Answer:
a = 22
Step-by-step explanation:
We know that YB is the altitude. An altitude basically just means a line straight through the middle. Since we know the measurements of both sides we can solve for a. XB=BZ, so if you solve a + 7 = 2a - 15, then you get your answer.
Answer:
A: ![v(t)=-t^2+4t-5](https://tex.z-dn.net/?f=v%28t%29%3D-t%5E2%2B4t-5)
Step-by-step explanation:
Acceleration is second derivative of position, velocity is first derivative. Therefore, the velocity is the integral of acceleration.
![v(t)=\int\ {2t+4} \, dt](https://tex.z-dn.net/?f=v%28t%29%3D%5Cint%5C%20%7B2t%2B4%7D%20%5C%2C%20dt)
Integrate:
![-t^2+4t+C](https://tex.z-dn.net/?f=-t%5E2%2B4t%2BC)
V(0)=-5:
![-0^2+4(0)+C=-5\\C=-5](https://tex.z-dn.net/?f=-0%5E2%2B4%280%29%2BC%3D-5%5C%5CC%3D-5)
Therefore, v(t):
![v(t)=-t^2+4t-5](https://tex.z-dn.net/?f=v%28t%29%3D-t%5E2%2B4t-5)
Answer:
1/4a+3/4b
Step-by-step explanation:
OP=OA+AP
OA=a
AP=3/4AB
AB=-a+b
thus AP=3/4(-a+b) which is-3/4a+3/4b
OP=a-3/4a+3/4b thus 1/4a+3/4b
Answer:
D. (0,5)
Step-by-step explanation:
We are given that,
The function f(x) represents the height of the object 'x' seconds after its launch in the air.
Also, the maximum height of the object is at 2 secs and the object hits the ground at 5 secs.
So, we get the following figure for the situation.
We have that,
<em>Domain of the function is represented by the time after the object is launched i.e. </em><em>the time when the object is in the air</em><em>.</em>
Since, the interval when the object is in the air is (0,5).
Hence, the domain of the given function should be (0,5).