1. A (3,6) = (1,9) B(-1,2) = (-3,5) C (1,-4) = (-1,-1) 2. K (5,7) = (-1,7) L (3,7) = (1,7) M (3,5) = (1,5) N (6,4) = (-2,4) 3. Q (5,-8) = (5,8) R(5,-4) = (5,4) S (2,-4) = (2,4) T (2,-8) = (2,8) I believe those are the correct points. I hope this helps :D
To find the perimeter, you need the lengths of all 3 legs,
and then you add them up.
Can you find the lengths of the other 2 sides of that triangle ?
Here are a couple of factoids I learned in 1954 and I never forgot:
In a 30°- 60°- 90° right triangle . . .
-- the side opposite the 30° angle (the shorter leg)
is 1/2 the length of the hypotenuse,
and
-- the side opposite the 60° angle (the longer leg)
is (1/2 the length of the hypotenuse) times (√3).
If you have a 30°- 60°- 90° triangle and you know the length
of any one side, then you can find the lengths of the other
two sides with these amazing factoids.
Answer:
76°
Step-by-step explanation:
![\because \: m \angle \: 1 = m \angle \: 12 \\ (exterior \: alternate \: \angle s) \\ \\ \therefore \: (5x + 44) \degree = (8x + 8) \degree \\ \\ \therefore \: 5x + 44 = 8x + 8 \\ \\ 5x - 8x = 8 - 44 \\ \\ - 3x = - 36 \\ \\ x = \frac{ - 36}{ - 3} \\ \\ \huge \red{x = 12} \\ \\ \because \: m \angle \: 12 = (8x + 8) \degree \\ \\ \therefore \: m \angle \: 12 = = (8 \times 12 + 8) \degree \\ \\ \therefore \: m \angle \: 12 = = (96+ 8) \degree \\ \\ \therefore \: \purple{m \angle \: 12 = 104 \degree} \\ \\ \because \: m \angle \: 10 + m \angle \: 12 = 180 \degree \\(linear \: pair \: \angle s) \\ \\ \therefore \: m \angle \: 10 + 104 \degree = 180 \degree \\ \\ \therefore \: m \angle \: 10 = 180 \degree - 104 \degree \\ \\ \huge \orange{ \boxed{\therefore \: m \angle \: 10 = 76 \degree }}](https://tex.z-dn.net/?f=%20%5Cbecause%20%5C%3A%20m%20%5Cangle%20%5C%3A%201%20%3D%20m%20%5Cangle%20%5C%3A%2012%20%5C%5C%20%28exterior%20%5C%3A%20alternate%20%5C%3A%20%20%5Cangle%20s%29%20%5C%5C%20%20%5C%5C%20%20%5Ctherefore%20%5C%3A%20%285x%20%2B%2044%29%20%5Cdegree%20%3D%20%288x%20%2B%208%29%20%5Cdegree%20%5C%5C%20%20%5C%5C%20%5Ctherefore%20%5C%3A%205x%20%2B%2044%20%3D%208x%20%2B%208%20%5C%5C%20%20%5C%5C%205x%20-%208x%20%3D%208%20-%2044%20%5C%5C%20%20%5C%5C%20%20-%203x%20%3D%20%20-%2036%20%5C%5C%20%20%5C%5C%20x%20%3D%20%20%5Cfrac%7B%20-%2036%7D%7B%20-%203%7D%20%20%5C%5C%20%20%5C%5C%20%20%5Chuge%20%5Cred%7Bx%20%3D%2012%7D%20%5C%5C%20%20%5C%5C%20%20%20%5Cbecause%20%5C%3A%20m%20%5Cangle%20%5C%3A%2012%20%20%3D%20%288x%20%2B%208%29%20%5Cdegree%20%5C%5C%20%20%5C%5C%20%20%5Ctherefore%20%5C%3A%20m%20%5Cangle%20%5C%3A%2012%20%20%3D%20%3D%20%288%20%5Ctimes%2012%20%2B%208%29%20%5Cdegree%20%20%5C%5C%20%20%5C%5C%20%5Ctherefore%20%5C%3A%20m%20%5Cangle%20%5C%3A%2012%20%20%3D%20%3D%20%2896%2B%208%29%20%5Cdegree%20%20%5C%5C%20%20%5C%5C%20%5Ctherefore%20%5C%3A%20%20%5Cpurple%7Bm%20%5Cangle%20%5C%3A%2012%20%20%3D%20104%20%5Cdegree%7D%20%20%5C%5C%20%20%5C%5C%20%20%5Cbecause%20%5C%3A%20m%20%5Cangle%20%5C%3A%2010%20%2B%20m%20%5Cangle%20%5C%3A%2012%20%20%3D%20180%20%5Cdegree%20%5C%5C%28linear%20%5C%3A%20pair%20%5C%3A%20%20%5Cangle%20s%29%20%20%5C%5C%20%20%5C%5C%20%20%5Ctherefore%20%5C%3A%20m%20%5Cangle%20%5C%3A%2010%20%2B%20104%20%5Cdegree%20%3D%20180%20%5Cdegree%20%5C%5C%20%20%5C%5C%20%5Ctherefore%20%5C%3A%20m%20%5Cangle%20%5C%3A%2010%20%3D%20180%20%5Cdegree%20%20-%20104%20%5Cdegree%20%5C%5C%20%20%5C%5C%20%20%5Chuge%20%5Corange%7B%20%5Cboxed%7B%5Ctherefore%20%5C%3A%20m%20%5Cangle%20%5C%3A%2010%20%3D%2076%20%5Cdegree%20%7D%7D)
It has to be 438 because any number that is 5 or higher must be rounded up so 438 could be rounded up to 440
Answer:
The circle on the far left is the sum of 4x + 3y and 2x - y so the answer is 4x + 3y + 2x - y = 6x + 2y. The circle on the far left is the sum of x + 4y and something. To find that "something" we can do 4x + 5y - (x + 4y) = 3x + y which is the value of the bottom right rectangle. This means that the value of the bottom circle is 2x - y + 3x + y = 5x.