Answer:
Step-by-step explanation:
The position function is
and if we are looking for the time(s) that the ball is 10 feet above the surface of the moon, we sub in a 10 for s(t) and solve for t:
and
and factor that however you are currently factoring quadratics in class to get
t = .07 sec and t = 18.45 sec
There are 2 times that the ball passes 10 feet above the surface of the moon, once going up (.07 sec) and then again coming down (18.45 sec).
For part B, we are looking for the time that the ball lands on the surface of the moon. Set the height equal to 0 because the height of something ON the ground is 0:
and factor that to get
t = -.129 sec and t = 18.65 sec
Since time can NEVER be negative, we know that it takes 18.65 seconds after launch for the ball to land on the surface of the moon.
Answer:
The answer in the procedure
Step-by-step explanation:
Let
A1 ------> the area of the first square painting
A2 ----> the area of the second square painting
D -----> the difference of the areas
we have
case 1) The area of the second square painting is greater than the area of the first square painting
The difference of the area of the paintings is equal to subtract the area of the first square painting from the area of the second square painting
D=A2-A1
case 2) The area of the first square painting is greater than the area of the second square painting
The difference of the area of the paintings is equal to subtract the area of the second square painting from the area of the first square painting
D=A1-A2
The answer is "MS and QS". Trust me I just took the test and made a 90
Answer:
Mean and standard deviation of the sampling distribution of the sample proportions are 76 and 0.0427 respectively.
Step-by-step explanation:
The mean for a sample proportion is given by μ = np
n = sample size = 100
p = fraction of the sample proportion that have what is being tested = 76% = 0.76
μ = 0.76 × 100 = 76.
Standard deviation of a sample proportion = σ = √[p(1-p)/n] = √(0.76×0.24/100) = 0.0427
Answer:
Step-by-step explanation:
Find the digram attached.
Perimeter of the track = perimeter of the rectangle + perimeter of the 2semicircles
Perimter of a rectangle = 2(x+r) where:
x is the length
2r is the width of the rectangle = diameter of the semicircle
Perimeter of semicircle = 2πr/2 = πr
Perimeter of 2semicirle = 2πr
Perimeter of the track = 2(x+2r) + 2πr
r is the radius if the semicircle
Expand
Perimeter of the track = 2x+4r + 2πr
Perimeter of the track = 2(x+2r+πr)
b) Given P = 2(x+2r+πr), we are to make x the subject of the formula.
P = 2x+4r+2πr
P-4r-2πr = 2x
Divide both sides by 2.
(P-4r-2πr)/2 = 2x
x = (P-4r-2πr)/2
c) Given
P = 600fr
r = 50ft
x = (600-4(50)-2π(50))/2
x = (600-200-100(3.14))/2
x = 400-314/2
x = 86/2
x = 43ft
Hence the value of x to nearest foot is 43ft
