Type the correct answer in each box. use numerals instead of words. if necessary, use / for the fraction bar(s).
consider the expression below.
(- 6)(x+ 2)
for (x - 6)(x + 2) to equal ,either (x - 6) or (x+2) must
the values of x that would result in the given expression being equal to 0, in order from least to

-2 and 6

Step-by-step explanation:

We need to find which values of x will make the following equation equal to zero:

According to the zero product property, if the product of two numbers is zero then either one of them is equal to zero or both of them are equal to zero. Since, product of (x - 6) and (x + 2) is equal to 0, we can write:

Either, x - 6 = 0 , this means x = 6

or

x + 2 = 0, this means x = -2

Thus, the complete statement will be:

The values of x that would result in the given expression being equal to 0, in order from least to greatest are -2  and 6

hello!

to begin with, we're given that on earth, a rock weighs 20 pounds, and while on jupiter, that same rock weighs 42 pounds. now, we're trying to find the weight of a rock that weighs 10.5 pounds on earth, on jupiter.

to solve this, first, set up a proportion, cross-multiply, and solve it algebraically.

setting up the proportion is the hardest and the most confusing part of the problem, but it's simple.

if on earth, the rock weighs 20 lbs, and on jupiter it weighs 42 lbs, our first part of our proportion is 20/42 lbs. the second part has to be equal to each other. this means that 10.5 has to be on the numerator of the fraction, and x, the value we're trying to find, has to be on the denominator. remember that the rock weighs 10.5 lbs on earth, and we're trying to find the weight of it on jupiter (which is x).

20/42 = 10.5/x

next, you would cross-multiply.

20x = 42(10.5)

20x = 441

now, solve it algebraically by using division.

x = 22.05 lbs

this can be rounded to 22.1 lbs.

therefore, the weight of the rock on jupiter is 22.1 lbs, the fourth choice.

yeah i know the problem you need. i am taking it right now. the answer is step 3

~mickey