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3 years ago

X divided by -2 = 6 ( what does x equal )

Mathematics

2 answers:

Semenov [28]3 years ago

5 0

Answer:

x = -12

Step-by-step explanation:

Equation:-

x/-2 = 6

=> x = 6(-2)

=> x = -12

klemol [59]3 years ago

3 0

Answer:

x = -12

Step-by-step explanation:

x/-2 =6

cross multiply : -2*6 = x

-12 = x

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Please use picture for the question not the writing

Alex787 [66]

$\tt \: f(x) = {x}^{2} + 4x - 1$

To complete the square, the same value needs to be added to both sides

$\tt \: f(x) + ? = {x}^{2} + 4x + ? - 1$

To complete the square, Add 4 on both sides of the equation

$\tt \: f(x) + 4 = {x}^{2} + 4x + 4 - 1$

Use a²+2ab+b²=(a+b)² to factor the expression

$\tt \: f(x) + 4 = (x + 2)^{2} - 1$

Move the constant to the right-hand side and change its sign

$\tt \: f(x) = {(x + 2)}^{2} - 1 - 4$

Calculate the difference

$\tt \: f(x) = {(x + 2)}^{2} -5$

And, We are done solving!!~

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The option that corresponds our final answer is option D...

7 0

3 years ago

Answer the photo please

belka [17]

Answer:

The answer to your question is Letter B.

7 0

3 years ago

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If a tank is filling up with water from the river at a rate of 1 foot per day and 1 foot of water is approximately 300 gallons,

Sveta_85 [38]

Answer:

4800 Galons, and 16 feet of water.

- Brainliest if this was helpful please :)

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3 years ago

Find the inverse of each function for problems 1–6. State the domain and range of both the function and its inverse. Restrict th

JulsSmile [24]

Answer:

Function:

$f(x)=-x^2$

Domain: (-∞,∞)

Range: (-∞,0]

Inverse Function:

$f^{-1}(x)=\sqrt{-x} ,and\\f^{-1}(x)=-\sqrt{-x}$

Domain: (-∞,0]

Range: (-∞,∞)

Function:

$f(x)=5x-1$

Domain: (-∞,∞)

Range: (-∞,∞)

Inverse Function:

$f^{-1}(x)=\frac{1}{5}x+\frac{1}{5}$

Domain: (-∞,∞)

Range: (-∞,∞)

Function:

$f(x)=-x+3$

Domain: (-∞,∞)

Range: (-∞,∞)

Inverse Function:

$f^{-1}(x)=-x+3$

Domain: (-∞,∞)

Range: (-∞,∞)

Function:

$f(x)=x^{2}+7$

Domain: (-∞,∞)

Range: [7,∞)

Inverse Function:

$f^{-1}(x)=\sqrt{x-7}, and\\f^{-1}(x)=-\sqrt{x-7}$

Domain: [7,∞)

Range: (-∞,∞)

Function:

$f(x)=14x-4$

Domain: (-∞,∞)

Range: (-∞,∞)

Inverse Function:

$f^{-1}(x)=\frac{1}{14}x+\frac{2}{7}$

Domain: (-∞,∞)

Range: (-∞,∞)

Function:

$f(x)=-3x+8$

Domain: (-∞,∞)

Range: (-∞,∞)

Inverse Function:

$f^{-1}(x)=-\frac{1}{3}x+\frac{8}{3}$

Domain: (-∞,∞)

Range: (-∞,∞)

Step-by-step explanation:

To find inverse of a function f(x), there are 4 steps we need to follow:

1. Replace f(x) with y

2. Interchange the y and x

3. Solve for the "new" y

4. Replace the "new" y with the notation for inverse function, $f^{-1}(x)$

Note: The domain of the original function f(x) is the range of the inverse and the range of the original function is the domain of the inverse function.

Let's calculate each of these.

$f(x)=-x^2$

Domain: There is no restriction on values of x we can put on it. Hence domain is (-∞,∞)

Range: No matter what we put into x, the y values will always be negative. And if we put 0, y value would be 0. So range is (-∞,0]

Finding the inverse:

$f(x)=-x^2\\y=-x^2\\x=-y^2\\y^2=-x\\y=+-\sqrt{-x} \\y=\sqrt{-x}, -\sqrt{-x}$

$f^{-1}(x)=\sqrt{-x} ,and\\f^{-1}(x)=-\sqrt{-x}$

Domain: this is the range of the original so domain is (-∞,0]

Range: this is the domain of the original so range is (-∞,∞)

$f(x)=5x-1$

Domain: There is no restriction on values of x we can put on it. Hence domain is (-∞,∞)

Range: All sorts of y values will occur, so the range is (-∞,∞)

Finding the inverse:

$f(x)=5x-1\\y=5x-1\\x=5y-1\\5y=x+1\\y=\frac{1}{5}x+\frac{1}{5}$

$f^{-1}(x)=\frac{1}{5}x+\frac{1}{5}$

Domain: this is the range of the original so domain is (-∞,∞)

Range: this is the domain of the original so range is (-∞,∞)

$f(x)=-x+3$

Domain: There is no restriction on values of x we can put on it. Hence domain is (-∞,∞)

Range: All sorts of y values will occur, so the range is (-∞,∞)

Finding the inverse:

$f(x)=-x+3\\y=-x+3\\x=-y+3\\y=-x+3$

$f^{-1}(x)=-x+3$

Domain: this is the range of the original so domain is (-∞,∞)

Range: this is the domain of the original so range is (-∞,∞)

$f(x)=x^{2}+7$

Domain: There is no restriction on values of x we can put on it. Hence domain is (-∞,∞)

Range: no matter what we put into x, it will always be a positive number greater than 7. Only when we put in 0, y will be 7. So 7 is the lowest number and it can go to infinity. Hence the range is [7,∞)

Finding the inverse:

$f(x)=x^2+7\\y=x^2+7\\x=y^2+7\\y^2=x-7\\y=+-\sqrt{x-7}$

$f^{-1}(x)=\sqrt{x-7}, and\\f^{-1}(x)=-\sqrt{x-7}$

Domain: this is the range of the original so domain is [7,∞)

Range: this is the domain of the original so range is (-∞,∞)

$f(x)=14x-4$

Domain: There is no restriction on values of x we can put on it. Hence domain is (-∞,∞)

Range: no matter what we put into x, we can get any y value from negative infinity to positive infinity. So range is (-∞,∞)

Finding the inverse:

$f(x)=14x-4\\y=14x-4\\x=14y-4\\14y=x+4\\y=\frac{1}{14}x+\frac{2}{7}$

$f^{-1}(x)=\frac{1}{14}x+\frac{2}{7}$

Domain: this is the range of the original so domain is (-∞,∞)

Range: this is the domain of the original so range is (-∞,∞)

$f(x)=-3x+8$

Domain: There is no restriction on values of x we can put on it. Hence domain is (-∞,∞)

Range: no matter what we put into x, we can get any y value from negative infinity to positive infinity. So range is (-∞,∞)

Finding the inverse:

$f(x)=-3x+8\\y=-3x+8\\x=-3y+8\\3y=-x+8\\y=-\frac{1}{3}x+\frac{8}{3}$

$f^{-1}(x)=-\frac{1}{3}x+\frac{8}{3}$

Domain: this is the range of the original so domain is (-∞,∞)

Range: this is the domain of the original so range is (-∞,∞)

8 0

4 years ago

Choose the correct simplification and demonstration of the closure property given: (3x3 + 2x2 − 5x) − (8x3 − 2x2). 5x3 + 2x2 + 5

cricket20 [7]

Answer:

Answer C

−5x3 + 4x2 − 5x; is a polynomial

Step-by-step explanation:

you just use distributive property to simplify, and then you know the closure property demonstrates that when you are adding, subtracting, multiplying, or dividing two polynomials, they will remain polynomials.

pls mark branliest

4 0

3 years ago

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