at what time does this equation reaches 10000? show steps. thanks
w(t)=(t-10)(t-20)+4500
Sorry the equation is wrong it should be
W(t)=5 (t-10(t-20)+4500
set 5(t-10)(t-20)+4500 = 10000
5(t-10)(t-20) = 5500
(t-10)(t-20) = 1100
t^2 - 30t + 200 = 1100
t^2 - 30 t = 900
I will complete the square..
t^2 - 30t + 225 = 1125
(t-15)^2 = 1125
t - 15 = ±√1125 = ±15√5
t = 15 ± 15√5 , but more than likely t > 0
so t = 15 + 15√5 or appr 48.541
Calculate 5*(t-10)(t-20)=5500
umhh,
We don't 'calculate' 5*(t-10)(t-20)=5500
we "solve" 5*(t-10)(t-20)=5500
and I believe I just did that for you.
Thanks Reiny :D
To find the time when the equation w(t) reaches 10000, we need to solve the equation:
w(t) = 10000
Let's substitute the equation for w(t):
(t-10)(t-20) + 4500 = 10000
Now, we can solve it step by step:
Step 1: Expand the equation
t^2 - 30t + 200 + 4500 = 10000
Step 2: Combine like terms
t^2 - 30t + 4700 = 10000
Step 3: Move 10000 to the left side of the equation
t^2 - 30t + 4700 - 10000 = 0
Step 4: Simplify
t^2 - 30t - 5300 = 0
Step 5: Factor the equation (if possible)
Unfortunately, this particular equation doesn't factor easily. Therefore, we'll need to use the quadratic formula to find the solutions.
The quadratic formula is given by:
t = (-b ± √(b^2 - 4ac)) / (2a)
For our equation, a = 1, b = -30, and c = -5300.
Using the quadratic formula, we have:
t = (-(-30) ± √((-30)^2 - 4(1)(-5300))) / (2(1))
Simplifying:
t = (30 ± √(900 + 21200)) / 2
t = (30 ± √(22100)) / 2
Step 6: Calculate the square root of 22100
√(22100) ≈ 148.55
Since we have a ± sign, we'll find two possible values for t:
t1 = (30 + 148.55) / 2 ≈ 89.28
t2 = (30 - 148.55) / 2 ≈ -59.28
We are only interested in the positive value of t since time cannot be negative. Therefore, the equation w(t) reaches 10000 approximately at t = 89.28.