Let
width,w
length, l = 1.6w ...eqn 1
area, a = l × w ...eqn 2
subst for l in eqn 2
a = 1.6 w × w = 4000
4000 = 1.6 w^2
w^2 = 2500
w = 50
subst for w in eqn 1
l = 50 x 1.6 = 80
length = 80 yds
width = 50 yds
Answer:
5 seconds
Step-by-step explanation:
In order to find the time when it landed, we will have to find the x-intercepts.
Equation given to us : -16t² + 48t + 160
Let's take the GCD, which is -16.
-16( t² - 3t - 10 )
Factoring what's inside the brackets, we get the x-intercepts.
What multiples to -10 but adds up to -3? The numbers are -5 and 2
-16 ( t - 5 ) ( t - 2 )
X-intercepts are t - 5 and t - 2
Which is 5 and 2 seconds. But one of this is an extraneous solution and that is 2.
If we substitute the value of 2 in the equation, we will not get 0.
During the x-intercept, the x has a value and y is 0. If we substitute 5 ans x. we will get y as 0.
Hence, the answer is 5 seconds.
Testing testing (ignore that)
Answer:
Only 4 hours i am doing for 6 hours