<img src="https://tex.z-dn.net/?f=%20%5Csqrt%5B3%5D%7B162%7D%20" id="TexFormula1" title=" \sqrt[3]{162} " alt=" \sqrt[3]{162} "

X+y=62 ; x-y=12 Solve for x in one equation and plug that value into the other equation. ===> x=12+y ; 12+y+y=62 Subtract 12 to both sides (2y=50), then divide by 2 to find y (y=25). Now, plug 25 as y into x=12+y, getting x=37. Your two numbers are 37 and 25.

6 0

2 years ago

Read 2 more answers

HELP ME ASAP PLEASE!!!!

Wewaii [24]

-(1/2)p<-16

Divide both sides by the coefficient of p. Flip the inequality sign because you are dividing a negative number.

p>32

Final answer: A

6 0

3 years ago

If x varies inversely as y 2, and x = 2 when y = 20, find x when y = 5.

Dominik [7]

Step-by-step explanation:

$x = \frac{k}{y}$

x varies inversely as y. This means that x is equal to the inverse of y which can be translated as x = 1/y but the problem said that x is not exactly equal to the inverse of y. Therefore, we need to introduce a factor k. Factor k will represent the word "varies" in the problem.

To illustrate:

$x = \frac{k}{y}$

Since we don't have the value of k, we have to solve for it by using the first values of x and y which was given as x=2 and y=20.

$x = \frac{k}{y} \\ 2 = \frac{k}{20} \\ k = 2(20) = 40$

Now that we have the value of k, we can now solve for the value of x when y=5.

$x = \frac{k}{y} \\ = \frac{40}{5} \\ = 8$

Therefore, when y is 5, the value of x is 8.

6 0

2 years ago