Answer:
1. x = -4y ---> y = (-1/4)x
slope = -1/4. y-intercept = (0,0)
2. y = -2x + 4
3. y = (1/3)x - 1
Step-by-step explanation:
1. Re-write your equation so that x is on the right and y is on the left:
x = -4y ---> y = (-1/4)x
slope = -1/4. y-intercept = (0,0)
2. y-intercept = (0,4) ----> P1
x-intercrpt = (2,0) ----> P2
slope m = (y2 - y1) / (x2 - x1)
= (0 - 4)/(2 - 0)
= -2
therefore, y - y1 = mx - x1 ---> y - 4 = -2x
or y = -2x + 4
3. y-intercept = (0,-1)
x-intercept = (3,0)
m = (0 - (-1)) / (3 -0) = 1/3
y - (-1) = (1/3)x - 0 ---> y = (1/3)x - 1
Hi, hope this helps you. Have a good day. :)
Answer:
A = $5,417.42 (balance of the account)
I = $2,617.42
The awnser is 60 because
It is a cube so it has the smae amout of sides
Point F is on line a, so it does represent Josiah's distance at a certain time. Also, point F is below line b, so it represents a distance that is less than Chana's distance. This is a distance-time graph problem.
<h3>
What is the proof for the above?</h3>
Recall that Josiah had a head start of 10 meters and he skates at 2 meters per second.
Since Y is the function that represents the distance in meters from the finished line, by observation, it is clear to see that all the factors that are related to his race are adequately represented in:
y = 10 + 2x
Where 10 is the head start in meters
2 is the rate at which he skates per second; and
x is the unknown amount of time in seconds.
Given that the point F sits over 25 seconds,
that is F(y) = 10 + 2 * 25
= 60 meters.
Hence, Point F is on line a, so it does represent Josiah's distance at exactly 25 seconds.
Learn more about distance-time graphs at:
brainly.com/question/4931057
#SPJ1
Answer:
b) 690 - 7.5*t
c) 0 < t < 92s time (t) is independent quantity
d) 0 < s < 690ft distance from bus stop (s) is dependent quantity
e) f(0) = 690 ft away from bus stop , f(60.25) = 238.125 ft away from bus stop
Step-by-step explanation:
Part a - see diagram
part b
initial distance from bus stop s0 = 690 ft
distance covered = 7.5*t
s = s0 - distance covered
s = 690 - 7.5*t = f(t)
part c
s = 0 or s = 690
0 = 690 -7.5*t
t = 92 s
Hence domain : 0 < t < 92s time (t) is independent quantity
part d
s = 0 or s = 690
Hence range : 0 < s < 690ft distance from bus stop (s) is dependent quantity because it depends on time (t)
part e
f(0) is s @t = 0
f(0) = 690 ft away from bus stop
f(60.25) is s @t = 60.25
f(60.25) = 690 - 7.5*60.25 = 238.125 ft away from bus stop.