Answer: 32°
Explanation:
180 - 88 = 92° (Adjacent Angles on a straight line)
180 - 124 = 56° (Adjacent Angles on a straight line)
180 - 92 - 56 = 32° (Sum of angles in a triangle)
Volume of a cone: 1/3
![\pi](https://tex.z-dn.net/?f=%20%5Cpi%20)
r²h
r is the radius (diameter ÷ 2) and h is the height.
Find the radius first using the diameter: 2
![\frac{1} {2}](https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%20%7B2%7D)
÷ 2 = 1.25
Now plug in the numbers into the formula.
1/3
![\pi](https://tex.z-dn.net/?f=%20%5Cpi%20)
1.25²10 = <span>16.4
</span>
The volume of the cone is 16.4 cubic feet.
Answer:
Step-by-step explanation:
0.5bh = 36
0.5(5b)(5h) = 25(0.5bh) = 25 × 36 = 900 sq cm
Answer:
r = 144 units
Step-by-step explanation:
The given curve corresponds to a parametric function in which the Cartesian coordinates are written in terms of a parameter "t". In that sense, any change in x can also change in y owing to this direct relationship with "t". To find the length of the curve is useful the following expression;
![r(t)=\int\limits^a_b ({r`)^2 \, dt =\int\limits^b_a \sqrt{((\frac{dx}{dt} )^2 +\frac{dy}{dt} )^2)} dt](https://tex.z-dn.net/?f=r%28t%29%3D%5Cint%5Climits%5Ea_b%20%28%7Br%60%29%5E2%20%5C%2C%20dt%20%3D%5Cint%5Climits%5Eb_a%20%5Csqrt%7B%28%28%5Cfrac%7Bdx%7D%7Bdt%7D%20%29%5E2%20%2B%5Cfrac%7Bdy%7D%7Bdt%7D%20%29%5E2%29%7D%20%20%20%20%20dt)
In agreement with the given data from the exercise, the length of the curve is found in between two points, namely 0 < t < 16. In that case a=0 and b=16. The concept of the integral involves the sum of different areas at between the interval points, although this technique is powerful, it would be more convenient to use the integral notation written above.
Substituting the terms of the equation and the derivative of r´, as follows,
![r(t)= \int\limits^b_a \sqrt{((\frac{d((1/2)t^2)}{dt} )^2 +\frac{d((1/3)(2t+1)^{3/2})}{dt} )^2)} dt](https://tex.z-dn.net/?f=r%28t%29%3D%20%5Cint%5Climits%5Eb_a%20%5Csqrt%7B%28%28%5Cfrac%7Bd%28%281%2F2%29t%5E2%29%7D%7Bdt%7D%20%29%5E2%20%2B%5Cfrac%7Bd%28%281%2F3%29%282t%2B1%29%5E%7B3%2F2%7D%29%7D%7Bdt%7D%20%29%5E2%29%7D%20%20%20%20%20dt)
Doing the operations inside of the brackets the derivatives are:
1 ) ![(\frac{d((1/2)t^2)}{dt} )^2= t^2](https://tex.z-dn.net/?f=%28%5Cfrac%7Bd%28%281%2F2%29t%5E2%29%7D%7Bdt%7D%20%29%5E2%3D%20t%5E2)
2) ![\frac{(d(1/3)(2t+1)^{3/2})}{dt} )^2=2t+1](https://tex.z-dn.net/?f=%5Cfrac%7B%28d%281%2F3%29%282t%2B1%29%5E%7B3%2F2%7D%29%7D%7Bdt%7D%20%29%5E2%3D2t%2B1)
Entering these values of the integral is
![r(t)= \int\limits^{16}_{0} \sqrt{t^2 +2t+1} dt](https://tex.z-dn.net/?f=r%28t%29%3D%20%5Cint%5Climits%5E%7B16%7D_%7B0%7D%20%20%5Csqrt%7Bt%5E2%20%2B2t%2B1%7D%20%20%20%20%20dt)
It is possible to factorize the quadratic function and the integral can reduced as,
![r(t)= \int\limits^{16}_{0} (t+1) dt= \frac{t^2}{2} + t](https://tex.z-dn.net/?f=r%28t%29%3D%20%5Cint%5Climits%5E%7B16%7D_%7B0%7D%20%28t%2B1%29%20%20dt%3D%20%5Cfrac%7Bt%5E2%7D%7B2%7D%20%2B%20t%20)
Thus, evaluate from 0 to 16
The value is ![r= 144 units](https://tex.z-dn.net/?f=%20r%3D%20144%20units)