Answer: y = 7
Step-by-step explanation:
Step 1) plug in 3 for y and 9 for x in the direct variation formula, which is y = kx
3 = k(9)
Step 2) Solve for k
3 = k(9) (Divide by 9 on both sides)
3/9 = k (Now simplify 3/9)
1/3 = k. So k = 1/3
*just FYI: “k” is the constant of variation.
Step 3) use the equation y = kx and plug in 1/3 for k
y = 1/3x
Step 4) using the equation y = 1/3x, plug in 21 for x.
y = 1/3(21) (1/3*21 is the same as 21/3)
y = 21/3 (21/3 = 7)
y = 7. So that’s you’re answer! When x=21, y=7 because 7 is 1/3 of 21!
Hope this helps! My best wishes to you as you go through school!
Answer:
2π² or 19.74
Step-by-step explanation:
- <em>Circumference of a circle is C = 2πr</em>
Given r = π
<u>Then</u>:
or
B is the answer because the value of x does not repeat
C Bill drove at a rate that was 10 mph faster than the rate Kevin drove
Perhaps the easiest way to find the midpoint between two given points is to average their coordinates: add them up and divide by 2.
A) The midpoint C' of AB is
.. (A +B)/2 = ((0, 0) +(m, n))/2 = ((0 +m)/2, (0 +n)/2) = (m/2, n/2) = C'
The midpoint B' is
.. (A +C)/2 = ((0, 0) +(p, 0))/2 = (p/2, 0) = B'
The midpoint A' is
.. (B +C)/2 = ((m, n) +(p, 0))/2 = ((m+p)/2, n/2) = A'
B) The slope of the line between (x1, y1) and (x2, y2) is given by
.. slope = (y2 -y1)/(x2 -x1)
Using the values for A and A', we have
.. slope = (n/2 -0)/((m+p)/2 -0) = n/(m+p)
C) We know the line goes through A = (0, 0), so we can write the point-slope form of the equation for AA' as
.. y -0 = (n/(m+p))*(x -0)
.. y = n*x/(m+p)
D) To show the point lies on the line, we can substitute its coordinates for x and y and see if we get something that looks true.
.. (x, y) = ((m+p)/3, n/3)
Putting these into our equation, we have
.. n/3 = n*((m+p)/3)/(m+p)
The expression on the right has factors of (m+p) that cancel*, so we end up with
.. n/3 = n/3 . . . . . . . true for any n
_____
* The only constraint is that (m+p) ≠ 0. Since m and p are both in the first quadrant, their sum must be non-zero and this constraint is satisfied.
The purpose of the exercise is to show that all three medians of a triangle intersect in a single point.
