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nika2105 [10]
3 years ago
9

A baby weighs 7 pounds at birth. the table shows the baby's weigh after each month of its birth, up to the sixth month

Mathematics
1 answer:
Alexus [3.1K]3 years ago
6 0
What’s the question?
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The LSU tigers won the 2019 Football National Championship. Joe Burrow threw for 60 touchdowns in the 15-game season. What is th
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what grade is this question?

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PLEASE HELP ILL GIVE YOU BRAINLIEST!!! Which number line represents the number 0.7?
Novay_Z [31]

Answer:

C

Step-by-step explanation:

We know 0.7 is in between 0 and 1

Thus the dot has to be in between 0 and 1

And it should be 7^t^h unit after 0

Observing all the number line , we see C is the absolute match!

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3 years ago
What is the square route of five
CaHeK987 [17]

<u><em>Answer: =2.23</em></u>

Step-by-step explanation:

Square root like this: √

5=2.23

You can also round up to the nearest hundredths and it's going to be stay. If the digit is 5 or more it's going to round up to the nearest tenths, hundredths, and thousandths.

Hope this helps!

Thank you!

Have a great day!


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3 years ago
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What is the value of x? show work
Alona [7]
2x+38=180
2x=142 (or x+x=142 from which 2x=142)
x=71 

The unknown angles are each 71

You know this answer is logical as, since two sides are 21 it hints that we are looking at an equilateral triangle, which means two of its angles will be the same
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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.

4 0
3 years ago
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