What is the question? Pythagorean is not too hard. a^2 + b^2 = c^2
Here, DH = HF
x+3 = 3y
x = 3y-3 ----(I)
GH = HE
4x-5 = 2y+3
4x = 2y+8
Substitute value of x,
4(3y-3) = 2y+8
12y-12 = 2y+8
12y-2y = 12+8
10y = 20
y = 2
Now, substitute it in equation 1,
x = 3(2)-3
x = 6-3 = 3
So, your final answer is x=3 & y=2
<span>The answer is true
Let's imagine that we have the following function function:
</span>

<span>We have to:
Independent variable: x
Dependent variable: y
For x = -1:
</span>

<span> For x = 1:
</span>

<span> We observe that the independent variable can only obtain one result.
Answer:
True</span>
Answer:
587
Step-by-step explanation: