A stuntman jumping off a 20-m-high building is modeled by the equation h = 20 – 5t2, where t is the time in seconds. A high-speed camera is ready to film him between 15 m and 10 m above the ground. For which interval of time should the camera film him?
Answer:
![1\leq t\geq \sqrt{2}](https://tex.z-dn.net/?f=1%5Cleq%20t%5Cgeq%20%5Csqrt%7B2%7D)
Step-by-step explanation:
Given:
A stuntman jumping off a 20-m-high building is modeled by the equation
-----------(1)
A high-speed camera is ready to making film between 15 m and 10 m above the ground
when the stuntman is 15m above the ground.
height
Put height value in equation 1
![15 =20-5t^{2}](https://tex.z-dn.net/?f=15%20%3D20-5t%5E%7B2%7D)
![5t^{2} =20-15](https://tex.z-dn.net/?f=5t%5E%7B2%7D%20%3D20-15)
![5t^{2} =5](https://tex.z-dn.net/?f=5t%5E%7B2%7D%20%3D5)
![t^{2} =1](https://tex.z-dn.net/?f=t%5E%7B2%7D%20%3D1)
![t =\pm1](https://tex.z-dn.net/?f=t%20%3D%5Cpm1)
We know that the time is always positive, therefore ![t=1](https://tex.z-dn.net/?f=t%3D1)
when the stuntman is 10m above the ground.
height
Put height value in equation 1
![10 =20-5t^{2}](https://tex.z-dn.net/?f=10%20%3D20-5t%5E%7B2%7D)
![5t^{2} =20-10](https://tex.z-dn.net/?f=5t%5E%7B2%7D%20%3D20-10)
![5t^{2} =10](https://tex.z-dn.net/?f=5t%5E%7B2%7D%20%3D10)
![t^{2} =\frac{10}{5}](https://tex.z-dn.net/?f=t%5E%7B2%7D%20%3D%5Cfrac%7B10%7D%7B5%7D)
![t^{2} =2](https://tex.z-dn.net/?f=t%5E%7B2%7D%20%3D2)
![t=\pm\sqrt{2}](https://tex.z-dn.net/?f=t%3D%5Cpm%5Csqrt%7B2%7D)
![t=\sqrt{2}](https://tex.z-dn.net/?f=t%3D%5Csqrt%7B2%7D)
Therefore ,time interval of camera film him is ![1\leq t\geq \sqrt{2}](https://tex.z-dn.net/?f=1%5Cleq%20t%5Cgeq%20%5Csqrt%7B2%7D)
The slope would be 0 because the line does not intersect the x - axis.
The angle of elevation is the degree at which you need to look up, therefore the angle between the dotted line, let's call it LM, and line LK is 5 degrees
Angle JLK and Angle KLM are also complimentary angles, meaning that they both add up to 90 degrees, therefore angle JLK = 90-55=35degrees
Now, SOHCAHTOA comes in handy to find JL. Since we know JLK and JK, this is going to be a breeze.
Since the sides we are using are opposite and adjacent to JLK, we are going to use tan.
So: tan = o/a
however, as we don't know a, we need to rearrange the equation so a is on its own
tan=o/a (times both sides by a)
tan a = o (divide both sides by tan)
a = 0/tan
Now we can solve the equation with the information given
a = 26/tan35
a = 26/0.7
a=37.13184818
=> 37.1
So, you can put down A as your answer :D
Answer:
Step-by-step explanation:
1) i think for the first question any probability greater than 0.4 has more likelihood that it will happen
2) mutually exclusive events are ones that cannot occur at the same time as it's impossible e.g. a single coin toss cannot result in you getting heads and tails
Hey there! :)
7. -4(2z + 6) - 12 = 4
Simplify.
-8z - 24 - 12 = 4
Simplify.
-8z - 36 = 4
Add 36 to both sides.
-8z - 36 + 36 = 4 + 36
Simplify.
-8z = 40
Divide both sides by -8.
z = -5
8. 3/2(x - 2) - 5 = 19
Simplify.
3/2x - 3 - 5 = 19
Simplify.
3/2x - 8 = 19
Add 8 to both side.
3/2x = 19 + 8
Simplify.
3/2x = 27
Multiply both sides by 2.
3/2x · 2 = 27 · 2
Simplify.
3x = 54
Divide both sides by 3.
x = 18
~hope I helped~