So, we know the center is at -1, -3, hmmm what's the radius anyway?
well, the radius will be the distance from the center to any point on the circle, it just so happen that we know -7, -5 is on it, thus
![\bf ~~~~~~~~~~~~\textit{distance between 2 points}\\\\ \begin{array}{ccccccccc} &&x_1&&y_1&&x_2&&y_2\\ % (a,b) &&(~ -1 &,& -3~) % (c,d) &&(~ -7 &,& -5~) \end{array} \\\\\\ d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ r=\sqrt{[-7-(-1)]^2+[-5-(-3)]^2}\implies r=\sqrt{(-7+1)^2+(-5+3)^2} \\\\\\ r=\sqrt{36+4}\implies r=\sqrt{40}\\\\ -------------------------------](https://tex.z-dn.net/?f=%5Cbf%20~~~~~~~~~~~~%5Ctextit%7Bdistance%20between%202%20points%7D%5C%5C%5C%5C%0A%5Cbegin%7Barray%7D%7Bccccccccc%7D%0A%26%26x_1%26%26y_1%26%26x_2%26%26y_2%5C%5C%0A%25%20%20%28a%2Cb%29%0A%26%26%28~%20-1%20%26%2C%26%20-3~%29%20%0A%25%20%20%28c%2Cd%29%0A%26%26%28~%20-7%20%26%2C%26%20-5~%29%0A%5Cend%7Barray%7D%0A%5C%5C%5C%5C%5C%5C%0Ad%20%3D%20%5Csqrt%7B%28%20x_2-%20x_1%29%5E2%20%2B%20%28%20y_2-%20y_1%29%5E2%7D%0A%5C%5C%5C%5C%5C%5C%0Ar%3D%5Csqrt%7B%5B-7-%28-1%29%5D%5E2%2B%5B-5-%28-3%29%5D%5E2%7D%5Cimplies%20r%3D%5Csqrt%7B%28-7%2B1%29%5E2%2B%28-5%2B3%29%5E2%7D%0A%5C%5C%5C%5C%5C%5C%0Ar%3D%5Csqrt%7B36%2B4%7D%5Cimplies%20r%3D%5Csqrt%7B40%7D%5C%5C%5C%5C%0A-------------------------------)
![\bf \textit{equation of a circle}\\\\ (x- h)^2+(y- k)^2= r^2 \qquad center~~(\stackrel{-1}{ h},\stackrel{-3}{ k})\qquad \qquad radius=\stackrel{\sqrt{40}}{ r} \\\\\\\ [x-(-1)]^2+[y-(-3)]^2=(\sqrt{40})^2\implies (x+1)^2+(y+3)^2=40](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Bequation%20of%20a%20circle%7D%5C%5C%5C%5C%20%0A%28x-%20h%29%5E2%2B%28y-%20k%29%5E2%3D%20r%5E2%0A%5Cqquad%20%0Acenter~~%28%5Cstackrel%7B-1%7D%7B%20h%7D%2C%5Cstackrel%7B-3%7D%7B%20k%7D%29%5Cqquad%20%5Cqquad%20%0Aradius%3D%5Cstackrel%7B%5Csqrt%7B40%7D%7D%7B%20r%7D%0A%5C%5C%5C%5C%5C%5C%5C%0A%5Bx-%28-1%29%5D%5E2%2B%5By-%28-3%29%5D%5E2%3D%28%5Csqrt%7B40%7D%29%5E2%5Cimplies%20%28x%2B1%29%5E2%2B%28y%2B3%29%5E2%3D40)
Y-2x=8
16+4x=2y
Rearrange the first equation equal to y
Y=2x+8
Sub the rearranged equation into the second equation and solve for x
16+4x=2(2x+8)
16+4x=4x+16
16-16=4x-4x
0=0
There would be no solution
Hope this helps!
Pemdas 1. parentheses: everything is already in parentheses. so next is exponents. 10×-2 is -20. and 10×6 is 60. so 4×-20 is the next step. it equals -80. and then 3×60 is 180. lastly,add -80 and 180 together. the answer should be 100 if i did that correctly.
Using proportions and the information given, it is found that:
- The class width is of 14.375.
- The lower class limits are: {19, 33.375, 47.750, 62.125, 76.500, 90.875, 105.250, 119.625}.
- The upper class limits are: {33.375, 47.750, 62.125, 76.500, 90.875, 105.250, 119.625, 134}.
-------------------------
- Minimum value is 19.
- Maximum value is of 134.
- There are 8 classes.
- The classes are all of equal width, thus the width is of:
-------------------------
The intervals will be of:
19 - 33.375
33.375 - 47.750
47.750 - 62.125
62.125 - 76.500
76.500 - 90.875
90.875 - 105.250
105.250 - 119.625
119.625 - 134.
- The lower class limits are: {19, 33.375, 47.750, 62.125, 76.500, 90.875, 105.250, 119.625}.
- The upper class limits are: {33.375, 47.750, 62.125, 76.500, 90.875, 105.250, 119.625, 134}.
A similar problem is given at brainly.com/question/16631975
