Yo sup??
To solve this question we have to apply trigonometric ratios
sin31=UT/TV
TV=UT/sin31
=7.76
=7.8
Hope this helps.
Domain means the values of independent variable(input) which will give defined output to the function.
Given:
The height h of a projectile is a function of the time t it is in the air. The height in feet for t seconds is given by the function

Solution:
To get defined output, the height h(t) need to be greater than or equal to zero. We need to set up an inequality and solve it to find the domain values.
![To \; find \; domain:\\\\h(t) \geq0\\\\-16t^2+96t \geq 0\\Factoring \; -16t \; in \; the \; left \; side \; of \; the \; inequality\\\\-16t(t-6) \geq 0\\Step \; 1: Find \; Boundary \; Points \; by \; setting \; up \; above \; inequality \; to \; zero.\\\\t(t-6)=0\\Use \; zero \; factor \; property \; to \; solve\\\\t=0 \; (or) \; t = 6\\\\Step \; 2: \; List \; the \; possible \; solution \; interval \; using \; boundary \; points\\(- \infty,0], \; [0, 6], \& [6, \infty)](https://tex.z-dn.net/?f=%20To%20%5C%3B%20find%20%5C%3B%20domain%3A%5C%5C%5C%5Ch%28t%29%20%5Cgeq0%5C%5C%5C%5C-16t%5E2%2B96t%20%5Cgeq%20%200%5C%5CFactoring%20%5C%3B%20-16t%20%5C%3B%20in%20%5C%3B%20the%20%5C%3B%20left%20%5C%3B%20side%20%5C%3B%20of%20%5C%3B%20the%20%5C%3B%20inequality%5C%5C%5C%5C-16t%28t-6%29%20%5Cgeq%20%200%5C%5CStep%20%5C%3B%201%3A%20Find%20%5C%3B%20Boundary%20%5C%3B%20Points%20%5C%3B%20by%20%5C%3B%20setting%20%5C%3B%20up%20%5C%3B%20above%20%5C%3B%20inequality%20%5C%3B%20to%20%5C%3B%20zero.%5C%5C%5C%5Ct%28t-6%29%3D0%5C%5CUse%20%5C%3B%20zero%20%5C%3B%20factor%20%5C%3B%20property%20%5C%3B%20to%20%5C%3B%20solve%5C%5C%5C%5Ct%3D0%20%5C%3B%20%28or%29%20%5C%3B%20t%20%3D%206%5C%5C%5C%5CStep%20%5C%3B%202%3A%20%5C%3B%20List%20%5C%3B%20the%20%5C%3B%20possible%20%20%5C%3B%20solution%20%5C%3B%20interval%20%5C%3B%20using%20%5C%3B%20boundary%20%5C%3B%20points%5C%5C%28-%20%5Cinfty%2C0%5D%2C%20%5C%3B%20%5B0%2C%206%5D%2C%20%5C%26%20%5B6%2C%20%5Cinfty%29%20)
![Step \; 3:Pick \; test \; point \; from \; each \; interval \; to \; check \; whether \\\; makes \; the \; inequality \; TRUE \; or \; FALSE\\\\When \; t = -1\\-16(-1)(-1-6) \geq 0\\-112 \geq 0 \; FALSE\\(-\infty, 0] \; is \; not \; solution\\Also \; Logically \; time \; t \; cannot \; be \; negative\\\\When \; t = 1\\-16(1)(1-6) \geq 0\\80 \geq 0 \; TRUE\\ \; [0, 6] \; is \; a \; solution\\\\When \; t = 7\\-16(7)(7-6) \geq 0\\-112 \geq 0 \; FALSE\\ \; [6, -\infty) \; is \; not \; solution](https://tex.z-dn.net/?f=%20Step%20%5C%3B%203%3APick%20%5C%3B%20test%20%5C%3B%20point%20%5C%3B%20from%20%5C%3B%20each%20%5C%3B%20interval%20%5C%3B%20to%20%5C%3B%20check%20%5C%3B%20whether%20%5C%5C%5C%3B%20makes%20%5C%3B%20the%20%5C%3B%20inequality%20%5C%3B%20TRUE%20%5C%3B%20or%20%5C%3B%20FALSE%5C%5C%5C%5CWhen%20%5C%3B%20t%20%3D%20-1%5C%5C-16%28-1%29%28-1-6%29%20%5Cgeq%20%200%5C%5C-112%20%5Cgeq%20%200%20%5C%3B%20FALSE%5C%5C%28-%5Cinfty%2C%200%5D%20%5C%3B%20is%20%5C%3B%20not%20%5C%3B%20solution%5C%5CAlso%20%5C%3B%20Logically%20%5C%3B%20time%20%5C%3B%20t%20%5C%3B%20cannot%20%5C%3B%20be%20%5C%3B%20negative%5C%5C%5C%5CWhen%20%5C%3B%20t%20%3D%201%5C%5C-16%281%29%281-6%29%20%5Cgeq%20%200%5C%5C80%20%5Cgeq%20%200%20%5C%3B%20TRUE%5C%5C%20%5C%3B%20%5B0%2C%206%5D%20%5C%3B%20is%20%5C%3B%20a%20%5C%3B%20solution%5C%5C%5C%5CWhen%20%5C%3B%20t%20%3D%207%5C%5C-16%287%29%287-6%29%20%5Cgeq%20%200%5C%5C-112%20%5Cgeq%20%200%20%5C%3B%20FALSE%5C%5C%20%5C%3B%20%5B6%2C%20-%5Cinfty%29%20%5C%3B%20is%20%5C%3B%20not%20%5C%3B%20solution%20)
Conclusion:
The domain of the function is the time in between 0 to 6 seconds

The height will be positive in the above interval.
So the car is moving at 651 revolutions per minute, with wheels of a radius of 18inches
so, one revolution, is just one go-around a circle, and thus 2π, 651 revolutions is just 2π * 651, or 1302π, the wheels are moving at that "angular velocity"
now, what's the linear velocity, namely, the arc covered per minute
well

now, how much is that in miles/hrs? well
let's keep in mind that, there are 12inches in 1foot, and 5280ft in 1mile, whilst 60mins in 1hr
thus

notice, after all the units cancellations, you're only left with mi/hrs
They give the formula as:
Surface Area =<span> (2 • <span>π <span>• r²) + (2 • <span>π • r • height)</span></span></span></span>
However the 2*PI*r^2 part of the formula is used to calculate the 2 "ends" of a cylinder. Since the problem states that you are NOT to count any of surface area of the "ends" then you only need the <span>(2 • <span>π • r • height) part of the formula.
So, r = 3 inches and height = 8 * 3 inches, the side area equals
2 * PI * r * height
2 * 3.14159 * 3 * 24 =
</span></span>
<span>
<span>
<span>
452.39 cubic inches which is the lateral area.</span></span></span>
Answer:




Step-by-step explanation:
We know that,

where,
A = Amount after time t,
P = Principle amount,
r = Rate of interest,
n = Number of times interest is compounded per year,
t = time period in year.
Investment of $25,000 for 4 years at an interest rate of 5% if the money is compounded semiannually
Here,
P = $25,000
r = 5% = 0.05
n = 2 (as compounded semiannually)
t = 4 years
Putting the values,




Investment of $25,000 for 4 years at an interest rate of 5% if the money is compounded quarterly.
Here,
P = $25,000
r = 5% = 0.05
n = 4 (as compounded quarterly)
t = 4 years
Putting the values,




Investment of $25,000 for 4 years at an interest rate of 5% if the money is compounded monthly.
Here,
P = $25,000
r = 5% = 0.05
n = 12 (as compounded monthly)
t = 4 years
Putting the values,



Investment of $25,000 for 4 years at an interest rate of 5% if the money is compounded continuously.

where,
A = Amount after time t,
P = Principle amount,
r = Rate of interest,
t = time period in year.
Putting all the values,

It can be observed that, the frequent we compound the amount, the more we get.