How the heck do I do this? And what’s the answer?

Mathematics

1 answer:

stira [4]2 years ago

5 0

Answer:

Option A

Step-by-step explanation:

We need to find two expressions that, when simplified, give the same results.

1) First, simplify the expression stated in the question. Multiply each of the terms in the parentheses by the number that is next to them. This would mean you have to multiply both 9x and -6 by $\frac{2}{3}$ . You also have to multiply $\frac{1}{2}x$ and $-\frac{1}{2}$ by 4. Then, simplify.

$\frac{2}{3}(9x-6) + 4 (\frac{1}{2} x - \frac{1}{2})\\\frac{18}{3}x- \frac{12}{3} + \frac{4}{2}x -\frac{4}{2} \\6x - 4 + 2x - 2$

2) Now, combine the like terms.

$6x - 4 + 2x - 2$

$8x - 6$

So, we need to find which of the expressions listed equal 8x - 6.

3) Let's try option A. Do the same as before. Multiply each of the terms in the parentheses by the number that is next to them. So, multiply 4x and -12 by $\frac{3}{4}$ . Also, multiply 30x and 18 by $\frac{1}{6}$ . Then, combine like terms and simplify.

$\frac{3}{4} (4x-12) + \frac{1}{6}(30x + 18)\\\frac{12}{4}x-\frac{36}{4} + \frac{30}{6} x + \frac{18}{6} \\3x - 9 + 5x + 3 \\8x - 6$

This also equals 8x - 6. Therefore, option A is the answer.

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For a function to begin to qualify as differentiable, it would need to be continuous, and to that end you would require that $a$ is such that

$\displaystyle\lim_{x\to0^-}g(x)=\lim_{x\to0^+}g(x)\iff\lim_{x\to0}ax=\lim_{x\to0}x^2-3x$

Obviously, both limits are 0, so $g$ is indeed continuous at $x=0$ .

Now, for $g$ to be differentiable everywhere, its derivative $g'$ must be continuous over its domain. So take the derivative, noting that we can't really say anything about the endpoints of the given intervals: