2 years ago

Helpqquuuiiiiccckkkkk

1 answer:

5 0

Here's some help. Hope this works.

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Sonja [21]

Answer:

1. -10

2. -11

Step-by-step explanation:

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2 years ago

Anarel [89]

Answer:

17/60

Step-by-step explanation:

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2 years ago

Kipish [7]

Answer:

The answer is C :) (x + 1)(x-3)(x-5)

Step-by-step explanation:

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2 years ago

GarryVolchara [31]

Rewrite the limit as

$\displaystyle\lim_{x\to0}x^2\log x^2=\lim_{x\to0}\frac{\log x^2}{\frac1{x^2}}$

Then both numerator and denominator approach infinity (with different signs, but that's not important). Applying L'Hopital's rule, we get

$\displaystyle\lim_{x\to0}\frac{\log x^2}{\frac1{x^2}}=\lim_{x\to0}\frac{\frac2x}{-\frac2{x^3}}=\lim_{x\to0}-x^2=\boxed{0}$

7 0

3 years ago

IgorLugansk [536]

6(2x + 5y)
6[2(3) + 5(4)]
6[6 + 20]
6(26)
156

The answer is 156

Hope this helps :)

4 0

3 years ago