The answer:
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x = (⅔)y ;
y = 3x/2.
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Given:
x + (⅓)y + x - (2/4)<span>y - x = (3/6)y ;
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Take the: x + x - x = 1x + 1x - 1x = 2x - 1x = 1x = x ;
and rewrite:
x + (⅓)y - (2/4)y = (3/6)y ;
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Note that: (2/4)y = (<span>½)y ;
and: (3/6)y = (</span><span>½)y ; so;
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Rewrite as:
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</span>x + (⅓)y - (½)y = (½)y ;
Add "(½)y" to EACH SIDE of the equation;
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x + (⅓)y - (½)y + (½)y = (½)y + (½)y ;
to get: x + (⅓)y = y ;
x = 1y - (⅓)y = (3/3) y - (1/3)y - [ (3-1)/3] y = (⅔)y ;
So: x = (<span>⅔)y ;
In terms of "y" ;
Given: </span>(⅔)y = x ; Multiply each side of the equation by "3" ;
3*[(⅔)y] = 3*x ;
to get: 2y = 3x ;
Now, divide EACH SIDE of the equation by "2" ; to isolate "y" on one side of the equation; and to solve for "y" (in terms of "x"):
2y / 2 = 3x / 2 ;
to get:
y = 3x/2 ;
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The answer is 26.36. If you were trying to figure out what is 25% of 6.59, you'd have to multiply, however you need to know what 6.59 is 25% of so you do the opposite, just divide 6.59 by .25 and you'll get 26.36 which is your answer.
Solve the first for y
y=13x-90 and the second is y=x^2-x-42
When there is/are solutions y=y so we can say:
x^2-x-42=13x-90 subtract 13x from both sides
x^2-14x-42=-90 add 90 to both sides
x^2-14x+48=0 expand so that it is easy to factor...
x^2-6x-8x+48=0 now factor 1st and 2nd pair of terms
x(x-6)-8(x-6)=0
(x-8)(x-6)=0
So there are two solutions, when x=6 and 8.
Using either original equation we can find the corresponding values for y. I'll use y=13x-90 because it is easier...
y(6)=-12, y(8)=14 so the two solutions are the points:
(6,-12) and (8,14)
This question basically asks for the LCM of 3 and 5, which is 3 x 5 =15
This is the answer: 15
Check : The first rectangle had 1/3 shaded; ( one in every three parts)
So for 15 parts altogether, 5 would be shaded ( 1/3 = 5/15)
The second rectangle had 2/5 shaded ( two for every five parts)
So for 15 parts altogether, 6 would be shaded (2/5 = 6/15)
Answer:
A
Step-by-step explanation:
From the first row, you can see the number of circles increase as you move from left to right therefore the same principle was applied for the row below it.
