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Darya [45]
3 years ago
9

What happens to the value of the expression 80/h as h increases from a small positive number to a large positive number?

Mathematics
2 answers:
Alex787 [66]3 years ago
6 0

Answer:

The numerator decreases when the denominator increases

Step-by-step explanation:

STatiana [176]3 years ago
4 0

For this case we have the following expression:

\frac {80} {h}

We substitute a small number h, that is, h = 0.05

So:

\frac {80} {0.05} = 1600

It is observed that the value of the expression is very large, that is, it increases with small values of h

We substitute a large number h, that is, h = 500

So:

\frac {80} {500} = 0.16

It is observed that the value of the expression decreases.

Answer:

The value of the given expression decreases when h increases from a small positive number to a large positive number.

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<h2>Step-by-step explanation:</h2>

A: f(x)=8x-2

1. Switch the f(x) to y

y=8x-2

2. Switch y to x and x to y

x=8y-2

3. Solve for y:

------------------------------

1. Add 2 to both sides.

 x=8y-2

+2     +2

2. Divide both sides by 8.

x+2=8y

---     ---

8       8

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3. Write in f-inverse form

f^-1(x)=1/8x+1/4  

B: g(x)=2/3x+6

1. Switch the g(x) to y

y=2/3x+6

2. Switch y to x and x to y

x=2/3y+6

3. Solve for y:

------------------------------

1. Subtract 6 from both sides.

x=2/3y+6

-6        -6

2. Multiply both sides by 3/2 (to cancel out the 2/3)

x-6=2/3y

2/3x-4=y

3. Write in f-inverse form

g^1(x)=2/3x-4

C: h(x)-4x-12

1. Switch the h(x) to y

y=-4x-12

2. Switch y to x and x to y

x=-4y-12

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------------------------------

1. Add 12 to both sides.

x=-4y-12

+12    +12

2. Divide both sides by -4

x+12=-4y

------    ---

-4        -4

-1/4x-3=y

3. Write in h-inverse form

h^2(x)=-1/4/x-3

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