Answer:
28
Step-by-step explanation:
4+6=10,10+6=16,16+6=22,22+6=28
Answer:
t>2.034s and t>0.154s
Step-by-step explanation:
Given the quadratic function h(t)=-16t^2+35t+5 which represents the height of the balloon, h, in feet t seconds after it is thrown.
To determine when the balloon is more than 10feet above the ground, we will substitute h = 10 into the expression to have;
h(t)=-16t^2+35t+5
-16t^2+35t+5>10
-16t^2+35t+5-10>0
16t^2-35t+5>0
Find t;
t > 35±√35²-4(16)(5)/2(16)
t>35±√1225-320/32
t>35±√905/32
t>35±30.08/32
t > 35+30.08/32 and t > 35-30.08/32
t>65.08/32 and t>4.92/32
t>2.034s and t>0.154s
Hence the balloon is more than 10feet above the ground at when t>2.034s and t>0.154s
Answer:
![\cos^{-1}(\frac{6.3}{9.8})](https://tex.z-dn.net/?f=%5Ccos%5E%7B-1%7D%28%5Cfrac%7B6.3%7D%7B9.8%7D%29)
![\sin^{-1}(\frac{7.5}{9.8})](https://tex.z-dn.net/?f=%5Csin%5E%7B-1%7D%28%5Cfrac%7B7.5%7D%7B9.8%7D%29)
Step-by-step explanation:
Recall the mnemonics SOH-CAH-TOA
The sine ratio is the length of the opposite side expressed over the hypotenuse.
![\sin m\angle ABC=\frac{7.5}{9.8}](https://tex.z-dn.net/?f=%5Csin%20m%5Cangle%20ABC%3D%5Cfrac%7B7.5%7D%7B9.8%7D)
Take the sine inverse of both sides to get;
![m\angle ABC=\sin^{-1}(\frac{7.5}{9.8})](https://tex.z-dn.net/?f=m%5Cangle%20ABC%3D%5Csin%5E%7B-1%7D%28%5Cfrac%7B7.5%7D%7B9.8%7D%29)
The cosine ratio is the length of the adjacent side expressed over the hypotenuse.
![\cos m\angle ABC=\frac{6.3}{9.8}](https://tex.z-dn.net/?f=%5Ccos%20m%5Cangle%20ABC%3D%5Cfrac%7B6.3%7D%7B9.8%7D)
Take the cosine inverse of both sides to get;
![m\angle ABC=\cos^{-1}(\frac{6.3}{9.8})](https://tex.z-dn.net/?f=m%5Cangle%20ABC%3D%5Ccos%5E%7B-1%7D%28%5Cfrac%7B6.3%7D%7B9.8%7D%29)