All categories

3 years ago

What is the value of f(–1) when f(x) = 2x + 2? f(–1) =

Mathematics

2 answers:

sweet [91]3 years ago

8 0

Answer: f(-1)= 0

Step-by-step explanation:

f(-1)

x=(-1)

f(-1) = 2x + 2

f(-1) = 2(-1) + 2

2 x (-1) = -2

-2 +2 = 0

EleoNora [17]3 years ago

4 0

Answer:

-3/2 = f

Step-by-step explanation:

f(-1) =2x+2

-1f=2x+2

-1f-2=2x

-3=2x

-3/2=x

You might be interested in

What is the first five nonzero multiples of 3 and 10

Masja [62]

Every multiples of 10s has a zero... you do know that... right?

5 0

3 years ago

Georgia spends $39 on gas each week. Approximately how much money does she spend on gas each week?

Pachacha [2.7K]

$40.........................

4 0

4 years ago

Read 2 more answers

Find the value of the variable y, where the sum of the fractions 6/(y+1) and y/(y-2) is equal to their product.

Blizzard [7]

Answer:

The answer is

$y = 3$

$y = - 4$

Step-by-step explanation:

We must find a solution where

$\frac{6}{y + 1} + \frac{y}{y - 2} = \frac{6}{y + 1} \times \frac{y}{y - 2}$

Consider the Left Side:

First, to add fraction multiply each fraction on the left by it corresponding denomiator and we should get

$\frac{6}{y + 1} \times \frac{y - 2}{y - 2} + \frac{y}{y - 2} \times \frac{y + 1}{y + 1}$

Which equals

$\frac{6y - 12}{(y -2) (y + 1)} + \frac{ {y}^{2} + y }{(y - 2)(y + 1)}$

Add the fractions

$\frac{y {}^{2} + 7y - 12 }{(y - 2)(y + 1)} = \frac{6}{y + 1} \times \frac{y}{y - 2}$

Simplify the right side by multiplying the fraction

$\frac{6y}{(y + 1)(y + 2)}$

Set both fractions equal to each other

$\frac{6y}{(y + 1)(y - 2)} = \frac{ {y}^{2} + 7y - 12}{(y + 1)(y - 2)}$

Since the denomiator are equal, we must set the numerator equal to each other

$6y = {y}^{2} + 7y - 12$

$= {y}^{2} + y - 12$

$(y + 4)(y - 3)$

$y = - 4$

$y = 3$

6 0

3 years ago

Read 2 more answers

JESUS LOVES YOU TvT . have a nice day ;P

Vsevolod [243]

answer:

(: thanks for the points

8 0

3 years ago

- 2x (a+h)^2+4(a+h)-12

hjlf

That’s the answer mama

7 0

3 years ago