Answer:
c
Step-by-step explanation:
i think it is c bec 1ft is 6m
Answer:
Option (B)
Step-by-step explanation:
There are two lines on the graph representing the system of equations.
First line passes through two points (-3, 1) and (-2, 3).
Slope of the line = 
= 
m = 2
Equation of the line passing through (x', y') and slope = m is,
y - y' = m(x - x')
Equation of the line passing through (-3, 1) and slope = 2 will be,
y - 1 = 2(x + 3)
y = 2x + 7 ----------(1)
Second line passes through (0, 1) and (-1, 4) and y-intercept 'b' of the line is 1.
Let the equation of this line is,
y = mx + b
Slope 'm' = 
= 
= -3
Here 'b' = 1
Therefore, equation of the line will be,
y = -3x + 1 ---------(2)
From equation (1) and (2),
2x + 7 = -3x + 1
5x = -6
x = 
x = 
From equation (1),
y = 2x + 7
y = 
= 
= 
= 
Therefore, exact solution of the system of equations is
.
Option (B) will be the answer.
Answer:
Step-by-step explanation:
You need to find out the circumference of the Ferris wheel
C=d*3.14=185*3.14=580.90 m
Error// Error// Error// Error// Error// Error//
A.) To find the maximum height, we can take the derivative of h(t). This will give us the rate at which the horse jumps (velocity) at time t.
h'(t) = -32t + 16
When the horse reaches its maximum height, its position on h(t) will be at the top of the parabola. The slope at this point will be zero because the line tangent to the peak of a parabola is a horizontal line. By setting h'(t) equal to 0, we can find the critical numbers which will be the maximum and minimum t values.
-32t + 16 = 0
-32t = -16
t = 0.5 seconds
b.) To find out if the horse can clear a fence that is 3.5 feet tall, we can plug 0.5 in for t in h(t) and solve for the maximum height.
h(0.5) = -16(0.5)^2 + 16(-0.5) = 4 feet
If 4 is the maximum height the horse can jump, then yes, it can clear a 3.5 foot tall fence.
c.) We know that the horse is in the air whenever h(t) is greater than 0.
-16t^2 + 16t = 0
-16t(t-1)=0
t = 0 and 1
So if the horse is on the ground at t = 0 and t = 1, then we know it was in the air for 1 second.