The area of the circle by definition is given by:
For f (r) = 154:
Clearing r we have:
For f (r) = 616:
Clearing r we have:
Answer:
a reasonable domain for f (r) is:
7 <r <14
(2300 / 4500) * 100 = 51.1%
Hope this helps! :)
Sure with what? u should have asked ur question
Answer:
Step-by-step explanation:
<u>Covert equations in standard form into slope-intercept form of y = mx + b</u>
<h3>#1</h3>
<h3>#2</h3>
- 5x + 3y = 6 ⇒ 3y = -5x + 6 ⇒ y = -5/3x + 2
<h3>#3</h3>
- x - 2y = 2 ⇒ 2y = x - 2 ⇒ y = 1/2x - 1
<h3>#4</h3>
- 3x + 2y = -8 ⇒ 2y = -3x - 8 ⇒ y = -3/2x - 4
![\bf 2sin(x)cos(x)=sin(x)\sqrt{2}\implies 2sin(x)cos(x)-sin(x)\sqrt{2}=0 \\\\\\ sin(x)~[2cos(x)-\sqrt{2}]=0\\\\ -------------------------------\\\\ sin(x)=0\implies \measuredangle x=0~~,~~\pi \\\\ -------------------------------\\\\ 2cos(x)-\sqrt{2}=0\implies 2cos(x)=\sqrt{2}\implies cos(x)=\cfrac{\sqrt{2}}{2} \\\\\\ \measuredangle x=\frac{\pi }{4}~~,~~\frac{7\pi }{4}](https://tex.z-dn.net/?f=%5Cbf%202sin%28x%29cos%28x%29%3Dsin%28x%29%5Csqrt%7B2%7D%5Cimplies%202sin%28x%29cos%28x%29-sin%28x%29%5Csqrt%7B2%7D%3D0%0A%5C%5C%5C%5C%5C%5C%0Asin%28x%29~%5B2cos%28x%29-%5Csqrt%7B2%7D%5D%3D0%5C%5C%5C%5C%0A-------------------------------%5C%5C%5C%5C%0Asin%28x%29%3D0%5Cimplies%20%5Cmeasuredangle%20x%3D0~~%2C~~%5Cpi%20%5C%5C%5C%5C%0A-------------------------------%5C%5C%5C%5C%0A2cos%28x%29-%5Csqrt%7B2%7D%3D0%5Cimplies%202cos%28x%29%3D%5Csqrt%7B2%7D%5Cimplies%20cos%28x%29%3D%5Ccfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D%0A%5C%5C%5C%5C%5C%5C%0A%5Cmeasuredangle%20x%3D%5Cfrac%7B%5Cpi%20%7D%7B4%7D~~%2C~~%5Cfrac%7B7%5Cpi%20%7D%7B4%7D)
now, we're not including the III and II quadrants, where the cosine has an angle of the same value, but is negative, because the exercise seems to be excluding the negative values of √(2).
