since we know the endpoints of the circle, we know then that distance from one to another is really the diameter, and half of that is its radius.
we can also find the midpoint of those two endpoints and we'll be landing right on the center of the circle.
![\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ (\stackrel{x_1}{-4}~,~\stackrel{y_1}{-7})\qquad (\stackrel{x_2}{-2}~,~\stackrel{y_2}{-5})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ \stackrel{diameter}{d}=\sqrt{[-2-(-4)]^2+[-5-(-7)]^2}\implies d=\sqrt{(-2+4)^2+(-5+7)^2} \\\\\\ d=\sqrt{2^2+2^2}\implies d=\sqrt{2\cdot 2^2}\implies d=2\sqrt{2}~\hfill \stackrel{~\hfill radius}{\cfrac{2\sqrt{2}}{2}\implies\boxed{ \sqrt{2}}} \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Cbf%20~~~~~~~~~~~~%5Ctextit%7Bdistance%20between%202%20points%7D%20%5C%5C%5C%5C%20%28%5Cstackrel%7Bx_1%7D%7B-4%7D~%2C~%5Cstackrel%7By_1%7D%7B-7%7D%29%5Cqquad%20%28%5Cstackrel%7Bx_2%7D%7B-2%7D~%2C~%5Cstackrel%7By_2%7D%7B-5%7D%29%5Cqquad%20%5Cqquad%20d%20%3D%20%5Csqrt%7B%28%20x_2-%20x_1%29%5E2%20%2B%20%28%20y_2-%20y_1%29%5E2%7D%20%5C%5C%5C%5C%5C%5C%20%5Cstackrel%7Bdiameter%7D%7Bd%7D%3D%5Csqrt%7B%5B-2-%28-4%29%5D%5E2%2B%5B-5-%28-7%29%5D%5E2%7D%5Cimplies%20d%3D%5Csqrt%7B%28-2%2B4%29%5E2%2B%28-5%2B7%29%5E2%7D%20%5C%5C%5C%5C%5C%5C%20d%3D%5Csqrt%7B2%5E2%2B2%5E2%7D%5Cimplies%20d%3D%5Csqrt%7B2%5Ccdot%202%5E2%7D%5Cimplies%20d%3D2%5Csqrt%7B2%7D~%5Chfill%20%5Cstackrel%7B~%5Chfill%20radius%7D%7B%5Ccfrac%7B2%5Csqrt%7B2%7D%7D%7B2%7D%5Cimplies%5Cboxed%7B%20%5Csqrt%7B2%7D%7D%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
![\bf ~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ (\stackrel{x_1}{-4}~,~\stackrel{y_1}{-7})\qquad (\stackrel{x_2}{-2}~,~\stackrel{y_2}{-5})\qquad \qquad \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left( \cfrac{-2-4}{2}~~,~~\cfrac{-5-7}{2} \right)\implies \left( \cfrac{-6}{2}~,~\cfrac{-12}{2} \right)\implies \stackrel{center}{\boxed{(-3,-6)}} \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Cbf%20~~~~~~~~~~~~%5Ctextit%7Bmiddle%20point%20of%202%20points%20%7D%20%5C%5C%5C%5C%20%28%5Cstackrel%7Bx_1%7D%7B-4%7D~%2C~%5Cstackrel%7By_1%7D%7B-7%7D%29%5Cqquad%20%28%5Cstackrel%7Bx_2%7D%7B-2%7D~%2C~%5Cstackrel%7By_2%7D%7B-5%7D%29%5Cqquad%20%5Cqquad%20%5Cqquad%20%5Cleft%28%5Ccfrac%7B%20x_2%20%2B%20x_1%7D%7B2%7D~~~%20%2C~~~%20%5Ccfrac%7B%20y_2%20%2B%20y_1%7D%7B2%7D%20%5Cright%29%20%5C%5C%5C%5C%5C%5C%20%5Cleft%28%20%5Ccfrac%7B-2-4%7D%7B2%7D~~%2C~~%5Ccfrac%7B-5-7%7D%7B2%7D%20%5Cright%29%5Cimplies%20%5Cleft%28%20%5Ccfrac%7B-6%7D%7B2%7D~%2C~%5Ccfrac%7B-12%7D%7B2%7D%20%5Cright%29%5Cimplies%20%5Cstackrel%7Bcenter%7D%7B%5Cboxed%7B%28-3%2C-6%29%7D%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
![\bf \textit{equation of a circle}\\\\ (x- h)^2+(y- k)^2= r^2 \qquad center~~(\stackrel{-3}{ h},\stackrel{-6}{ k})\qquad \qquad radius=\stackrel{\sqrt{2}}{ r} \\[2em] [x-(-3)]^2+[y-(-6)]^2=(\sqrt{2})^2\implies (x+3)^2+(y+6)^2=2](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Bequation%20of%20a%20circle%7D%5C%5C%5C%5C%20%28x-%20h%29%5E2%2B%28y-%20k%29%5E2%3D%20r%5E2%20%5Cqquad%20center~~%28%5Cstackrel%7B-3%7D%7B%20h%7D%2C%5Cstackrel%7B-6%7D%7B%20k%7D%29%5Cqquad%20%5Cqquad%20radius%3D%5Cstackrel%7B%5Csqrt%7B2%7D%7D%7B%20r%7D%20%5C%5C%5B2em%5D%20%5Bx-%28-3%29%5D%5E2%2B%5By-%28-6%29%5D%5E2%3D%28%5Csqrt%7B2%7D%29%5E2%5Cimplies%20%28x%2B3%29%5E2%2B%28y%2B6%29%5E2%3D2)
Answer:
Step-by-step explanation:
13 = 6+7
276*13 = 276*(7+6)
= 276*7 + 276 *6 {distributive property upod addition}
= 1932 + 1656
= 3588
Answer:Find the distance between the parallel lines m and n whose equations are y = x + 4 and y = x - 6, respectively.
There are several ways to do this...here's one
Let (0, 4) be a point on the first line
Then.......a line with a negative reciprocal slope going through this point will have the equation :
y = -x + 4........so......we can find the intersection of this line with y = x - 6....set both equations equal
-x + 4 = x - 6 add x, 6 to both sides
10 = 2x divide both sides by 2
5 = x
So...using -x + 4, the y value at intersection = -1.......
So...we just need to find the distance from (0,4) to ( 5, -1) =
√[ (5)^2 + (4 + 1)^2 ] = 5√2 ≈ 7.07 units
Here's a pic....AB is the distance with A = (0,4) and B = (5, -1)
Step-by-step explanation:
Answer:
Steak sandwich = x = 4
Cheese fries = y = 1.75
Taco salad = z = 2
Step-by-step explanation:
Let :
Steak sandwich = x
Cheese fries = y
Taco salad = z
This week:
x + 2y + 2z = 11.50 - - - - (1)
Last week :
2x + 3y + z = 15.25 - - - (2)
Two weeks ago :
x + 4y + z = 13 - - - - - (3)
Taking (1) and (2)
x + 2y + 2z = 11.50 ___(1)
2x + 3y + z = 15.25 ___(2)
Multiply (1) by 2 and (2) by 1 and subtract
2x + 4y + 4z = 23
2x + 3y + z = 15.25
_______________
y + 3z = 7.75 - - - - (4)
Taking (2) and (3)
2x + 3y + z = 15.25 - - (2)
x + 4y + z = 13 - - - - - (3)
Multiply (2) by 1 and (3) by 2 and subtract
2x + 3y + z = 15.25
2x + 8y + 2z = 26
______________
-5y - z = - 10.75 - - - - - (5)
Lets solve (4) and (5)
y + 3z = 7.75 - - - - (4) - - - multiply by 5
-5y - z = - 10.75 - - - - - (5) - - - multiply by 1
Then add the result :
5y + 15z = 38.75
-5y - z = - 10.75
____________
14z = 28
z = 28 / 14
z = 2
To find y ; put z = 2 in (4)
y + 3z = 7.75
y + 3(2) = 7.75
y + 6 = 7.75
y = 7.75 - 6
y = 1.75
From equation 3 ;
x + 4y + z = 13 - - - - - (3)
x + 4(1.75) + 2 = 13
x + 7 + 2 = 13
x + 9 = 13
x = 13 - 9
x = 4
Hence,
Steak sandwich = x = 4
Cheese fries = y = 1.75
Taco salad = z = 2
