Answer:
The answer to your question is ∠3 = 63°
Step-by-step explanation:
Data
∠2 = 117°
∠3 = x
Process
1.- Angles 2 and 3 are supplementary which means that their addition gives 180°.
So,
∠2 + ∠3 = 180
Substitution
117° + ∠3 = 180°
Solve for ∠3
∠3 = 180° - 117°
Simplification and result
∠3 = 63°
Answer:
10% would be the answer
Step-by-step explanation:
Answer:
c five hours
Step-by-step explanation:
i did the math
Answer:
8x^2+14x^2+11x+2/4x+1
22x^2+11x+2/4x+1 ( It's 22x^2 cause 14x^2+*x^2 is 22x^2 and they are like terms.)
22x^2+11x+2/4x+1 ( The answer stays same cause there is no like terms so we can divide them)
Answer:
The maximum height of the particle is 484 m.
Step-by-step explanation:
Given that,
A particle is moving along a projectile path at an initial height of 160 feet with an initial speed of 144 feet per second. This can be represented by the function :
....(1)
We need to find the maximum height of the particle. For maximum height put ![\dfrac{dH}{dt}=0](https://tex.z-dn.net/?f=%5Cdfrac%7BdH%7D%7Bdt%7D%3D0)
So,
![\dfrac{d(-16t^2+144t+160)}{dt}=0\\\\-32t+144=0\\\\t=4.5\ s](https://tex.z-dn.net/?f=%5Cdfrac%7Bd%28-16t%5E2%2B144t%2B160%29%7D%7Bdt%7D%3D0%5C%5C%5C%5C-32t%2B144%3D0%5C%5C%5C%5Ct%3D4.5%5C%20s)
Put t = 4.5 s in equation (1) as :
![H(t) = -16(4.5)^2 + 144(4.5)+ 160\\\\H(t)=484\ m](https://tex.z-dn.net/?f=H%28t%29%20%3D%20-16%284.5%29%5E2%20%2B%20144%284.5%29%2B%20160%5C%5C%5C%5CH%28t%29%3D484%5C%20m)
So, the maximum height of the particle is 484 m.