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Arte-miy333 [17]
2 years ago
14

Subtract. Add to check: 735,249 - 575,388

Mathematics
2 answers:
Ksivusya [100]2 years ago
4 0
735,249 - 575,388 = 159,861

example for checking:
1. 159,861 + 575,388 = 735,249
Art [367]2 years ago
3 0
THE ANSWER WOULD BE 159,861
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A metal rod is 2/5 meters long. It will be cut into pieces that are each 1/30 meters long. How many pieces will be made from the
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namely, how many times does 1/30 go into 2/5?

\bf \cfrac{2}{5}\div \cfrac{1}{30}\implies \cfrac{2}{5}\cdot \cfrac{30}{1}\implies \cfrac{2}{1}\cdot \cfrac{30}{5}\implies 2\cdot 6\implies 12

3 0
2 years ago
In Andrew’s Furniture Shop, he builds bookshelves and tables. Each type of furniture takes him about the same time to make. He f
Paul [167]
A. You may set the variables in either order. But for argument sake, let's set as follows:

x = Amount of bookshelves
y = Amount of tables

B. Because of the amount of things you need to make, the following is an inequality using those variables.

x + y > 25

Plus you can determine a second inequality based on the amount of money that you have to spend. 

20x + 45y < 675

Finally you may also add in that each value must be greater than or equal to zero, since they cannot have negative tables. 

C. By solving the system and looking at basic constraints when graphed, you can see the feasible region has 4 vertices. 

(0,0)
(18, 7)
(0, 15)
(33.75, 0) or (33, 0) if you insist on rounding. 
8 0
2 years ago
Given: circle k(O), ED diameter, m∠OEF=32°, m EF=(2x+10)°. Find: x
Snezhnost [94]

Answer:

The value of x is 53\°

Step-by-step explanation:

see the attached figure to better understand the problem

step 1

Find the measure of arc DF

we know that

The inscribed angle measures half that of the arc comprising

so

m

we have

m

substitute

32\°=\frac{1}{2}(arc\ DF)\\ arc\ DF=64\°

step 2

Find the measure of x

we know that

arc\ DF+arc\ EF=180\° ---> is a semi circle

we have

arc\ DF=64\°\\ arc\ EF=(2x+10)\°

substitute

64\°+(2x+10)\°=180\°\\ 2x\°=180\°-74\°\\ 2x\°=106\°\\ x=53\°

4 0
2 years ago
The children from a football club are put into rows in the sports hall. When put into rows of 9 children, there are 2 children l
g100num [7]

We are told that the children from a football club are put into rows in the sports hall. When put into rows of 9 children there are 2 children left over. When put into rows of 12 children there are 2 children left over.

We will find least number of children in football club by finding LCM of 9 and 12.

Multiples of 9 are: 9, 18, 27, 36, 45, 54, 63,...

Multiples of 12 are: 12, 24, 36, 48, 60, 72, 84,...

We can see that least common multiple of 9 and 12 is 36.

We are told that 2 children left over from putting them into 9 and 12 children per row. To find least number we will add 2 to 36.

9\cdot4+2=36+2=38

12\cdot3+2=36+2=38      

\text{Least number of children in football club}=36+2=38

Therefore, the least number of children in the football club is 38.

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Answer:

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Step-by-step explanation:

The explanation for this question is given in the attachment below.

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2 years ago
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