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3 years ago

What is the square root of 144?

Mathematics

2 answers:

zloy xaker [14]3 years ago

8 0

Answer:

Step-by-step explanation:

This is true because 12x12 is 144, or 12 squared is 144.

choli [55]3 years ago

6 0

Answer:

$12$

Step-by-step explanation:

$\sqrt{144 } = \sqrt{12 \times 12} = 12$

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What is the value of the 2 in 3.25?

gizmo_the_mogwai [7]

20 because when you do it that way when it is behind the decimal the numbers are always the opposite thousand hundred tens ones . tens hundreds thousands

3 0

3 years ago

A father is 32 years older than his son. In four years,

sergiy2304 [10]

Answer:

father: 36
son: 4

Step-by-step explanation:

Let s represent the son's age now. Then s+32 is the father's age. In 4 years, we have ...

5(s+4) = (s+32)+4

5s +20 = s +36 . . . . . eliminate parentheses

4s = 16 . . . . . . . . . . . . subtract s+20

s = 4

The son is now 4 years old; the father, 36.

_____

Alternate solution

In 4 years, the ratio of ages is ...

father : son = 5 : 1

The difference of their ages at that time is 5-1 = 4 "ratio units". Since the difference in ages is 32 years, each ratio unit must stand for 32/4 = 8 years. That is, the future age ratio is ...

father : son = 40 : 8

So, now (4 years earlier), the ages must be ...

father: 36; son: 4.

5 0

3 years ago

13. The least common multiple of two non-zero integers a and b is the unique positive integer m such that (i) m is a common mult

Vlad [161]

Answer:

[a,b] divides n

Step-by-step explanation:

Let us denote the least common multiple of a and b [a,b]=m.

We want to prove that m divides n, where n is a multiple of a and b.

We suppose m does not divide n, then by the Division Theorem, there exists q and r integers such that:

(1) ... n=mq+r, where 0<r<m

As n is a multiple of a and b, there exists s and t integers such that:

sa=n and tb=n

Same thing happens to m as it is the least common multiple, there exists u and v such that:

ua=m and vb=m

So (1) has the following form:

n=mq+r ⇒ sa=uaq+r ⇒sa-uaq=r⇒(s-uq)a=r and

n=mq+r ⇒ tb=vbq+r ⇒ tb-vbq=r⇒ (t-vq)b=r

So r is a multiple of a and b, but r<m which is a contradiction as, m is the least common multiple of a and b. So this concludes the proof.

So this means that $\frac{ab}{m}$ is and integer.

As m= vb, then $\frac{m}{b}$ is an integer, lets say $\frac{m}{b}$ =v; and as m=ua, then $\frac{m}{a}$ =u.

So $\frac{ab}{m}$ v= $\frac{ab}{m}$ $\frac{m}{b}$ =a, so $\frac{ab}{m}$ divides a; on the other hand, $\frac{ab}{m}$ u= $\frac{ab}{m}$ $\frac{m}{a}$ =b, so $\frac{ab}{m}$ divides b. From this we can conclude that $\frac{ab}{m}$ is a common divisor of a and b.