Answer:
The ball will take 1 sec
Step-by-step explanation:
∵ The distance from the ground when the ball leaves = 6 ft
∵ The distance from the ground when the ball catches = 10 ft
∴ The difference between two position = 10 - 6 = 4 ft
∵ s = ut - 1/2 at² ⇒ s is the distance , u is the initial velocity , a is acceleration of gravity and t is the time
∵ s = 4 feet , u = 20 feet/sec and a ≅ 32 feet/sec²
∴ 4 = 20t -32/2 t²
∴ 16t² - 20t + 4 = 0
∴ (4t - 4)(4t - 1) = 0
∴ 4t - 4 = 0 ⇒ t = 1 and 4t - 1 = 0 ⇒ t = 1/4
∴ The ball will take 1 sec to go from first player to the second player
Note: 1/4 sec is the time of the ball after leaves the position of the first player by 4 feet above the initial position.
Answer:
I believe the answer would be 12
Step-by-step explanation:
Answer:
x=1
Step-by-step explanation:
log_4(x + 3) + log_4x = 1
We know that loga(b) + loga(c) = loga(bc)
log_4(x + 3)x = 1
Raise each side to the base of 4
4^log_4(x + 3)x = 4^1
(x+3)x = 4
x^2 +3x = 4
Subtract 4 from each side
x^2 +3x -4 = 0
Factor
(x+4) (x-1) =0
Using the zero product property
x= -4 x=1
But x cannot be negative since logs cannot be negative
x=1
Answer:
c
Step-by-step explanation:
-8n+7=31
-8n+7-7=31-7
-8n=31-7
-8n=24
-8n÷(-8)=24÷(-8)
n=24÷(-8)
n=-24÷8
n=-3
