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2 years ago

PLEASE HELP ME WITH THIS ASAP AND SHOW WORK PLEASE

Mathematics

1 answer:

Shkiper50 [21]2 years ago

4 0

<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

1/8x+1/4=y

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

3. Solve for y:

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

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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marusya05 [52]

<h3>Answer: 1</h3>

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

Explanation:

The rule is

x^0 = 1

where x is nonzero (ie, not equal to zero). So whatever number you raise to the zeroth power, the answer is always 1.

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

Let's see why this rule works

x^1 = x ... raising something to the first power is itself

(x^1)/x = x/x ... divide both sides by x

(x^1)/x = 1 .... turn the 'x/x' into 1

(x^1)/(x^1) = 1 .... rewrite the second 'x' as 'x^1'

x^(1-1) = 1 ... subtract exponents

x^0 = 1

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3 years ago

a square red rug has a purple square in the center. the side length of the purple square is x inches. the width of the red band

Bumek [7]

Answer:

Area of red band is $(20x + 100) inch^{2}$

Step-by-step explanation:

We know that area of square = $side^{2}$

Please refer to the attached figure, we have to calculate the area of red band.

Required area = Area of square <em>ABCD </em>- Area of <em>purple square </em>

Side of purple square = $x$

So, area of purple square = $x^{2}$

<em>ABCD </em>is a square with purple square at the center and there is symmetry in the figure.

So, width of red band towards both the end is 5 inches.

$\Rightarrow \text{side of }ABCD = 5 + 5 +x = (x+10) inches$

Required area = $(x+10)^{2} - x^{2}$

Using formula $a^{2} - b^{2} = (a-b)\times (a+b)$

$\Rightarrow (x+10-x)(x+10+x)\\\Rightarrow (10)(2x+10)\\\Rightarrow (20x+100)$

Hence, Area of red band is $(20x + 100) inch^{2}$

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Answer:

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Step-by-step explanation:

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3 years ago

Read 2 more answers

Please help me solve this, and show your work

zlopas [31]

$6r+5=11r\ \ \ \ |\text{subtract 6r from both sides}\\\\5=5r\ \ \ \ |\text{divide both sides by 5}\\\\\boxed{r=1}\\-------------------------\\\dfrac{2x-4}{2}=6x+3\\\\\dfrac{2x}{2}-\dfrac{4}{2}=6x+3\\\\x-2=6x+3\ \ \ \ \ |\text{add 2 to both sides}\\\\x=6x+5\ \ \ \ |\text{subtract 6x from both sides}\\\\-5x=5\ \ \ \ \ |\text{divide both sides by -5}\\\\\boxed{x=-1}$

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