Hey there,
100%=50
x%=20
Problem: 100%/x%=50/20
(100/x)*x=(50/20)*x
100=2.5*x
100/2.5=x
40=x
x=40
Answer: 20 is 40% of 50.
They told me to answer a bunch of questions best of luck tho
Let
be the coordinates of point P.
The distance from the x-axis is exactly the y coordinates of point P.
The distance from the point (3,-3) is given by
![\sqrt{(x-3)^2+(x+3)^2}](https://tex.z-dn.net/?f=%5Csqrt%7B%28x-3%29%5E2%2B%28x%2B3%29%5E2%7D)
So, we want
![y = 3\sqrt{(x-3)^2+(y+3)^2}](https://tex.z-dn.net/?f=y%20%3D%203%5Csqrt%7B%28x-3%29%5E2%2B%28y%2B3%29%5E2%7D)
We can square both sides to get
![y^2=9((x-3)^2+(y+3)^2)](https://tex.z-dn.net/?f=y%5E2%3D9%28%28x-3%29%5E2%2B%28y%2B3%29%5E2%29)
Expanding the squares, we have
![y^2=9(x^2-6x+9+y^2+6y+9) = 9(x^2+y^2-6x+6y+18)=9x^2+9y^2-54x+54y+162](https://tex.z-dn.net/?f=y%5E2%3D9%28x%5E2-6x%2B9%2By%5E2%2B6y%2B9%29%20%3D%209%28x%5E2%2By%5E2-6x%2B6y%2B18%29%3D9x%5E2%2B9y%5E2-54x%2B54y%2B162)
So, the final equation is
![y^2=9x^2+9y^2-54x+54y+162 \iff 9x^2+8y^2-54x+54y+162=0](https://tex.z-dn.net/?f=y%5E2%3D9x%5E2%2B9y%5E2-54x%2B54y%2B162%20%5Ciff%209x%5E2%2B8y%5E2-54x%2B54y%2B162%3D0)
Answer:
x = 9
y = 10.8
okay..
Step-by-step explanation:
see further explanation in attachment.
Step-by-step explanation:
7x-40=3x+3x-20 {exterior angle of triangle is equal to the sum of its opposite interior angle}
hope it helps
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