## function has the function rule y = -7x – 2. If the input is -4, what is the output?

## Related Questions

What is the range of the function f(x) = 3×2 + 6x – 9

f(x)=3x^2-6x+1

My solution is:

The domain is all real numbers–there are no restrictions like a square root or variable in the denominator

this is a U shape parabola and you want to find the bottom point, it is at the vertex

x=-b/2a =- (-6) /2 (3) =1

f(1) =3(1)^2 -6(1) +1 =3-6+1 =-2

the minimum is (1, -2)

so the range is all real numbers >= -2

By examining my solution, you could just answer the problem on your own! I hope it helps!

Solve the equation for y 2x+6y=13

So here is how we are going to solve for y for the given equation above:

2x+6y=13

Next, transfer 2x to the right side and it will look like this:

6y=13-2x

Now, divide both sides by 6, and it will look like this:

y=13-2x

——–

6

So, this is the final answer for y.

Hope this answers your question. Have a great day!

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► For instance, the quadratic equation:

x² – (6 + 4i)x + (9 + 12i) = 0

has for discriminant:

Δ = (6 + 4i)² – 4(9 + 12i) = 36 – 16 + 48i – 36 – 48i = -16

which is indeed negative.

Its solutions will then be:

x₁ = [(6 + 4i) + 4i]/2 = 3 + 4i

x₂ = [(6 + 4i) – 4i]/2 = 3

And the other solution here is 3.

► If you are not convinced, the quadratic equation:

x² – (6 + 5i)x + (5 + 15i) = 0

has for discriminant:

Δ = (6 + 5i)² – 4(5 + 15i) = 36 – 25 + 60i – 20 – 60i = -9

which is indeed negative.

Its solutions will then be:

x₁ = [(6 + 5i) + 3i]/2 = 3 + 4i

x₂ = [(6 + 5i) – 3i]/2 = 3 + i

And the other solution here is 3+i.

► In fact, every quadratic equation of the form:

x² – [6 + (4 + α)i]x + (3 + 4i)(3 + αi) = 0

where α is any real, has for discriminant:

Δ = [6 + (4 + α)i]² – 4(3 + 4i)(3 + αi)

= 36 – (4 + α)² + 12(4 + α)i – 36 + 16α – 12(4 + α)i

= 16α – (4 + α)²

= 16α – 16 – 8α – α²

= -16 + 8α – α²

= -(α – 4)²

WILL be negative.

Their solutions will then be:

x₁ = [ [6 + (4 + α)i] – (α – 4)i ]/2 = 3 + 4i

x₂ = [ [6 + (4 + α)i] + (α – 4)i ]/2 = 3 + αi

And the other solution will then be is 3+αi.

Since α can take any real value, you’ll obtain an infinity of solutions of the form 3+αi.

► So conclusively:

If the discriminant of a quadratic is negative AND one of the solutions is 3+4i, the only thing we can say about the other solution is that its real part must be 3.

First one looks like you are squaring the number, then multiplying the result by 4, i.e.

y=4x2

second one is similar, but instead of squaring and multiplying by 4, you are squaring and then dividing by 2

I hope my answer has come to your help. God bless and have a nice day ahead!

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**1) The answer is: [B]: r = 5 .****__________________________****Explanation:****___________**_______________

Given: 7r − 7 = 2r + 18 ; Round your answer to the nearest tenth, if necessary.

____________________________

Since “r” is the only variable given, let us assume we want to solve for “r” (instead of “x”).

___________________________

→ Subtract “2r” from EACH SIDE of the equation; and & add “7” to EACH SIDE of the equation:

_____________

→ 7r − 7 − 2r + 7 = 2r + 18 − 2r + 7 ; to get: → 5r = 25 ;

_____________

→ Now, divide EACH SIDE of the equation by “5”; to isolate “r” on one side of the equation; and to solve for “r” :

______________

→ 5r / 5 = 25 / 5 → r = 5 → which is: “Answer choice: [B]”.

_________________**Let us check our answer, by plugging in “5” for “r” in the original equation:**

_________________

→ 7r − 7 = 2r + 18 ; → 7(5) − 7 =? 2(5) + 18? ;

______________________

→ 35 − 7 =? 10 + 18 ?; → 28 =? 28? Yes!

______________________** 2) The answer is: [D]: x = 2 .****_____________****Explanation: **

_____________

Given: 2x + 12 = 18 − x ; Solve for “x” (round to nearest tenth, if necessary).

_______________

→ Add “x” to EACH SIDE of the equation, & subtract “12” from EACH SIDE of the equation: → 2x + 12 + x − 12 = 18 − x + x − 12 ;

______________

→ To get: 3x = 6 ; → Divide EACH SIDE of the equation by “3”;

to isolate “x” on one side of the equation; and to solve for “x”:

_____________

→ 3x / 3 = 6 / 3 ; → x = 2 ; which is: “Answer choice: [D]”.

______________**Let us check our answer, by plugging in “2” for “x” in the original equation:**

________________

→ 2x + 12 = 18 − x ; → 2(2) + 12 =? 18 − 2 ?

________________

→ 4 + 12 =? 18 − 2 ? ; → 16 =? 16? Yes!

________________________________**3) The answer is: [A]: x = -3 . ****_____________****Explanation:**

________________

Given: 8x − 3 = 15x + 18 ; Solve for “x”. Round your answer to the nearest tenth, if necessary.

_________________

→ Subtract “8x” from EACH SIDE of the equation, & add “3” to EACH SIDE of the equation:

_______________

→ 8x − 3 − 8x + 3 = 15x + 18 − 8x + 3 ; to get:

_______________

→ 0 = 7x + 21 ; → Subtract “21” from EACH SIDE of the equation;

_______________

→ 0 − 21 = 7x + 21 − 21 ; to get:

_______________

→ -21 = 7x ; Now divide EACH SIDE of the equation by “7”;

to isolate “x” on one side of the equation; & to solve for “x”:

_______________

→ = -21 / 7 = 7x / 7 ; → -3 = x ; which is “Answer choice: [A].”**_________________****Let us check our answer, by plugging in “-3” for “x” in the original equation:**

________________

→ 8x − 3 = 15x + 18 ; → 8(-3) − 3 =? 15(-3) + 18 ?;

________________________

→ -24 − 3 =? -45 + 18 ? ; → -27 =? -27? Yes!**___________________________****4) The answer is: [C]: y = 11 .****_____________****Explanation:****____________**

Given: 6y − 6 = 4y + 16 ; Solve for “y”; Round to the nearest tenth, if necessary.

____________

(Note: Since “y” is the only variable given; assume we are to solve for “y” instead of “x”).

____________

→ Subtract “4y” from EACH SIDE of the equation, & add “6” to EACH SIDE of the equation; → 6y − 6 − 4y + 6 = 4y + 16 − 4y + 6 ; to get:

_______________

→ 2y = 22 ; Now, divide EACH SIDE of the equation by “2”; to isolate “y” one side of the equation; and to solve for “y” ;

_______________

→ 2y / 2 = 22 / 2 ; → y = 11 → which is “Answer choice: [C]”.**_______________________________****Let us check our answer, by plugging in “11” for “y” in the original equation:****___________________**

→ 6y − 6 = 4y + 16 ; → 6(11) − 6 =? 4(11) + 16 ?

_______________________

→ 66 − 6 =? 44 + 16 ? → 60 =? 60 ? Yes!**__________________****5) The answer is: [B]: x = -11 .****_____________________****Explanation:****_________________**

Given: 3(x − 4) = 5(x + 2) ; Solve for “x”. Round to the nearest tenth, if necessary.**___________****→Note the “distributive property of multiplication”: ****_____________****a*(b + c) = ab + ac ; and: a*(b − c) = ab − ac ;****_______________**

→ Let us expand EACH SIDE of our given equation.

→Start with the “left-hand side”:

____________

3(x − 4) = (3*x) − (3*4) = 3x − 12;

______________________________

→Now let us expand the “right-hand side” of the given equation:

____________

→ 5(x + 2) = (5*x) + (5*2) = 5x + 10 ;

______________

→Now, we can rewrite the original equation:

_______________

→ 3(x − 4) = 5(x + 2) ; by substituting the expanded values for EACH SIDE of the question: → 3x − 12 = 5x + 10 ;

__________________

→ Subtract “3x” from EACH SIDE of the equation; and add “12” to EACH SIDE of the equation: → 3x − 12 − 3x + 12 = 5x + 10 − 3x + 12 ; to get:

________________

→ 0 = 2x + 22; → Now subtract “22” from EACH SIDE of the equation:

______________

→ 0 − 22 = 2x + 22 − 22 ; to get: → -22 = 2x ;

__________

→ Divide EACH SIDE of the equation by “2”; to isolate “x” on one side of the equation; & to solve for “x” ;

_____________

→ -22 / 2 = 2x /2 ; → -11 = x ; which is “Answer choice: [B]”.

______________**Let us check our answer, by plugging in “-11” for “x” in the original equation:****___________**

→ 3(x − 4) = 5(x + 2) ; → 3(-11 − 4) =? 5(-11 + 2) ? ;

_______________________

→3(-15) =? 5(-9) ? ; → -45 =? -45 ? Yes!**_____________________________________________****Hope these answers and explanations are helpful. Best of luck!**

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Each vertex of the hexagon is translated 8 units left and 7 units down. So, the x-coordinate is 8 units smaller (x – 8), and the y-coordinate is 7 units smaller (y – 7).

Therefore (x, y) → (x – 8, y -7)

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