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3 years ago

BRAINLIESTTT ASAP!!! PLEASE HELP ME :) Explain your answer please and show your steps/work.

Mathematics

1 answer:

algol [13]3 years ago

4 0

Answer:

√x

Step-by-step explanation:

First, simplify -3⁄6 to -½. Next, according to the Negative Exponential Rule, move the denominator to the numerator to get x¹\²:

bⁿ = 1\b⁻ⁿ

Then, according to the Definition of Rational Exponents [part I], rewrite this as a radical, which in this case would be √x. This is when your numerator is 1:

ⁿ√aᵐ = aᵐ\ⁿ

If you are ever in need of assistance, do not hesitate to let me know by subscribing to my You-Tube channel [USERNAME: MATHEMATICS WIZARD], and as always, I am joyous to assist anyone at any time.

**I have a video that explains all six exponential rules. It is titled "Six Exponential Procedures". I encourage you to watch the video and gain alot more knowledge from it, so you can better understand the concept.

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3 years ago

For the graphed exponential equation, calculate the average rate of change from x = -3 to x = 0

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3 years ago

If f(x) = 62.7"), what is 13)?

Lemur [1.5K]

Answer:

f(x)=|x−13|+11

To find the x

coordinate of the vertex, set the inside of the absolute value x−13 equal to 0. In this case, x−13=0

x−13=0

Add 13

to both sides of the equation.

x=13

Replace the variable x

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y=|(13)−13|+11

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hope this helps ;)

6 0

3 years ago

All four B points are equidistant from points A and C. Are they all midpoints of AC? Justify your answer.

Rina8888 [55]

Answer:

Step-by-step explanation:

For a point to be the midpoint of a line segment, it must bisect it into two equal segments and be on the line segment (hence, colinear with the endpoints). All four B points are equidistant from points A and C, but aren’t colinear with A and C. Therefore, they aren’t all midpoints of line segment AC.

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5 0

2 years ago

why write a quadratic function whose graph has the given characteristics vertex :(2,3) point on graph (0,-1)

GaryK [48]

<h3>Answer:</h3>

f(x) = -(x -2)² +3

<h3>Step-by-step explanation:</h3>

We can fill in the vertex (h, k) values immediately in the vertex form ...

... f(x) = a(x -h)² +k

To find the value of a, we solve the equation for a at some point other than the vertex. The given point is (0, -1), so we can use that:

... -1 = a(0 -2)² +3

... -4 = 4a . . . . . . . . . subtract 3, simplify

... -1 = a . . . . . . . . . . . divide by 4

Now, we know the function is ...

... f(x) = -(x -2)² +3

5 0

3 years ago