Direct variation is y = kx, where k is the constant of variation.
But now it says y varies directly with x2 (or 2x), so now the x in the equation is 2x.
The equation is y = k(2x)
Now you find k.
y = 96 when x = 4.
(96) = k(2*4)
96 = k(8)
k = 12
The equation is now y = 12(2x)
To find the value of y when x=2, plug 2 into the equation you made.
y = 12(2*2)
y = 48
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Now it's with a "quadratic variation," which is the same thing except x is squared.
The equation is y = kx^2
But y varies directly with x2 (same thing as 2x), so now it's y = k(2x)^2.
Now you find k by substituting y and x values that were given.
y = 180 when x = 6
(180) = k(2*6)^2
180 = k(12)^2
180 = k(144)
k = 1.25
k, 1.25, is the constant of variation.
C,
{(-1, -3),(0,-1),(1,1),(2,3),(3,5)}
Answer:
2,5 b 1 c 0,5 d -1 e 2,5
Step-by-step explanation:
nsweres el dodeel s Step-by-step explanationen
Answer:
23. x = 4; DE = 44
24. x = 25; DS = 28
Step-by-step explanation:
23. Point S is the midpoint of DE, so ...
DS = SE
3x +10 = 6x -2
12 = 3x . . . . . . . . . add 2-3x
4 = x . . . . . . . . . . . divide by 3
Then DS has length ...
DS = 3x +10 = 12 +10 = 22
and DE is twice that length, so ...
DE = 44
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24. DS is half the length of DE, so is ...
DS = DE/2 = 56/2
DS = 28
Then x can be found from ...
DS = x +3
28 -3 = x = 25 . . . . . substitute value for DS
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<em>Comment on problem 24</em>
Sometimes it is easier to work parts of a problem out of sequence. Here, finding DS first makes finding x easier.
