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Julli [10]
3 years ago
6

Solve for x: 12=-2(x+8)

Mathematics
2 answers:
liberstina [14]3 years ago
7 0
Your answer would be : x=-14
erastova [34]3 years ago
6 0
<span>12=-2(x+8)
-----------
Simplify and flip
</span>12=-<span>2<span>(x+8<span>)
</span></span></span>12=<span>(-2)(x)+<span>(-2)<span>(8<span>)
</span></span></span></span>12=-2x+-16
12=-<span>2x-<span>16
</span></span>-2x-16=<span>12
</span>-------------
Add 16 on each side
<span><span>-2x-16</span>+16</span>=<span>12+16
</span><span><span>-2x</span></span>=<span>28
</span>-----------------
Now, divide each side by -2
-2x ÷ -2 = 28 ÷ -2
x = -14
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Problem:
Molodets [167]

Answer:

a)

We know that:

a, b > 0

a < b

With this, we want to prove that a^2 < b^2

Well, we start with:

a < b

If we multiply both sides by a, we get:

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now let's go back to the initial inequality.

a < b

if we now multiply both sides by b, we get:

a*b < b*b

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Then we have the two inequalities:

a^2 < b*a

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Then we can rewrite this as:

a^2 < b*a < b^2

This means that:

a^2 < b^2

b) Now we know that a.b > 0, and a^2 < b^2

With this, we want to prove that a < b

So let's start with:

a^2 < b^2

only with this, we can know that a*b will be between these two numbers.

Then:

a^2 < a*b < b^2

Now just divide all the sides by a or b.

if we divide all of them by a, we get:

a^2/a < a*b/a < b^2/a

a < b < b^2/a

In the first part, we have a < b, this is what we wanted to get.

Another way can be:

a^2 < b^2

divide both sides by a^2

1 < b^2/a^2

Let's apply the square root in both sides:

√1 < √( b^2/a^2)

1 < b/a

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a < b

7 0
3 years ago
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2 years ago
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