Answer:
idk
Step-by-step explanation:
Answer:
![\mathbf{\dfrac{\pi}{6}[5 \sqrt{5}-1]}](https://tex.z-dn.net/?f=%5Cmathbf%7B%5Cdfrac%7B%5Cpi%7D%7B6%7D%5B5%20%5Csqrt%7B5%7D-1%5D%7D)
Step-by-step explanation:
Given that:
The surface area (S.A) 
Hence the S.A is of form z = f(x,y)
Then the S.A can be represented with the equation

where :
D = cylinder 
In polar co-ordinates:
D = {(r, θ): 0≤ r ≤ 1, 0 ≤ θ ≤ 2π)
Similarly,
and 
Therefore;



![= [\theta]^{2 \pi}_{0} \dfrac{1}{8}\times \dfrac{2}{3}\begin {bmatrix} (1+4r^2)^{\dfrac{3}{2}}\end {bmatrix}^1_0](https://tex.z-dn.net/?f=%3D%20%5B%5Ctheta%5D%5E%7B2%20%5Cpi%7D_%7B0%7D%20%5Cdfrac%7B1%7D%7B8%7D%5Ctimes%20%5Cdfrac%7B2%7D%7B3%7D%5Cbegin%20%7Bbmatrix%7D%20%281%2B4r%5E2%29%5E%7B%5Cdfrac%7B3%7D%7B2%7D%7D%5Cend%20%7Bbmatrix%7D%5E1_0)
![= 2 \pi \times \dfrac{1}{12}[5^{\dfrac{3}{2}} - 1]](https://tex.z-dn.net/?f=%3D%202%20%5Cpi%20%5Ctimes%20%5Cdfrac%7B1%7D%7B12%7D%5B5%5E%7B%5Cdfrac%7B3%7D%7B2%7D%7D%20-%201%5D)
![\mathbf{=\dfrac{\pi}{6}[5 \sqrt{5}-1]}](https://tex.z-dn.net/?f=%5Cmathbf%7B%3D%5Cdfrac%7B%5Cpi%7D%7B6%7D%5B5%20%5Csqrt%7B5%7D-1%5D%7D)
Answer:
14 km
Step-by-step explanation:
<u>1) look at the information you are given:</u>
140 km total (<em>to and from work</em>)
Note that this number includes both the distance to work and the distance from work.
He works 5 days a week.
<u>2)Figure out the distance he travels within one day</u>
This means that you should take the 140 km and divide it by 5 in order to find out the amount of km he travels each day.
<h2>
140 km / 5 days = 28 km</h2>
28 km is the amount of KM he travels each DAY.
<u>3) Figure out the distance from work to home </u>
From this point, it is important to note the question is asking for <u>the distance his work is from home.</u>
Each day, he must travel 2 times: he must go to work and go from work to home, which are both the same distance.
This means that in order to get only the distance from work to home, you must divide by 2.
<h2>
28 km / 2 = 14 km</h2>
Answer:
8.60
Step-by-step explanation:
A^2+B^2=C^2
5*5=25 7*7=49 49+25= 74
√ 74= 8.602...
1 ) the formula used to calculate the simple interest is
I = Ci*n, f his formula you obtain n , that is the time in years n = I/ Ci, where c is the initial capital, i is the earned interest , and i is he rate of interest,
n = 120/600 (0,025) = 120 / 15 = 8 , at 8 years, you will obtain $ 120 of INTEREST