First question:
interest rate is 5% annually. After 4 years relative value of interest will be 1.05^4
Now we multiply that with starting funds and get
4000* 1.05^4 = 4862.03
Second question:
Because interest is 8% semiannually that means that after 35 years relative value of interest will be 1.04^(35*2)
6000*1.04^70 = 93,429.71
Third question:
after first year tractor's value will be 0.86*15450. After 3 years its value will be:
(0.86)^3*15,450 = 9,827.07
Answer:
1 and 9/28
Step-by-step explanation:
you need to put everything over the same denominator
21/28+16/28=37/28
this can be written as 1 9/28
Answer:
<u>The answer is option C. 6a-7</u>
Step-by-step explanation:
Given that
5(3a-1)-2(3a-2)=3(a+2)+v
Solve for v
∴ v = 5(3a-1)-2(3a-2) - 3(a+2)
∴ v = 15a - 5 - 6a + 4 - 3a - 6
∴ v = 15a - 6a - 3a - 5 + 4 - 6
∴ v = 6a - 7
<u>So the answer is option C. 6a-7</u>
Answer:
(-1,4)(1,4)
2
Step-by-step explanation:
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,
