Answer:
w= −16/t
Step-by-step explanation:
Let's solve for w.
wt = −21+5
Step 1: Divide both sides by t.
tw/t = −16/t
w= −16/t
Convert the mixed number into a decimal and multiply on a calculator (or by hand).
Answer:
Step-by-step explanation:
Method 1: Taking the log of both sides...
So take the log of both sides...
5^(2x + 1) = 25
log 5^(2x + 1) = log 25 <-- use property: log (a^x) = x log a...
(2x + 1)log 5 = log 25 <-- distribute log 5 inside the brackets...
(2x)log 5 + log 5 = log 25 <-- subtract log 5 both sides of the equation...
(2x)log 5 + log 5 - log 5 = log 25 - log 5
(2x)log 5 = log (25/5) <-- use property: log a - log b = log (a/b)
(2x)log 5 = log 5 <-- divide both sides by log 5
(2x)log 5 / log 5 = log 5 / log 5 <--- this equals 1..
2x = 1
x=1/2
Method 2
5^(2x+1)=5^2
2x+1=2
2x=1
x=1/2
For question number 1:The plot H = H(t) is the parabola and it reaches its maximum in the moment when exactly at midpoint between the roots t = 0 and t = 23. At that moment t = 23/2 or 11.5 seconds.
For question number 2:To find the maximal height, just simply substitute t = 11.5 into the quadratic equation. The answer would be 22.9.
For question number 3:H(t) = 0, or, which is the same as -16t^2 + 368t = 0.Factor the left side to get -16*t*(t - 23) = 0.t = 0, relates to the very start of the process, when the ash started its way up.The other root is t = 23 seconds, and it is precisely the time moment when the bit of ash will go back to the ground.
#1) 4.445 ft
#2) yes
Explanation
#1) The maximum height is the y-coordinate of the vertex. We first find the axis of symmetry, given by x=-b/2a:
x=-0.17/2(-0.005) = -0.17/-0.01 = 17
Plugging this into the equation,
y=-0.005(17²)+0.17(17)+3 = 4.445
#2) Substituting 30 into the equation,
y=-0.005(30²) + 0.17(30) + 3 = 3.6
The ball will 3.6 in the air, so yes, it will clear the 3 ft tall net.
