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emmainna [20.7K]
3 years ago
15

Please help me please please help please

Mathematics
1 answer:
poizon [28]3 years ago
6 0

Answer:

its 2 blurry so i cant help take a better pic and i should be able to help you out

Step-by-step explanation:

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Separate the number 41 and the two parts so that the first number is eight more than twice the second number what are the two nu
vesna_86 [32]

The two numbers are 30 and 11

<em><u>Solution:</u></em>

Given that we have to separate the number 41 into two parts

Let the second number be "x"

<em><u>Given that first number is eight more than twice the second number</u></em>

first number = eight more than twice the second number

first number = 8 + twice the "x"

first number = 8 + 2x

So we can say first number added with second number ends up in 41

first number + second number = 41

8 + 2x + x = 41

8 + 3x = 41

3x = 41 - 8

3x = 33

x = 11

first number = 8 + 2x = 8 + 2(11) = 8 + 22 = 30

Thus the two numbers are 30 and 11

7 0
2 years ago
Solve for x. Enter your answer in interval notation using grouping symbols <br> X^2 + 5x &lt; 24
icang [17]
I think the correct answer would be : (-8,3)
I hope this helps !
7 0
3 years ago
How do you do this problem? 3x+1=10
Vlad1618 [11]
This should help 10-1=9 9÷3=3 x=3
5 0
3 years ago
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What is the value of the expression |-20+5|
Scrat [10]
25, is it absolute value, the answer to this expression in this case

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3 years ago
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7¼× 8/9×3/4<br>find the sum ​
larisa86 [58]

Answer:

Step-by-step explanation:

7\frac{1}{4}*\frac{8}{9}*\frac{3}{4}=\frac{29}{7}*\frac{8}{9}*\frac{3}{4}\\\\=\frac{29*8*3}{7*9*3}\\\\=\frac{29*2*1}{7*3*1}\\\\=\frac{58}{21}\\\\=2\frac{16}{21}\\

7 0
2 years ago
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