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Annette [7]
3 years ago
12

A sofa is 12 years oldvand a table is 36 years omd.In how many years will the table be twice as old as the sofa ?​

Mathematics
1 answer:
maw [93]3 years ago
4 0

Answer:

12 years later.

Step-by-step explanation:

Let assume in X years, the table will be twice as old as the sofa.

36+X=2(12+X), X=12

Or think it the other way. The difference of ages between two things never change. So they must have a difference(of age) of 24 forever. When will this difference(24) be as same as the younger's age so that the older one is exactly twice as old as the younger? 12 years later when the sofa is 24.

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Rudiy27
Sandy's cost equation is y = 5x + 10 where y = cost and x = number of children.
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4 years ago
1 2 3 4 5 6 7 8 9 10
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x^{2} + 2

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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:

a*a < b*a

a^2 < b*a

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

a*b < b^2

Then we have the two inequalities:

a^2 < b*a

a*b < b^2

a*b = b*a

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.

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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

Now we multiply both sides by a:

a < b

7 0
3 years ago
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