[tex]1 imes rac{3}{4} imes 111.8 imes 860 \ [/tex] me.

Mathematics, 22.07.2019 20:30 stormserena

$1 \times \frac{3}{4} \times 111.8 \times 860 \\$ me.

Answers: 1

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Answer from: rallen1158

The answer is:  " 386 " .
____________________________________________________
Explanation:
____________________________________________________

Eliminate the "1" ; since "1" multiplied by any value, results in that same value.

We have:

" (3/4) * (11.8) * (360) " ;

↔ "(3/4) * (360) * (1.8)" ;

Start with: "(3/4) * (360)" ;

→ " (3/4) * (360) = $\frac{(3*360)}{4}$ ;

→ The "4" cancels out to a "1" ; and the "360" cancels out to a "90" ;

→ since: "(360 ÷ 4 = 90)" ; and since: "(4 ÷ 4 = 1)" ;

And we have:
____________________________________________________

→ $\frac{(3 * 90) }{1} = (3 * 90) = 270$ .

Now, multiply: "1.8 * 270" ;
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→ 1.8
* 270
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= (18 * 27) ; → Take that value and add a "0" to the end of that value;
and then: → take "this obtained value" ; and move it back ONE DECIMAL place;
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So, "1.8 * 27 = ? " ;
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Step 1)

  ₅
18
* 27
_________________________
126
+ 36
_________________________
3 8 6
_________________________

Step 2: Add a "0" to the end of that value:
__________________________________________________
"386 " ; → " 3860 " ;
__________________________________________________

Step 3: Remove this "obtained value" one decimal place back:
__________________________________________________
"3860" ; → "386.0" ; write as: → " 386 " .
__________________________________________________
The answer is: " 386 " .
__________________________________________________

Answer from: Quest

mr. flanders’ class sold 47 doughnuts and mr. rodriquez’s class sold 38 cartons of milk

step-by-step explanation:

let x be the number of doughnuts sold by mr. flanders’ class and y be the number of cartons sold by mr. rodriquez’s class.

1. together, the classes sold 85 items, then

x+y=85.

2. both classes earn earned $104.49 for their school. if doughnuts were sold for $1.35 each, then x doughnuts costed $1.35x. if cartons of milk were sold for $1.08 each, then y cartons costed $1.08y. hence,

1.35x+1.08y=104.49.

3. solve the system of two equations:

$\left\{\begin{array}{l}x+y=85\\1.35x+1.08y=104.49\end{array}\right.\rightarrow \left\{\begin{array}{l}x=85-y\\1.35(85-y)+1.08y=104.49\end{array}\right.$

$114.75-1.35y+1.08y=104.49,\\ \\114.75-104.49=1.35y-1.08y,\\ \\10.26=0.27y,\\ \\y=38,\\ \\x=85-38=47.$

Answer from: Quest

Letter b sorry i’m i’m wrong xx

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[tex]1 \times \frac{3}{4} \times 111.8 \times 860 \\ [/tex] me....