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miss Akunina [59]
3 years ago
13

HELP!! 15 POINTS  Evaluate the expression when a = 4. 2a^3 A. 14 B. 24 C. 128 D. 512

Mathematics
2 answers:
spayn [35]3 years ago
7 0
2a^3 when a = 4:

First, to start out, our step should always be to substitute the value of the variable (a) into the problem. Below is what our new problem should look like with the value of the variable. Keep in mind that parentheses mean multiplication.
2(4)^{3}

Second, our next step is to simplify 4^3. Four raised to the third power equals 64, so 4^3 = 64. 
2 \times 64

Third, this last step is quite easy. All you need to do is (2 times 64). Best to use a calculator on this or take the hard way and do long multiplication. Either way, the answer for 2 times 64 is 128.

Answer: \fbox {C: 128}
Minchanka [31]3 years ago
4 0
C is the awseyto that pro
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It’s A because 2+3=5 and 5x7=35
8 0
3 years ago
WILL MARK BRAINLIEST <br><br> what is the domain of (f/g) (x)?
tatyana61 [14]

Given that f(x) = \sqrt{7-x} and g(x) = \sqrt{x + 2}, we can say the following:

\Bigg(\dfrac{f}{g}\Bigg)(x) = \dfrac{f (x)}{g(x)} = \dfrac{\sqrt{7 - x}}{\sqrt{x+2}}


Now, remember what happens if we have a negative square root: it becomes an imaginary number. We don't want this, so we want to make sure whatever is under a square root is greater than 0 (given we are talking about real numbers only).


Thus, let's set what is under both square roots to be greater than 0:

\sqrt{7 - x} \Rightarrow 7 - x \geq 0 \Rightarrow x \leq 7

\sqrt{x + 2} \Rightarrow x + 2 \geq 0 \Rightarrow x \geq -2


Since both of the square roots are in the same function, we want to take the union of the domains of the individual square roots to find the domain of the overall function.

x \leq 7 \,\,\cup x \geq -2 = \boxed{-2 \leq x \leq 7}


Now, let's look back at the function entirely, which is:

\Bigg( \dfrac{f}{g} \Bigg)(x) = \dfrac{\sqrt{7 - x}}{\sqrt{x+2}}

Since \sqrt{x + 2} is on the bottom of the fraction, we must say that \sqrt{x + 2} \neq 0, since the denominator can't equal 0. Thus, we must exclude \sqrt{x + 2} = 0 \Rightarrow x + 2 = 0 \Rightarrow x = -2 from the domain.


Thus, our answer is Choice C, or \boxed{ \{ x | -2 < x \leq 7 \}}.


<em>If you are wondering why the choices begin with the x | symbol, it is because this is a way of representing that x lies within a particular set.</em>

6 0
3 years ago
NEED HELP ASAP WILL MARK BRANLIEST!!!!!!
Blababa [14]
The heck, what math are you in?
7 0
3 years ago
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