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mash [69]
3 years ago
15

Round 3.06735995694 to the nearest millionth.

Mathematics
1 answer:
makvit [3.9K]3 years ago
8 0
The "th" family is a little confusing but you look and find the millionth place.  It's basically backwards.  

0 is tenths
6 is hundredths
7 is thousandths
3 is ten-thousandths
5 is hundred-thousands 
9 is millionths 

the number next to it is a 9 (rule round up if it's greater than 5)  

since the number is a 9 you have to round up even one more making the 5 a 6 

3.06736
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Huong is organizing paperback and hardback books for her club's used books sale. The number of paperbacks is 12 less than three
TiliK225 [7]

Answer:

The number of hardback books in book sale   = 58

Step-by-step explanation:

Let the number of hardback books  = m

So, the number of paperback books = 3 ( number of Hardbacks)  - 12

                                                             = 3 m - 12

⇒the number of paperback books   = 3 m -12

Now, total number of  paperback books  = 162

⇒ (3m -12)  = 162

or, 3m = 162 + 12

or, 3m  = 174     ⇒  m = 174 / 3

or, m = 58

Hence, the number of hardback books  = 58

4 0
3 years ago
Help me on #13 and please explain
hichkok12 [17]

Answer:

56.4°

Step-by-step explanation:

Let θ = the angle of elevation. Then

tanθ = 199/132 = 1.508

     θ = arctan1.508

     θ = 56.4°

The angle of elevation is 56.4°.

4 0
3 years ago
Evaluate A2 for A = -4
ratelena [41]
The answer is -8 because -4x2 is -8
6 0
3 years ago
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If n is a positive integer, how many 5-tuples of integers from 1 through n can be formed in which the elements of the 5-tuple ar
erma4kov [3.2K]

Answer:

\frac{(n+4)*(n+3)*(n+2)*(n+1)*n}{120}

Step-by-step explanation:

Given

5 tuples implies that:

n = 5

(h,i,j,k,m) implies that:

r = 5

Required

How many 5-tuples of integers (h, i, j, k,m) are there such thatn\ge h\ge i\ge j\ge k\ge m\ge 1

From the question, the order of the integers h, i, j, k and m does not matter. This implies that, we make use of combination to solve this problem.

Also considering that repetition is allowed:  This implies that, a number can be repeated in more than 1 location

So, there are n + 4 items to make selection from

The selection becomes:

^{n}C_r => ^{n + 4}C_5

^{n + 4}C_5 = \frac{(n+4)!}{(n+4-5)!5!}

^{n + 4}C_5 = \frac{(n+4)!}{(n-1)!5!}

Expand the numerator

^{n + 4}C_5 = \frac{(n+4)!(n+3)*(n+2)*(n+1)*n*(n-1)!}{(n-1)!5!}

^{n + 4}C_5 = \frac{(n+4)*(n+3)*(n+2)*(n+1)*n}{5!}

^{n + 4}C_5 = \frac{(n+4)*(n+3)*(n+2)*(n+1)*n}{5*4*3*2*1}

^{n + 4}C_5 = \frac{(n+4)*(n+3)*(n+2)*(n+1)*n}{120}

<u><em>Solved</em></u>

6 0
3 years ago
Hey I'm not sure what -x=9-3y is​
mars1129 [50]

Answer:

x=0 y=3

Step-by-step explanation:

6 0
3 years ago
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