Answer:
1330.808 euros
Step-by-step explanation:
Let's first calculate the total amount of kilograms,
for the first 60, we have to:
50 kg
4 hg = 0.5 kg
60 dag = 0.6
5 g = 0.005 kg
adding: 50 + 0.5 +0.6 + 0.005 = 51.105 kg per tree, but they are 60
51.105 * 60 = 3066.3
now for the other 20:
53 kg
2 hg = 0.2 kg
16 dag = 0.16
5 g = 0.005
adding: 53 + 0.2 +0.16 + 0.005 = 53.365 kg per tree, but they are 20
53,365 * 20 = 1067.3
in total it would be: 3066.3 + 1067.3 = 4133.6
they tell us that 1/5 is the quantity in liters:
4133.6 * 1/5 = 826.72 liters
so to calculate the earnings it would be:
2.65 * 826.72 - 860 = 1330.808
which means that in net earnings are 1330.808 euross
Answer:
452km
Step-by-step explanation:
Answer:

or

Step-by-step explanation:
Given


![[a,b] = [0,2]](https://tex.z-dn.net/?f=%5Ba%2Cb%5D%20%3D%20%5B0%2C2%5D)
Required
The volume of the solid formed
Rotating about the x-axis.
Using the washer method to calculate the volume, we have:

Integrate


Substitute values for a, b, f(x) and g(x)

Evaluate the exponents

Simplify like terms

Factor out 8

Integrate
![v = 8\pi [ \frac{2x^{2+1}}{2+1} - \frac{x^{3+1}}{3+1} ]|\limit^2_0](https://tex.z-dn.net/?f=v%20%3D%208%5Cpi%20%5B%20%5Cfrac%7B2x%5E%7B2%2B1%7D%7D%7B2%2B1%7D%20-%20%5Cfrac%7Bx%5E%7B3%2B1%7D%7D%7B3%2B1%7D%20%5D%7C%5Climit%5E2_0)
![v = 8\pi [ \frac{2x^{3}}{3} - \frac{x^{4}}{4} ]|\limit^2_0](https://tex.z-dn.net/?f=v%20%3D%208%5Cpi%20%5B%20%5Cfrac%7B2x%5E%7B3%7D%7D%7B3%7D%20-%20%5Cfrac%7Bx%5E%7B4%7D%7D%7B4%7D%20%5D%7C%5Climit%5E2_0)
Substitute 2 and 0 for x, respectively
![v = 8\pi ([ \frac{2*2^{3}}{3} - \frac{2^{4}}{4} ] - [ \frac{2*0^{3}}{3} - \frac{0^{4}}{4} ])](https://tex.z-dn.net/?f=v%20%3D%208%5Cpi%20%28%5B%20%5Cfrac%7B2%2A2%5E%7B3%7D%7D%7B3%7D%20-%20%5Cfrac%7B2%5E%7B4%7D%7D%7B4%7D%20%5D%20-%20%5B%20%5Cfrac%7B2%2A0%5E%7B3%7D%7D%7B3%7D%20-%20%5Cfrac%7B0%5E%7B4%7D%7D%7B4%7D%20%5D%29)
![v = 8\pi ([ \frac{2*2^{3}}{3} - \frac{2^{4}}{4} ] - [ 0 - 0])](https://tex.z-dn.net/?f=v%20%3D%208%5Cpi%20%28%5B%20%5Cfrac%7B2%2A2%5E%7B3%7D%7D%7B3%7D%20-%20%5Cfrac%7B2%5E%7B4%7D%7D%7B4%7D%20%5D%20-%20%5B%200%20-%200%5D%29)
![v = 8\pi [ \frac{2*2^{3}}{3} - \frac{2^{4}}{4} ]](https://tex.z-dn.net/?f=v%20%3D%208%5Cpi%20%5B%20%5Cfrac%7B2%2A2%5E%7B3%7D%7D%7B3%7D%20-%20%5Cfrac%7B2%5E%7B4%7D%7D%7B4%7D%20%5D)
![v = 8\pi [ \frac{16}{3} - \frac{16}{4} ]](https://tex.z-dn.net/?f=v%20%3D%208%5Cpi%20%5B%20%5Cfrac%7B16%7D%7B3%7D%20-%20%5Cfrac%7B16%7D%7B4%7D%20%5D)
Take LCM
![v = 8\pi [ \frac{16*4- 16 * 3}{12}]](https://tex.z-dn.net/?f=v%20%3D%208%5Cpi%20%5B%20%5Cfrac%7B16%2A4-%2016%20%2A%203%7D%7B12%7D%5D)
![v = 8\pi [ \frac{64- 48}{12}]](https://tex.z-dn.net/?f=v%20%3D%208%5Cpi%20%5B%20%5Cfrac%7B64-%2048%7D%7B12%7D%5D)

Simplify


or




When dilation is about the origin, as it is here in every case, the image point coordinates are the original (pre-image) coordinates multiplied by the scale factor.
1. Multiply every coordinate value by 5:
... W' = (-5, 10), X' = (-15, -5), Y' = (25, -5), Z' = (15, 10)
2. Multiply every coordinate value by 1/3:
... A' = (-2, 5), B' = (0, 5/3), C' = (1, 10/3)
3. A' = (2, 8), B' = (6, 2), C' = (2, 2)
4. The image coordinates are 5 times the original coordinates, so ...
... the scale factor of the dilation is 5.