Yes, this is a right triangle. B'(-2,0)
1) Given those coordinates let's plot that triangle:
Let's find the legs:
![\begin{gathered} d_{AC}=\sqrt[]{(8-2)^2+(-5-3)^2}=10 \\ d_{BC}=\sqrt[]{(8-6)^2+(-5-6)^2}=5\sqrt[]{5} \\ d_{AB}=\sqrt[]{(6-2)^2+(6-3)^2}=5 \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20d_%7BAC%7D%3D%5Csqrt%5B%5D%7B%288-2%29%5E2%2B%28-5-3%29%5E2%7D%3D10%20%5C%5C%20d_%7BBC%7D%3D%5Csqrt%5B%5D%7B%288-6%29%5E2%2B%28-5-6%29%5E2%7D%3D5%5Csqrt%5B%5D%7B5%7D%20%5C%5C%20d_%7BAB%7D%3D%5Csqrt%5B%5D%7B%286-2%29%5E2%2B%286-3%29%5E2%7D%3D5%20%5Cend%7Bgathered%7D)
Let's test whether this is or not a Right Triangle, by using the Pythagorean Theorem:
![\begin{gathered} (5\sqrt[]{5})^2=10^2+5^2 \\ 125=100+25 \\ 125=125\text{ TRUE} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20%285%5Csqrt%5B%5D%7B5%7D%29%5E2%3D10%5E2%2B5%5E2%20%5C%5C%20125%3D100%2B25%20%5C%5C%20125%3D125%5Ctext%7B%20TRUE%7D%20%5Cend%7Bgathered%7D)
Hence, this is a right triangle.
b) Reflecting that triangle across leg AC, we have:
Counting to the left of Vertix A, 2 units to the left we have the Vertix B' and since its a reflection across AC, A = A' and C=C'
Therefore
B'(-2,0)
3.5
Keep flip change
7/8 x 4/1 = 3 1/2
Convert to a decimal
3 1/2 = 3.5
Answer:
x=28
y= -4
Step-by-step explanation:
9x+6y= -12
-x-4y= -12
multiply -x-4y=-12 by 9 so the x's are equal
9x+6y= -12
-9x-36y= -108
add the equations so the x's cancel out
-30y=120
y= -4
plug y into an equation to find x
-x-4(-4)=-12
-x+16= -12
-x= -28
x=28
6k + 2 = 8
Isolate the 8. Do the opposite of PEMDAS.
Note the equal sign. What you do to one side you do to the other.
Subtract 2 from both sides
6k + 2 (-2) = 8 (-2)
6k = 8 - 2
6k = 6
Isolate the k, divide6 from both sides.
6k/6 = 6/6
k = 6/6
k = 1
1 is your answer.
Hope this helps
Answer:
no
Step-by-step explanation:
