20a2b2 - 15ab3. Please help

Mathematics
1 answer:
Blizzard [7]3 years ago
8 0

Answer:

The GCF for the variable part is ab².

GCF<em>Variable</em>=ab²

Multiply the GCF of the numerical part 5 and the GCF of the variable part ab².

5ab²

Step-by-step explanation:

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1/6

Step-by-step explanation:

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Evaluate the limit of tan 4x/ 4tan3x​
Brut [27]

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  1/3

Step-by-step explanation:

The ratio is undefined at x=0, so we presume that's where we're interested in the limit. Both numerator and denominator are zero at x=0, so L'Hôpital's rule applies. According to that rule, we replace numerator and denominator with their respective derivatives.

  \displaystyle\lim\limits_{x\to 0}\dfrac{\tan{(4x)}}{4\tan{(3x)}}=\lim\limits_{x\to 0}\dfrac{\tan'{(4x)}}{4\tan'{(3x)}}=\lim\limits_{x\to 0}\dfrac{4\sec{(4x)^2}}{12\sec{(3x)^2}}=\dfrac{4}{12}\\\\\boxed{\lim\limits_{x\to 0}\dfrac{\tan{(4x)}}{4\tan{(3x)}}=\dfrac{1}{3}}

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3 years ago
Consider the expansion of (5p + 2q)^6. Determine the coefficients for the terms with the powers of p and q shown.
Step2247 [10]

Answer:

Remember, the expansion of (x+y)^n is (x+y)^n=\sum_{k=0}^n \binom{n}{k}x^{n-k}y^k, where \binom{n}{k}=\frac{n!}{(n-k)!k!}.

Then,

(5p+2q)^6=\sum_{k=0}^6\binom{6}{k}(5p)^{6-k}(2q)^k=\sum_{k=0}^6\binom{6}{k}5^{6-k}2^k p^{6-k}q^k

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