Let 'x' be the problems worth 2 points.
Let 'y' be the problems worth 3 points.
Since, there are 38 total problems.
So, 
(equation 1)
x = 38-y
Since, a perfect score is 100 points.
So, 
(equation 2)
Substituting the value of 'x', we get
y = 24
x+y = 38
x = 38-24 = 14
So, 14 problems are worth 2 points and 24 problems are worth 3 points.
1.) D
2.) A................
59 is D
because with the point (-3,7) you substitute it into the equation, making it: 7=4x+b. solve for b. then you have y=4x+19. work out the algebra in the possible choices and whatever equals y=4x+19 will be the answer. in this case, its D.
60 is C
same as above, you do the algebra of the equation. bring the one over after doing distribution with the 4 and voila!
61 is A
a relatively easy one, all you do is the the slope -4 where m goes, and 3 where b goes. y= -4x+3
62 is C.
this one requires more work.
chose one of the points, in this case (2,7) and put them into the equation.
but wait, you need a slope!
you get that use the formula (y2-y1)/(x2-x1) which will be
(7-5)/(2-3) which will be
-2.
now you have y-7= -2(x-2)
voila!
63 is C. y= 1/2x+3
64 is B. (3, -5)
66 is B. negative. the line goes \ ( not / which is positive)
67 would be A. because it is positive and the I and the E are in the right places.
70 is C. 2/3. as before, remember we can but the points into this equation and have (6-4)/(3-0) which = 2/3
71 is D. y= 3x+10
72 is C. a third degree monomial
73 can't read
74 can't read
75 can't read.
Answer:
So estimate.
Really depends on what needs to be estimated, the answer or the overall equation.
So doing the real one first.
283/11 = 25 8/11
You can’t have “part” of a student so rounding this you can go up since 8 is more then half of 11
So 26 students
If you were to round the overall equation then you’d get
280 / 10 which is 28 students per class
So two options.
650 * 3456 = 2,246,400 is the answer.
