"The value of y varies directly with x" ----> y = kx
when x=3, y=21
21 = k(3)
k = 7
so we have y = 7x
plug in y= 105 and solve for x
Pls help I been working on this for forever now.
when x=3, y=21
21 = k(3)
k = 7
so we have y = 7x
plug in y= 105 and solve for x
Why are you confused. What I did is the standard way of handling direct variations.
You could set up a ratio .....
y1/y2 = x1/x2
in your case x1=3 and y1 =21
find x2 when x2 = 105
21/105 = 3/x2
cross-multiply
21x2 = 315
x2 = 315/21 = 15
My answer is um....B
Now use the point-slope form for your desired line:
y-5 = 1/2 (x-0)
So, not B.
3/21=x/105
m in this equation (y = 1.2x + 3.8) is 1.2,
It is not equation A since it has a slope of 1.3,
It is not equation B since it has a slope of -1.2 which makes it cross over line y = 1.2x + 3.8. (perpendicular)
C is the answer. Take the equations y = 1.2x + 5 and y = 1.2x + 3.8 and make tables with them. As you can see the y values line up.
Equation C (y = 1.2x + 5) Equatioion y = 1.2x + 3.8
x y x y
-1 2.6 -2 2.6
0 3.8 -1 3.8
1 5 0 5
2 6.2 1 6.2
3 7.4 2 7.4
Finding Points on a Line
To find points on the line y = mx + b,
Choose x and solve the equation for y,
Example,
Find two points on the line y = 1.2x + 3.8:
Let x =1. Substitute x = 1 into y = 21.x + 3.8 and solve for y,
y = 1.2(1) + 3.8 = 1.2 + 3.8 = 5
(1,5)