The Y-intercept is when the line touches or crosses the y-axis and so the answer is (0,-24)
hope this helps!
Answer:
The answer to your question is 2 hours
Step-by-step explanation:
Data
Juan speed = 4 mi/h
Cathy speed = 7 mi/h
t = x
Formula
speed = distance/time
Solve for distance
distance = speed x time
d = s x t
Distance travelled by Juan
d = 4t
Distance travelled by Juan in 1.5 h
d = 4(1.5)
d = 6 mi
total distance d = 6 + 4t
Distance travelled by Cathy
d = vt
d = 7t
Equal both equations
6 + 4t = 7t
Solve for t
6 = 7t - 4t
6 = 3t
6/3 = t
2 = t
Cathy wll catch up with Juan in 2 hours
Answer:
Step-by-step explanation:
15x + 12 + 10 - 20 = 12
15x + 22 - 20 = 12
15x + 2 = 12
15x = 12 - 2
15x = 10
x = 10/15 which reduces to 2/3
X=number of senior tickets
y=number of student tickets
1st day
7x+1y=101
2nd day
7x+12y=134
easy, we can eliminate the x's
multiply first equation by -1
-7x-1y=-101
<u>7x+12y=134 +</u>
0x+11y=33
11y=33
divide by 11 both sides
y=3
sub back
7x+y=101
7x+3=101
minus 3 both sides
7x=98
divide both sides by 7
x=14
senior tickets cost $14
student tickets cost $3
that sounds unfair
9514 1404 393
Answer:
see attached
Step-by-step explanation:
One way to approximate the derivative at a point is by finding the slope of the secant line between points on either side. That is what is done in the attached spreadsheet.
f'(0.1) ≈ (f(0.2) -f(0.0))/(0.2 -0.0) = -5 . . . for example
__
Another way to approximate the derivative is to write a polynomial function that goes through the points (all, or some subset around the point of interest), and use the derivative of that polynomial function.
These points are reasonably approximated by a cubic polynomial. The derivative of that polynomial at the points of interest is given in the table in the second attachment. (f1 is a rounding of the derivative function f')
