Answer: The numbers are: " 21 " and " 105 " .
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Explanation:
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Let "x" be the "one positive number:
Let "y" be the "[an]othyer number".
x = 1/5 (y)
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Given that the difference of the two number is "84" ; and that "x" is (1/5) of "y" ; we determine that "x" is smaller than "y".
So, y − x = 84 .
Add "x" to each side of this equation; to solve for "y" in terms of "x" ;
y − x + x = 84 + x ;
y = 84 + x ;
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So, we have:
x = (1/5) y ;
and: y = 84 + x ;
Substitute "(1/5)y" for "x" ; in "y = 84 + x " ; to solve for "y" ;
y = 84 + [ (1/5)y ]
Subtract " [ (1/5)y ] " from EACH SIDE of the equation ;
y − [ (1/5)y ] = 84 + [ (1/5)y ] − [ (1/5)y ] ;
to get:
[ (4/5)y ] = 84 ;
↔ (4y) / 5 = 84 ;
→ 4y = 5 * 84 ;
Divide EACH SIDE of the equation by "4" ;
to isolate "y" on one side of the equation; and to solve for "y" ;
4y / 4 = (5 * 84) / 4 ;
y = 5 * (84/4) = 5 * 21 = 105 .
y = 105 .
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Now, plug "105" for "y" into:
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Either:
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x = (1/5) y ;
OR:
y = 84 + x ;
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to solve for "x" ;
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Let us do so in BOTH equations; to see if we get the same value for "x" ; which is a method to "double check" our answer ;
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Start with:
x = (1/5)y
→ (1/5)*(105) = 105 / 5 = 21 ; x = 21 ;
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So, x = 21; y = 105 .
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Now, let us see if this values hold true in the other equation:
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y = 84 + x ;
105 = ? 84 + 21 ?
105 = ? 105 ? Yes!
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The numbers are: " 21 " and "105 " .
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Answer:
I dont really know but all i can say is that it is more then 200
<h2>
Answer: 14</h2>
Step-by-step explanation:
If we factorize the numbers, we may be able to surmise the GCF:
28 = 2 × 2 × 7
42 = 2 × 3 × 7
56 = 2 × 2 × 2 × 7
∴ the GCF is 2 × 7 = 14
um is this all one equation?
like -2.2×(-2)÷(-14)×5-2.2×(-2)÷(-41)×5?
