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Dmitriy789 [7]
3 years ago
11

What is the minimum y-value after which the exponential function will always be greater than the linear function? y = 1 y = 3 y

= 4 y = 5
Mathematics
1 answer:
Stells [14]3 years ago
5 0

Answer:

y = 4.

Step-by-step explanation:

I suppose that this question relates to the image that can be seen below.

In the image, the green line represents the exponential function and the blue line represents the linear function.

The y-value after which the exponential function will always be greater than the linear function is the y-value where bot graphs intersect, such that after that point, the blue line starts increasing fast, and is always above the green line.

In this case, this point is the second intersection, and we can see that this intersection happens in the point (2, 4)

Remember that the usual notation for points is (x, y).

Then the y-value after which the exponential function will always be greater than the linear function is y = 4.

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Help me solve this! Can’t figure out how to set up the equation.
Katarina [22]

Answer: GF = 7. Answer choice C

Step-by-step explanation:

The triangles are similar, meaning their angles and sides are proportional. You can use ratios to solve this. Find out what all the sides are. I used variables to represent the side lengths because we don’t know how long they are.

Step 1) Find side IG. Because we don’t know it, we can just use a variable. I used “y”.

Step 2) Find side GH. GH is the same length as IG, so we can use the same variable.

Step 3) find side IH by adding IG to GH. This gives us y + y, which equals 2y. We can label side IH with 2y.

Step 4) find side IF. I used the variable “a” because we don’t know side IF. Side FJ is the same length as IF, so we can label it “a” too.

Step 5) find side IJ.

IJ is IF + FJ.

IF = a. FJ = a.

So IJ = a + a. IJ = 2a.

Now we know the all the sides of both triangles.

Step 6) set up fractions of the sides of the triangles. Put a side from triangle IHJ in the numerator and a side from triangle IGF on the denominator. You have to match up the sides, though.

Side IH should go over IG. IJ/IF. HJ/GF

IH/IG = 2y/y

IJ/IF = 2a/a

HJ/GF = (4x-2)/(x + 3)

Step 7) make an equation if these fractions. They are all equal to each other.

2y/y = 2a/a = (4x-2)/(x + 3)

Step 8) simplify the fractions

2 = 2 = (4x-2)/(x+3)

FYI you don’t have to write 2=2, you can just write one 2. So it would be: 2 = (4x-2)/(x+3)

Step 9) factor (4x-2)

4x and -2 are both divisible by 2, so we can factor out 2.

This gives us: 2(2x-1)

We can’t factor (x+3).

Now the fraction is: 2(2x-1)/(x+3)

Step 10) rewrite the equation

2(2x-1)/(x+3) = 2

Step 11) Solve this equation.

You will end up with: x = 4

Step 12) Find side GF

GF = x + 3

Plug in 4 for x (because x = 4)

GF = 4 + 3

GF = 7

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