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allsm [11]
3 years ago
6

D" id="TexFormula1" title="\frac{3}{5} times \frac{7}{12} divided by 2\frac{7}{10}" alt="\frac{3}{5} times \frac{7}{12} divided by 2\frac{7}{10}" align="absmiddle" class="latex-formula"> helpppppppppppppp i need this for 2morrow
Mathematics
1 answer:
Rasek [7]3 years ago
3 0

let's first convert the mixed fraction to improper, and then proceed.


\bf \stackrel{mixed}{2\frac{7}{10}}\implies \cfrac{2\cdot 10+7}{10}\implies \stackrel{improper}{\cfrac{27}{10}}
\\\\[-0.35em]
~\dotfill


\bf \cfrac{3}{5}\cdot \cfrac{7}{12}\div \cfrac{27}{10}\implies \cfrac{3}{5}\cdot \cfrac{7}{12}\cdot \stackrel{\stackrel{notice}{\downarrow }}{\cfrac{10}{27}}\implies \cfrac{3}{27}\cdot \cfrac{7}{12}\cdot \cfrac{10}{5}\implies \cfrac{1}{9}\cdot \cfrac{7}{12}\cdot \cfrac{2}{1}
\\\\\\
\cfrac{1}{9}\cdot \cfrac{7}{1}\cdot \cfrac{2}{12}\implies \cfrac{1}{9}\cdot \cfrac{7}{1}\cdot \cfrac{1}{6}\implies \cfrac{7}{54}


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Stadium has 48,000 seats seats sell for $30 in section 8 $24 in section B and $18 in section C the number of seats in section a
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Step-by-step explanation:

Step 1

let  the number of seats in section B = x

and the number of seats in section C.=y

Given that the  number of seats in section A is the sum of seats in sections B and C, we can then say A =x + y

From the given question,  We have that

seatsA + seatsB + seatsC = 48,000 which also equals

(x+y)  + x      + y      = 48,000

2x + 2y = $48,000

2(x+y) = 48,000

x + y = 48,000/2 =24,000  which also equals

B+C =24,000

Now the number of seats in section A =24,000 and the sum of seats in sections B and C is 24000.

We now have

B+C = 24,000------ eqn  (1)

Also

  since we already have the number of seats for Section A, We will find Section  B and C  

A +B+C =1,239,600,

$30 x 24,000 + 24B + 18C= 1,239,600

720,000 + 24B + 18C= 1,239,600

 24B + 18C= 1,239,600-720,000

24B + 18C= 519,600--------- eqn 2

Step 2 We will now solve with the following equations

B+C = 24,000------ eqn  (1)

24B + 18C= 519,600--------- eqn 2

Multiply eq(1) by 24 (both sides). Then subtract eq(2) from it:

24B + 24C = 576,000-------eqn    (3)

24B+ 18C =   519,600-- ---eqn(4)

6C = 576,000- 519,600

6C=56,400

C= 9,400

TO get B.

B+C = 24,000

B= 24,000 - 9,400

B=14,600

Therefore, there are 24,000 sears in section A,  14,600 in section B  and  19,400 in section C.

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3 years ago
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