PreCalculus Question: If you don't know how to do college-level math, don't give me a useless "this is hard" or "look at your no
tes." I've looked through my notes and my textbook and genuinely need help. So if you've passed college trig or precalculus, please help.
Please look at the images attatched and see if you can help me. I've already found the speed of the merry-go-round to be 13.963 ft/s, and I know for sure the graph of the shadow's motion is sinusoidal. However, I'm not sure how to make that graph or where to go from here.
Please look at the images attatched and see if you can help me. I've already found the speed of the merry-go-round to be 13.963 ft/s, and I know for sure the graph of the shadow's motion is sinusoidal. However, I'm not sure how to make that graph or where to go from here.
1 answer:
You might be interested in
- ∆ABD is right angled hence area:-
There is only one option containing 6x^2 i.e Option D.
Hence without calculating further
Option D is correct
Every confidence interval has associated z value. As confidence interval increases so do the z value associated with it.
The confidence interval can be calculated using following formula:
Where
is the mean value, z is the associated z value, s is the standard deviation and n is the number of samples.
We know that standard deviation is simply a square root of variance:
The confidence interval of 95% has associated z value of <span>1.960.
</span>Now we can calculate the confidence interval for our income:
The confidence interval can be calculated using following formula:
Where
We know that standard deviation is simply a square root of variance:
The confidence interval of 95% has associated z value of <span>1.960.
</span>Now we can calculate the confidence interval for our income:
The braking distance is the distance the car travels before coming to a stop after the brakes are applied
a. The braking distances are as follows;
- The braking distance at 25 mph, is approximately <u>63.7 ft.</u>
- The braking distance at 55 mph, is approximately <u>298.35 ft.</u>
- The braking distance at 85 mph, is approximately <u>708.92 ft.</u>
b. If the car takes 450 feet to brake, it was traveling with a speed of 98.211 ft./s
Reason:
The given function for the braking distance is D = 2.6 + v²/22
a. The braking distance if the car is going 25 mph is therefore;
25 mph = 36.66339 ft./s
At 25 mph, the braking distance is approximately <u>63.7 ft.</u>
At 55 mph, the braking distance is given as follows;
55 mph = 80.65945 ft.s
At 55 mph, the braking distance is approximately <u>298.35 ft.</u>
At 85 mph, the braking distance is given as follows;
85 mph = 124.6555 ft.s
At 85 mph, the braking distance is approximately <u>708.92 ft.</u>
b. The speed of the car when the braking distance is 450 feet is given as follows;
v² = (450 - 2.6) × 22 = 9842.8
v = √(9842.2) ≈ 98.211 ft./s
The car was moving at v ≈ <u>98.211 ft./s</u>
Learn more here: