step 1
Subtract 14 from 50
50-14=36
so
![p+14\leq50](https://tex.z-dn.net/?f=p%2B14%5Cleq50)
If you solve for p
![p\leq36](https://tex.z-dn.net/?f=p%5Cleq36)
the answer is the second option
Answer:
a) ![\left(x,y\right)=\left(4.95,-4.95\right)](https://tex.z-dn.net/?f=%5Cleft%28x%2Cy%5Cright%29%3D%5Cleft%284.95%2C-4.95%5Cright%29)
b) ![r\angle\theta = 7\angle0.5236\,\text{radians}](https://tex.z-dn.net/?f=r%5Cangle%5Ctheta%20%3D%207%5Cangle0.5236%5C%2C%5Ctext%7Bradians%7D)
Step-by-step explanation:
Polar coordinates are represented as:
, where 'r' is the length (or magnitude) of the line, and '
' is the angle measured from the positive x-axis.
in our case:
![7\angle\dfrac{3\pi}{4}](https://tex.z-dn.net/?f=7%5Cangle%5Cdfrac%7B3%5Cpi%7D%7B4%7D)
to covert the polar to cartesian:
![x = r\cos{\theta}](https://tex.z-dn.net/?f=x%20%3D%20r%5Ccos%7B%5Ctheta%7D)
![y = r\sin{\theta}](https://tex.z-dn.net/?f=y%20%3D%20r%5Csin%7B%5Ctheta%7D)
we can plug in our values:
![x = 7\cos{\dfrac{3\pi}{4}} = -7\dfrac{\sqrt{2}}{2}](https://tex.z-dn.net/?f=x%20%3D%207%5Ccos%7B%5Cdfrac%7B3%5Cpi%7D%7B4%7D%7D%20%3D%20-7%5Cdfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D)
![y = 7\sin{\dfrac{3\pi}{4}} = 7\dfrac{\sqrt{2}}{2}](https://tex.z-dn.net/?f=y%20%3D%207%5Csin%7B%5Cdfrac%7B3%5Cpi%7D%7B4%7D%7D%20%3D%207%5Cdfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D)
the Cartesian coordinates are:
![\left(x,y\right)=\left(-7\dfrac{\sqrt{2}}{2},7\dfrac{\sqrt{2}}{2}\right)](https://tex.z-dn.net/?f=%5Cleft%28x%2Cy%5Cright%29%3D%5Cleft%28-7%5Cdfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D%2C7%5Cdfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D%5Cright%29)
![\left(x,y\right)=\left(4.95,-4.95\right)](https://tex.z-dn.net/?f=%5Cleft%28x%2Cy%5Cright%29%3D%5Cleft%284.95%2C-4.95%5Cright%29)
(b) to convert (x,y) = (6.06,-3.5)
we'll use the pythagoras theorem to find 'r'
![r^2 = x^2+y^2](https://tex.z-dn.net/?f=r%5E2%20%3D%20x%5E2%2By%5E2)
![r^2 = (6.06)^2+(-3.5)^2](https://tex.z-dn.net/?f=r%5E2%20%3D%20%286.06%29%5E2%2B%28-3.5%29%5E2)
![r = \sqrt{48.97} \approx 7](https://tex.z-dn.net/?f=r%20%3D%20%5Csqrt%7B48.97%7D%20%5Capprox%207)
the angle can be found by:
![\tan{\theta} = \dfrac{y}{x}](https://tex.z-dn.net/?f=%5Ctan%7B%5Ctheta%7D%20%3D%20%5Cdfrac%7By%7D%7Bx%7D)
![\tan{\theta} = \dfrac{3.5}{6.06}](https://tex.z-dn.net/?f=%5Ctan%7B%5Ctheta%7D%20%3D%20%5Cdfrac%7B3.5%7D%7B6.06%7D)
![\theta = \arctan{left(\dfrac{3.5}{6.06}\right)}](https://tex.z-dn.net/?f=%5Ctheta%20%3D%20%5Carctan%7Bleft%28%5Cdfrac%7B3.5%7D%7B6.06%7D%5Cright%29%7D)
![\theta = 0.5236 \text{radians}](https://tex.z-dn.net/?f=%5Ctheta%20%3D%200.5236%20%5Ctext%7Bradians%7D)
to convert radians to degrees:
![\theta = 0.5236 \times \dfrac{180}{\pi} \approx 30^\circ](https://tex.z-dn.net/?f=%5Ctheta%20%3D%200.5236%20%5Ctimes%20%5Cdfrac%7B180%7D%7B%5Cpi%7D%20%5Capprox%2030%5E%5Ccirc)
writing in polar coordinates:
![r\angle\theta = 7\angle30^\circ\,\,\text{OR}\,\,7\angle0.5236\,\text{radians}](https://tex.z-dn.net/?f=r%5Cangle%5Ctheta%20%3D%207%5Cangle30%5E%5Ccirc%5C%2C%5C%2C%5Ctext%7BOR%7D%5C%2C%5C%2C7%5Cangle0.5236%5C%2C%5Ctext%7Bradians%7D)
Answer:
D
Step-by-step explanation:
Answer:
Weight of wastage=179.775kg
weight of rice cultivated= 585.225 kg
percentage of rice cultivated=76.5%
Step-by-step explanation:
Area of land=100 square meters
Cultivated rice=765kg
Wastage=23.5%
1) Weight of the wastage=23.5% of 765kg
=23.5/100 × 765
=17977.5 / 100
=179.775 kg
2) Weight and percentage of rice cultivated.
weight of rice cultivated = 765 kg - 179. 775 kg
= 585.225 kg
percentage of rice cultivated = 100 - 23.5
= 76.5%
3) if area is increased 40 times in size
New area=1000 square meters × 40
=40,000 square meters
Cultivated rice= 765kg × 40
=30,600 kg
Cultivated rice excluding wastage=585.225 kg × 40
=23,409 kg
You could do -1/-2 times -5/-4 which would be 5/8