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3 years ago

Explain why √3x+2 = -16 has no solution.

2 answers:

8 0

The equation you wrote has no solutions since the left hand side is equal to 6x-10 and the right hand side is equal to 6x+8. Setting them equal we get -10=8 which is obviously wrong.

Artist 52 [7]3 years ago

3 0

Answer:

Step-by-step explanation:

If we subtract 2 from both sides we have √3x= -18

It's impossible to take a square root of a number and have it be negative

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Review the graph of complex number z.<br> What is the modulus of<br> Z.

Serggg [28]

Answer: square root of 10

Step-by-step explanation:

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3 years ago

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PLS HELP ME AS SOON AS POSSIBLE

Veseljchak [2.6K]

Answer:

7a. D)

7b. $78

Step-by-step explanation:

7a: We're given the information that there's a daily fee of $50 and $0.35 per mile driven, which is m. We can write "$0.35 per mile driven" as 0.35m. Then, the total amount of money it costs represented as an expression is 50 + 0.35m, thus, making it answer D).

7b: In this problem, we have to take our previous expression 50 + 0.35m, plug in the value of m, and simplify.

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Thus, the answer is $78.

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8 0

3 years ago

Analyze the expression and then choose the sentences describing whether the binomial theorem can be used to factor it. Check all

daser333 [38]

Answer:

Step-by-step explanation:

N is 3 bc #of terms minus 1

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Can be factored using binomial theorem

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3 years ago

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Answer:

$y=\pm 2\sqrt{17}$

Step-by-step explanation:

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4 0

2 years ago

Which of these limits evaluate to 0?

Vikentia [17]

<h3>Answer: C) I and II only</h3>

===============================================

Work Shown:

Part I

$\displaystyle \lim_{x \to 2}\frac{x-2}{x+2} = \frac{2-2}{2+2}\\\\\\\displaystyle \lim_{x \to 2}\frac{x-2}{x+2} = \frac{0}{4}\\\\\\\displaystyle \lim_{x \to 2}\frac{x-2}{x+2} = 0\\\\\\$

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Part II

$\displaystyle \lim_{x \to 0}\frac{\sin(x)}{x+2} = \frac{\sin(0)}{0+2}\\\\\\\displaystyle \lim_{x \to 0}\frac{\sin(x)}{x+2} = \frac{0}{2}\\\\\\\displaystyle \lim_{x \to 0}\frac{\sin(x)}{x+2} = 0\\\\\\$

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Part III

$\displaystyle \lim_{x \to 5}\frac{x}{x} = \lim_{x \to 5}1\\\\\\\displaystyle \lim_{x \to 5}\frac{x}{x} = 1\\\\\\$

7 0

3 years ago