-4x + 7 = -21 whats x AHSAJ I NEEDD HELPPP

Mathematics

2 answers:

yKpoI14uk [10]3 years ago

7 0

Answer:

-4x+7= -21

-4x= -21-7( this positive 7 will change to negative as we take in other side )

-4x = -28

Both the negative will be cancelled

4x = 28

x = 28/4

X = 7

therefore the X value is 7

Arturiano [62]3 years ago

4 0

- 4x+7=-21

-4x=-21-7

-4x=-28

x=7

Step-by-step explanation:

Like terms are collected first

The +7 moves to the other side of the equality sign making its sign negative (-7)

Two negatives makes a positive but the sign remains negative (when you have two numbers with a negative sign you sum them up but the sign of the sum remains negative)

Then divide both side by the coefficient of x which is - 4

-4x/4 =-28/4

The negative signs cancel out

x=7

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A=1/2bh. A=200ft. H=10. What is b

Svetllana [295]

Answer:

B= $\frac{2A}{H}$

Step-by-step explanation:

If you noticed, A= $\frac{1}{2}$ BH is the formula for finding the area for a triangle. Your goal is to get B by it's self. Your first step will be to clear of the fraction first, so you will multiply both sides by 2. 2(A)=2( $\frac{1}{2}$ BH). On the left, you have 2×A= On the right side, you have 2( $\frac{1}{2}$ BH), but since you have a number in the equation, you will only use 2× $\frac{1}{2}$ . To solve 2× $\frac{1}{2}$ , you will cnacel out both 2's and you have 1. 1×BH will still equal BH, so you are now left will B×H.

(Your new equation looks like this by the way). 2A=BH

Since you need to get B by its self, the way to clear the H away from the B is by dividing. You will now divide the B and H aswell as 2A and H. (It will look like this) $\frac{2A}{H} = \frac{BxH}{H}$ . (Again when you have the same number or letter, you cross it out. When you divide, you won't change anything on the left side, and all you have to do on the right id to cross out the H next to the B and cross out the H on the bottom of the equation). You should be left with $\frac{2A}{H}$ = B. Now you can turn it around for your final answer. B= $\frac{2A}{H}$ .