13 units from 20 could either be 33 or 7; 15 units from -8 could either be 7 or 23.Since 7 satisfy both of the above requirements, Sarah's favorite number is 7. (You could show your work using a number line)

The two possibilities of Neil's favorite number are 26 and -12.

7 0

2 years ago

Read 2 more answers

The equation v/2 -21=-15 is solved in several steps below.

Arada [10]

Answer:

V/2 - 21 = -15

V/2 - 21 + 21 = -15 + 21

V/2 + 0 = 6

V/2 * 2 = 6 * 2

V = 12

8 0

3 years ago

If you bought 28 shares of stock at $30/sh and sold them at $25/sh. And How much did you net after selling the shares? Was it a

Strike441 [17]

Answer:

selling price$840

it's a loss.

Step-by-step explanation:

cost price = 28× 30 = $840

selling prince =25×28=$700

since cost price is bigger than selling price, it's a loss.

5 0

2 years ago

Evaluate the difference quotient for the given function. Simplify your answer.

nasty-shy [4]

I suppose you mean

$f(x) = \dfrac{x+5}{x+1}$

Then

$f(3) = \dfrac{3+5}{3+1} = \dfrac84 = 2$

and the difference quotient is

$\dfrac{f(x)-f(3)}{x-3} = \dfrac{\frac{x+5}{x+1}-2}{x-3} \\\\ \dfrac{f(x)-f(3)}{x-3} = \dfrac{\frac{x+5-2(x+1)}{x+1}}{x-3} \\\\ \dfrac{f(x)-f(3)}{x-3} = \dfrac{-x+3}{(x+1)(x-3)} \\\\ \dfrac{f(x)-f(3)}{x-3} = \boxed{-\dfrac{x-3}{(x+1)(x-3)}}$

If it's the case that x ≠ 3, then (x - 3)/(x - 3) reduces to 1, and you would be left with

$\dfrac{f(x)-f(3)}{x-3}\bigg|_{x\neq3} = -\dfrac1{x+1}$

4 0

2 years ago