avagracegirlp17zx2
18.09.2019 •
Mathematics
In teds class of 45 students, the average on the math exam was 90%. in andrews class of 15 students, the average of the class was 82%. what is the average of the two classes
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Answers on questions: Mathematics
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Ответ:
86% Average of the two classes
Step-by-step explanation:
90+82=172
172/2= 86
Ответ:
y = (3/4)x
6–10: There are many useful forms of the equation for a line. It can be worthwhile to learn a few of them, or at least keep them where you have ready access.
6. The point-slope form is a good place to start. It can be simplified to slope-intercept form. For slope m and point (h, k), the point-slope equation can be written as
y = m(x -h) +k
For point (4, 7) this is
y = m(x -4) +7
When you substitute your value for m (not given here), you can find the slope-intercept equation to be
y = mx +(7-4m)
7. The 2-point form of the equation for a line is a good place to start. For points (x1, y1) and (x2, y2), the equation of the line is
y = (y2 -y1)/(x2 -x1)*(x -x1) +y1
y = (3-(-5))/(-1-(-5))*(x -(-5)) +(-5)
y = (8/4)(x +5) -5
y = 2x + 5 . . . . . . . . slope-intercept form
8. You can start with the form
mx -y = mh -k . . . . . . . . line with slope m through point (h, k)
To put this into standard form, you need to make sure the coefficient of x is a positive integer. You do that by multiplying the equation by the negative of the denominator of "m".
For your point, this is
mx - y = 6 -3m
When you substitute your value of m (not given here), you will have to finishe the arithmetic on the right side of the equation.
9. There are several ways to write the standard form of equation for a line through two points. One place to start for points (x1, y1) and (x2, y2) could be ...
(y2 -y1)x -(x2 -x1)y = (x1)(y2 -y1) -(y1)(x2 -x1)
For your points, this will be
(3-0)x -(-5-1)y = 1*(3-0) -0(-5-1)
3x +6y = 3
A factor of 3 must be removed to put this into standard form.
x +2y = 1
10. To find the slope of the given line, solve for y.
-5y = -4x -10
y = (-4/-5)x +(-10/-5) = (4/5)x +2
The perpendicular line will have a slope that is the negative reciprocal of the slope of this line. That is,
m = -1/(4/5) = -5/4
As above, it is convenient to start with the point-slope form, then simplify to the slope-intercept form.
y = (-5/4)(x -2) +3
y = (-5/4)x + 11/2