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3 years ago

2p-7make p the subject

Mathematics

1 answer:

n200080 [17]3 years ago

4 0

Answer:

making letters subject is only possible

with equations but not expression.

So the question has no solution.

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Function P represents the perimeter, in inches, of a square with the side length x inches.

tatyana61 [14]

Answer:

x =0,1,2,3,4,5,6

P(x) =0,4,8,12,16,20,24

Step-by-step explanation:

P(x)= 4x

4 x 1 = 4

4 x 2 = 8

4 x 3 =12

4 x 4 = 16

4 x 5 = 20

4 x 6 = 24

7 0

3 years ago

4TH TIME ASKING THIS!!! Please help me! Someone pleaseeee. I need the correct answers. I don’t want to fail

Alexeev081 [22]

Answer:

The functions are inverses; f(g(x)) = x ⇒ answer D

$h^{-1}(x)=\sqrt{\frac{x+1}{3}}$ ⇒ answer D

Step-by-step explanation:

* Lets explain how to find the inverse of a function

- Let f(x) = y

- Exchange x and y

- Solve to find the new y

- The new y = $f^{-1}(x)$

* Lets use these steps to solve the problems

∵ $f(x)=\sqrt{x-3}$

∵ f(x) = y

∴ $y=\sqrt{x-3}$

- Exchange x and y

∴ $x=\sqrt{y-3}$

- Square the two sides

∴ x² = y - 3

- Add 3 to both sides

∴ x² + 3 = y

- Change y by $f^{-1}(x)$

∴ $f^{-1}(x)=x^{2}+3$

∵ g(x) = x² + 3

∴ $f^{-1}(x)=g(x)$

∴ The functions are inverses to each other

* Now lets find f(g(x))

- To find f(g(x)) substitute x in f(x) by g(x)

∵ $f(x)=\sqrt{x-3}$

∵ g(x) = x² + 3

∴ $f(g(x))=\sqrt{(x^{2}+3)-3}=\sqrt{x^{2}+3-3}=\sqrt{x^{2}}=x$

∴ f(g(x)) = x

∴ The functions are inverses; f(g(x)) = x

* Lets find the inverse of h(x)

∵ h(x) = 3x² - 1 where x ≥ 0

- Let h(x) = y

∴ y = 3x² - 1

- Exchange x and y

∴ x = 3y² - 1

- Add 1 to both sides

∴ x + 1 = 3y²

- Divide both sides by 3

∴ $\frac{x + 1}{3}=y^{2}$

- Take √ for both sides

∴ ± $\sqrt{\frac{x+1}{3}}=y$

∵ x ≥ 0

∴ We will chose the positive value of the square root

∴ $\sqrt{\frac{x+1}{3}}=y$

- replace y by $h^{-1}(x)$

∴ $h^{-1}(x)=\sqrt{\frac{x+1}{3}}$

4 0

3 years ago

How can you use the DISTRIBUTIVE PROPERTY solve the equation: 3(x-4) = 24 ? *

guapka [62]

By using the Distributive Property, you would multiply 3 into the parenthesis (3 · x and 3 · -4)

3(x - 4) = 24

3x - 12 = 24

Then, you would solve like any normal one-step equation.

3x - 12 = 24

+12 +12

__________

3x 36

__ = __

3 3

x = 12, which is your final solution.

If you need any further explanations, please ask me :)

7 0

3 years ago

Read 2 more answers

If f(x) = 5x2+2, then what is the slope of f(x) at x = 4?

REY [17]

The slope of a function at a point is the value of its derivative there.

... f'(x) = 5·(2x) + 0 = 10x . . . . . . using the power rule: (d/dx)(xⁿ) = n·xⁿ⁻¹

Then

... f'(4) = 10·4 = 40 . . . . . the slope at x=4

6 0

3 years ago

Step 1: Place the graph paper in landscape orientation. Measure from the top left hand corner 6 inches right and 5 inches down.

cluponka [151]

<h3>How to determine the hypotenuse?</h3>

Step 1 and 2: Draw an isosceles right triangle

See attachment (figure 1) for this triangle

The legs of this triangle have a length of 1 inch

Step 3: The hypotenuse

This is calculated using the following Pythagoras theorem

$h^2 = 1^2 + 1^2$

This gives

$h = \sqrt 2$

Step 4: Draw another isosceles right triangle

Add 1 inch to one of the legs

See attachment (figure 2) for this triangle

The legs of this triangle have lengths of 1 inch and 2 inches, respectively

This hypotenuse is calculated using the following Pythagoras theorem

$h^2 = 2^2 + 1^2$

This gives

$h = \sqrt 5$

Step 5: Draw another isosceles right triangle

Add 1 inch to one of the legs

See attachment (figure 3) for this triangle

The legs of this triangle have lengths of 1 inch and 3 inches, respectively

This hypotenuse is calculated using the following Pythagoras theorem

$h^2 = 3^2 + 1^2$

This gives

$h = \sqrt {10$

Step 6: Draw another isosceles right triangle

Add 1 inch to one of the legs

See attachment (figure 4) for this triangle

The legs of this triangle have lengths of 1 inch and 4 inches, respectively

This hypotenuse is calculated using the following Pythagoras theorem

$h^2 = 4^2 + 1^2$

This gives

$h = \sqrt{[17$

See that the hypotenuse is the square root of 17

Hence, the right triangle whose legs are 1 inch and 4 inches has an hypotenuse of √17

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