Major league baseball's fastest recorded pitch is 105 miles per hours. The distance between the pitcher's mound and home plate i
s 60 feet, 6 inches. How long did it take the ball to travel from the pitcher to the batter?
1 answer:
Answer:
0.4 sec
Step-by-step explanation:
Remember that
1 mile=5,280 feet
1 foot=12 inches
1 hour=3,600 seconds
Let
s -----> the speed in ft/sec
d ----> the distance in ft
t -----> the time in sec
s=d/t
Solve for t
t=d/s
step 1
Convert miles/hour to ft/sec
105 mi/h=105(5,280/3,600)=154 ft/sec
Convert 60 ft 6 in to ft
60 ft 6 in=60+(6/12)=60.5 ft
step 2
Find the time
t=d/s
we have
s=154 ft/sec
d=60.5 ft
substitute
t=60.5/154
t=0.39 sec
Round to the nearest tenth
t=0.4 sec
You might be interested in
1 Metre is the length of each side of the cube.
To example, here is the Volume Of Cube Equation:
V = a^3
We are given the value for V, so just take the cube root of 1, which equates to 1.
Summer and Winter is the answer
Y = mx + b
slope(m) = 4
(-3,-1)...x = -3 and y = -1
now we sub...we r looking for b, the y int.
-1 = 4(-3) + b
-1 = -12 + b
-1 + 12 = b
11 = b
so ur equation is : y = 4x + 11
Answer:
(0,-5). y=mx+b or -5=0 × 0+b, or solving for b: b=-5-(0)(0). b=-5.
(2,-5). y=mx+b or -5=0 × 2+b, or solving for b: b=-5-(0)(2). b=-5.
y=-5
A. = logb (3/7)^1/2
= logb 3^1/2 - logb 7^1/2
= 1/2(logb 3 - logb 7)
b. = log 7x^3 + log y^5
= log 7 + log x^3 + log y^5
= log 7 + 3 log x + 5 log y