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AfilCa [17]
2 years ago
12

On a coordinate plane, 2 exponential fuctions are shown. Function f (x) decreases from quadrant 2 into quadrant 1 and approaches

y = 0. It crosses the y-axis at (0, 6) and goes through (1, 2). Function g (x) approaches y = 0 in quadrant 2 and increases into quadrant 1. It goes through (negative 1, 2) and crosses the y-axis at (0, 6).
Which function represents g(x), a reflection of f(x) = 6(one-third) Superscript x across the y-axis?



g(x) = −6(one-third) Superscript x


g(x) = −6(one-third) Superscript negative x


g(x) = 6(3)x


g(x) = 6(3)−x

Mathematics
2 answers:
Gala2k [10]2 years ago
8 0

Answer:

g(x)=6(3)^x

Step-by-step explanation:

We are given  that

f(x)=6(\frac{1}{3})^x

Function f decreases from quadrant  2 to quadrant 1 and approaches  y=0

It cut the y- axis at (0,6) and passing through the point (1,2).

Function g(x) approaches y=0 in quadrant 2 and increases into quadrant 1.

It passing through the point (-1,2) and cut the y-axis at point (0,6).

Reflection across y- axis:

Rule of transformation is given by

(x,y)\rightarrow (-x,y)

Using the rule then we get

g(x)=6(\frac{1}{3})^{-x}=6(3)^x

By using

x^{-a}=\frac{1}{x^a}

Substitute x=-1

g(-1)=6\times (\frac{1}{3})=2

Substitute x=0

g(0)=6

Therefore,g(x)=6(3)^x is true.

SVEN [57.7K]2 years ago
6 0

Answer:

Its A on edg

Step-by-step explanation:

follow my ifunny "dankmemehistory"

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Mr harrles worked 40 houres this week he also got a bounuse of 180$ hemr.Abdulla took a takie from his home to the air port. The
Volgvan

Question:

Abdulla took a takie from his home to the air port. The taxi driver chared an initial fee of 12AED pluse 2.50AED per km. How much change should the taxi driver give mr abdulla bach if he gave him 100AED at the end of his trip

Answer:

88 - 2.50x

Step-by-step explanation:

Given :

Initial charge = 12

Charge per kilometer = 2.50

Amount given to driver = 100

Let distance from home to airport = x

Total charge = (initial charge + (distance * charge per km)

Total charge = 12 + 2.50x

Change = (Amount given to driver - total charge)

Change = (100 - (12 + 2.50x))

100 - 12 - 2.50x

88 - 2.50x

3 0
1 year ago
Consider the graphs of f (x) = StartAbsoluteValue x EndAbsoluteValue + 1 and g (x) = StartFraction 1 Over x cubed EndFraction. T
Westkost [7]

Answer:

We have the functions:

f(x) = IxI + 1

g(x) = 1/x^3.

Now, we know that the composite functions do not permute.

How we can prove this?

First, two composite functions are commutative if:

f(g(x)) = g(f(x))

Well, you could use brute force (just replace the values and see if the composite functions are commutative or not)

But i will use a more elegant way.

We can notice two things:

g(x) has a discontinuity at x = 0.

so:

f(g(x)) = I 1/x^3 I + 1

still has a discontinuty at x = 0, but:

g(f(x)) = 1/( IxI + 1)^3

here the denominator is IxI + 1, is never equal to zero.

So now we do not have a discontinuity.

Then the composite functions can not be commutative.

7 0
2 years ago
Read 2 more answers
A baker makes apple tarts and apple pies each day. Each tart, t, requires 1 apple, and each pie, p, requires 8 apples. The baker
topjm [15]
My guess is C.........
6 0
2 years ago
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Explain whether 13t−3t is equivalent to 3t−13t. Support your answer by evaluating the expressions for t=2.
djverab [1.8K]

Answer:

  • Not equivalent

Step-by-step explanation:

  • 13t - 3t = (13 - 3)t = 10t
  • 3t - 13t = (3 - 13)t = -10t
  • 10t ≠ -10t

They are not equivalent

<u>When t = 2</u>

  • 13t - 3t = 13*2 - 3*2 = 26 - 6 = 20
  • 3t - 13t = 3*2 - 13*2 = 6 - 26 = -20
6 0
2 years ago
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What’s the 38th term of the arithmetic sequence 31,40,49,58?
Trava [24]

Answer:

a_3_8=364

Step-by-step explanation:

we know that

In an <u><em>Arithmetic Sequence</em></u> the difference between one term and the next is a constant, and this constant is called the common difference

we have

31,40,49,58,...

Let

a_1=31\\a_2=40\\a_3=49\\a_4=58

we have that

a_2-a_1=40-31=9

a_3-a_2=49-40=9

a_4-a_3=58-49=9

so

The common difference is equal to 9

We can write an Arithmetic Sequence as a rule:

a_n=a_1+d(n-1)

where

a_n is the nth term                                                              

a_1 is the first term

d is the common difference                        

n is the number of terms

Find the 38th term of the arithmetic sequence

we have                  

a_1=31\\d=9\\n=38              

substitute the values

a_3_8=31+9(38-1)

a_3_8=31+9(37)

a_3_8=364

3 0
2 years ago
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