Answer: A≈1519.76³ Units
Step-by-step explanation:
A=4πr²
r = 11
11²=121
121 * 3.14 = 379.94
379.94(4) = 1519.76
A≈1519.76³ Units OR
A=4πr2=4·π·112≈1520.53084
General formula for circles at the origin is x^2+y^2=R^2 where R is the radius.So R^2=4225. Solve for R.
Answer:
15000(1.003425)^12t ;
4.11%
4.188%
Step-by-step explanation:
Given that:
Loan amount = principal = $15000
Interest rate, r = 4.11% = 0.0411
n = number of times compounded per period, monthly = 12 (number of months in a year)
Total amount, F owed, after t years in college ;
F(t) = P(1 + r/n)^nt
F(t) = 15000(1 + 0.0411/12)^12t
F(t) = 15000(1.003425)^12t
2.) The annual percentage rate is the interest rate without compounding = 4.11%
3.)
The APY
APY = (1 + APR/n)^n - 1
APY = (1 + 0.0411/12)^12 - 1
APY = (1.003425)^12 - 1
APY = 1.04188 - 1
APY = 0.04188
APY = 0.04188 * 100% = 4.188%
Answer:
'Twice the sum of a number and 5' can be translated into variable expression such as:
Step-by-step explanation:
Given the phrase
<em>Twice the sum of a number and 5</em>
First, let us breakdown the English Phrase
Let the number be = n
The sum of a number 'n' and 5 = n + 5
now, twice the sum of a number and 5 can be determined by multiplying (n+5) with 5.
Thus,
'<em>Twice the sum of a number and 5</em>' can be translated into variable expression such as:
Given:
The diameter of the right cylinder is 2x cm.
The total surface area is 96 cm cube.
The radius is calculated as,
The total surface area is,
Volume is,
Now, differentiate with respect to x,
Now,
![\begin{gathered} \frac{dV}{dx}=0 \\ 84-3\pi(x^2)=0 \\ x^2=\frac{16}{\pi} \\ x=\sqrt[]{\frac{16}{\pi}} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20%5Cfrac%7BdV%7D%7Bdx%7D%3D0%20%5C%5C%2084-3%5Cpi%28x%5E2%29%3D0%20%5C%5C%20x%5E2%3D%5Cfrac%7B16%7D%7B%5Cpi%7D%20%5C%5C%20x%3D%5Csqrt%5B%5D%7B%5Cfrac%7B16%7D%7B%5Cpi%7D%7D%20%5Cend%7Bgathered%7D)
Now, differentiate (1) with respect to x again,
![\begin{gathered} \frac{d^2V}{dx^2}=\frac{d}{dx}(84-3\pi(x^2)) \\ =-6\pi x \\ At\text{ x=}\sqrt[]{\frac{16}{\pi}} \\ \frac{d^2V}{dx^2}=-6\pi\sqrt[]{\frac{16}{\pi}}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20%5Cfrac%7Bd%5E2V%7D%7Bdx%5E2%7D%3D%5Cfrac%7Bd%7D%7Bdx%7D%2884-3%5Cpi%28x%5E2%29%29%20%5C%5C%20%3D-6%5Cpi%20x%20%5C%5C%20At%5Ctext%7B%20x%3D%7D%5Csqrt%5B%5D%7B%5Cfrac%7B16%7D%7B%5Cpi%7D%7D%20%5C%5C%20%5Cfrac%7Bd%5E2V%7D%7Bdx%5E2%7D%3D-6%5Cpi%5Csqrt%5B%5D%7B%5Cfrac%7B16%7D%7B%5Cpi%7D%7D%3C0%20%5C%5C%20%20%5Cend%7Bgathered%7D)
Since, the double derivative is negative.
![So,\text{ the volume is maximum at }\sqrt[]{\frac{16}{\pi}}](https://tex.z-dn.net/?f=So%2C%5Ctext%7B%20the%20volume%20is%20maximum%20at%20%7D%5Csqrt%5B%5D%7B%5Cfrac%7B16%7D%7B%5Cpi%7D%7D)
So, the volume becomes,
![\begin{gathered} V=\pi(x^2)h \\ V=\pi(\sqrt[]{\frac{16}{\pi}})^2h \\ V=\frac{16h}{\pi} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20V%3D%5Cpi%28x%5E2%29h%20%5C%5C%20V%3D%5Cpi%28%5Csqrt%5B%5D%7B%5Cfrac%7B16%7D%7B%5Cpi%7D%7D%29%5E2h%20%5C%5C%20V%3D%5Cfrac%7B16h%7D%7B%5Cpi%7D%20%5Cend%7Bgathered%7D)
Answer: maximum volume of the cylinder is,
