Answer:
A. (0, -2) and (4, 1)
B. Slope (m) = ¾
C. y - 1 = ¾(x - 4)
D. y = ¾x - 2
E. -¾x + y = -2
Step-by-step explanation:
A. Two points on the line from the graph are: (0, -2) and (4, 1)
B. The slope can be calculated using two points, (0, -2) and (4, 1):
Slope (m) = ¾
C. Equation in point-slope form is represented as y - b = m(x - a). Where,
(a, b) = any point on the graph.
m = slope.
Substitute (a, b) = (4, 1), and m = ¾ into the point-slope equation, y - b = m(x - a).
Thus:
y - 1 = ¾(x - 4)
D. Equation in slope-intercept form, can be written as y = mx + b.
Thus, using the equation in (C), rewrite to get the equation in slope-intercept form.
y - 1 = ¾(x - 4)
4(y - 1) = 3(x - 4)
4y - 4 = 3x - 12
4y = 3x - 12 + 4
4y = 3x - 8
y = ¾x - 8/4
y = ¾x - 2
E. Convert the equation in (D) to standard form:
y = ¾x - 2
-¾x + y = -2
Answer:
35
Step-by-step explanation:
if 1 man is 37
and 1 man is 43
the 3rd man is 45
they add to 115
to average 40, we gotta add 35
this old mans giving u that!
8 x 2 = 16
Square root of 16 = 4
One side of the square is 4 meters.
Answer:
400 and 500
Step-by-step explanation:
the ratios are 8:10
total ratio = 8+10 = 18
share for the ratio is 8 = 8/18×900
= 0.444444444×900= 400
therefore, share for the ratio of 8= 400
share for the ratio of 10= 10/18×900
= 0.555555555×900=500
therefore, share for the ratio of 10= 500
Answer:
school building, so the fourth side does not need Fencing. As shown below, one of the sides has length J.‘ (in meters}. Side along school building E (a) Find a function that gives the area A (I) of the playground {in square meters) in
terms or'x. 2 24(15): 320; - 2.x (b) What side length I gives the maximum area that the playground can have? Side length x : [1] meters (c) What is the maximum area that the playground can have? Maximum area: I: square meters
Step-by-step explanation:
