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iragen [17]
3 years ago
5

Explain how an estimate helps you place the decimal point when multiplying 3.9 × 5.3

Mathematics
1 answer:
4vir4ik [10]3 years ago
4 0
An estimate helps you place the decimal point when multiplying because if you find out how many decimals there are and once you are done doing you estimated multiplication you put two decimal points after that.


Hope I helped some what and didn't make you confused.

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5 and (2,6)<br> Pls answer
Gekata [30.6K]

Answer:

Step-by-step explanation:

I do not understand  what you mean

or maybe you mean .

5 * (2,6)

(10,30)

5 0
3 years ago
What is the element a^23 in matrix A? <br> A=8 -4 -3<br> 3 -9 -5<br> -8 8 8
labwork [276]
 8 -4 -3
 3 -9 -5-8  8  8

This is a matrix of 3 rows and 3 columns (it's called square matrix because the number of rows = the number of columns)
 
To localise an element of a matrix we use indices R and C, the first index being ALWAYS the row and the second, ALWAYS the column.

Hence:
A₂₃ = the element in row 2 and column 3, that is - 5

                  COLUMN
                1st   2nd   3rd
--------------------------------
          1st  | 8     -4      -3
ROW 2nd | 3     -9      -5           3rd |-8      8       8

 
4 0
3 years ago
Read 2 more answers
2v-4^9v+5^2v-4^9v+5<br> simpify your answer
Gennadij [26K]

Answer:

2v^1 -   2^(18v+1)  + 5 ^ 2v  +5

Step-by-step explanation:

2v-4^9v+5^2v-4^9v+5

Combine like terms

2v  - 4 ^ 9v  - 4 ^ 9v  + 5 ^ 2v+5

We know a^b - a^b = 2 a^b

2v - 2*4^9v  + 5 ^ 2v+5

Rewriting 4 as 2^2  and v as v^1

2v^1 - 2 * 2^2^9v  + 5 ^ 2v+5

We know that a^b^c = a^ b*c

2v^1 - 2 * 2^18v  + 5 ^ 2v+5

2v^1 - 2^1 * 2^18v  + 5 ^ 2v+5

We know that a^1 * a^b = a^ b+1

2v^1 -   2^(18v+1)  + 5 ^ 2v+5

3 0
4 years ago
Please explain as well!!
Tanzania [10]
√8 x √32 

<span>√(8x32) </span>

<span>√256 </span>

<span>= 16</span>
3 0
4 years ago
Decimals from 4 to 5 with an interval of 0.25
iragen [17]
There are 3 numbers: 4.25,4.50,and 4.75.
3 0
3 years ago
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