Use the following formula to find the answer to this question:
S * Q + M * Q = P
Representations:
S = The amount of points for short-answer questions.
M = The a<span>mount of points for m</span>ultiple-choice questions.
Q = Number of questions for each of the type of questions that add up to the total number of questions on the test (20 questions).
P = Total points on the test.
Since there is a total of 50 points on the 20 question test, you would need to divide up the amount of questions there are for each of the type there are on the test, short-answer, and multiple-choice.
3 * Q + 1 * Q = P
Now find how many of the type of questions there are out of 20 questions on the test that would add up to the total number of points on the test.
3 * Q + 1 * Q = 50
(15 + 5 = 20 questions on the test, 15 and 5 can be used for the number of questions for each type of question on the test).
3 * 15 + 1 * 5 = 50
45 + 5 = 50
50 = 50
So, your total number of multiple choice questions are: 5.
And, your total number of short answer questions are: 15.
I would go with D. for your answer. I may be wrong.
<em>I hope this helps. </em>
<em>~ Notorious Sovereign</em>
width = w
length = w +2
perimeter = 2L + 2w
40 = 2(w+2) + 2w
40 = 2w+4 +2w
40 = 4w+4
36=4w
w=36/4 = 9
with = 9
length = 9+2 = 11
area = l x w = 9*11 = 99 square cm
Answer:
y=0x+-6
Step-by-step explanation:
y=mx+b
With what? I don’t see anything
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.
