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erma4kov [3.2K]
3 years ago
10

The graph of f(x) = 6(0.25)x and its reflection across the y-axis, g(x), are shown.

Mathematics
2 answers:
Alborosie3 years ago
8 0

Answer:The domain of g(x) is all real numbers.

Step-by-step explanation:

jeka57 [31]3 years ago
6 0

<u>Answer-</u>

The domain of g(x) is all real numbers.

<u>Solution-</u>

The given function is,

f(x)=6(0.25)^x

The reflection across y-axis of a point (x, y) will be (-x, y)

So, the reflection of the given function across y-axis will be,

g(x)=6(0.25)^{-x}

It is an exponential function. The domain of exponential functions is all real numbers and the range is all real numbers greater than zero.

Therefore, first option is correct.




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Jeannette has $5 and $10 bills in her wallet. the number of fives is three more than six times the number of tens. let t represe
sergejj [24]

Use the expression:

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3 0
3 years ago
Select all expressions that are equivalent to (2n+6)(n+3)
eduard

Answer:

(2n+6)(n+3)=

(2*n+2*3)(n+3)=

2(n+3)(n+3)=.....(1)

2(n+3)^2=....(2)

2(n^2+2*n*3+3^2) =

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2n^2+2*6n+2*9=

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6 0
3 years ago
In order to take honors-level math, Lisa must have an average of 85 on her math tests. Her scores so far are 85, 70, and 98. Wha
Mila [183]

Answer:

d) 87

Step-by-step explanation:

to solve this lets set up the equation

\frac{85+70+98+x}{4} =85\\\frac{253+x}{4} =85

then we multiply 4 on both sides

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5 0
2 years ago
Someone help please
Alla [95]

Answer:  Choice A

\tan(\alpha)*\cot^2(\alpha)\\\\

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

Explanation:

Recall that \tan(x) = \frac{\sin(x)}{\cos(x)} and \cot(x) = \frac{\cos(x)}{\sin(x)}. The connection between tangent and cotangent is simply involving the reciprocal

From this, we can say,

\tan(\alpha)*\cot^2(\alpha)\\\\\\\frac{\sin(\alpha)}{\cos(\alpha)}*\left(\frac{\cos(\alpha)}{\sin(\alpha)}\right)^2\\\\\\\frac{\sin(\alpha)}{\cos(\alpha)}*\frac{\cos^2(\alpha)}{\sin^2(\alpha)}\\\\\\\frac{\sin(\alpha)*\cos^2(\alpha)}{\cos(\alpha)*\sin^2(\alpha)}\\\\\\\frac{\cos^2(\alpha)}{\cos(\alpha)*\sin(\alpha)}\\\\\\\frac{\cos(\alpha)}{\sin(\alpha)}\\\\

In the second to last step, a pair of sine terms cancel. In the last step, a pair of cosine terms cancel.

All of this shows why \tan(\alpha)*\cot^2(\alpha)\\\\ is identical to \frac{\cos(\alpha)}{\sin(\alpha)}\\\\

Therefore, \tan(\alpha)*\cot^2(\alpha)=\frac{\cos(\alpha)}{\sin(\alpha)}\\\\ is an identity. In mathematics, an identity is when both sides are the same thing for any allowed input in the domain.

You can visually confirm that \tan(\alpha)*\cot^2(\alpha)\\\\ is the same as \frac{\cos(\alpha)}{\sin(\alpha)}\\\\ by graphing each function (use x instead of alpha). You should note that both curves use the exact same set of points to form them. In other words, one curve is perfectly on top of the other. I recommend making the curves different colors so you can distinguish them a bit better.

6 0
2 years ago
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