2. -3n(2n – 5) = 0

2 answers:

8 0

$-3n(2n-5)=0\iff -3n=0\ \vee\ 2n-5=0\\\\-3n=0\ \ \ |:(-3)\\n=0\\\\2n-5=0\ \ \ \ |+5\\2n=5\ \ \ |:2\\n=\dfrac{5}{2}$

$Answer:\ c.\ n=0\ or\ n=\dfrac{5}{2}$

____ [38]3 years ago

7 0

Altho' it's rather obvious what you're supposed to do here, I'd suggest you always include all of the instructions.

-3n(2n – 5) = 0 is easily solved. Ignore the coefficient -3. Then you have n(2n-5)=0 Then either n=0 or 2n=5, which results in n=0 and n = 5/2 (solutions).

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Answer:

A. $f(x)=3(\frac{1}{2})^x$

Step-by-step explanation:

The options are:

$A. f(x)=3(\frac{1}{2})^x\\\\B. f(x)=\frac{1}{2}(3)^x\\\\C. f(x)=(3)^{2x}\\\\ D. f(x)=3^{(\frac{1}{2}x)}$

For this exercise it is important to remember that, by definition, the Exponential parent functions have the form shown below:

$f(x) = a^x$

Where "a" is the base.

There are several transformations for a function f(x), some of those transformations are shown below:

1. If $bf(x)$ and $b>1$ , then the function is stretched vertically by a factor of "b".

2. If $bf(x)$ and $0$ , then the function is compressed vertically by a factor of "b"

Therefore, based on the information given above, you can identify that the function that represents a vertical stretch of an Exponential function, is the one given in the Option A. This is:

$f(x)=3(\frac{1}{2})^x$

Where the factor is:

$b=3$