Answer:
ΔABC≅ΔDEC by AAS
Step-by-step explanation:
You can use the AAS method of congruency.
Since you already have <BAC and <EDC congruent to eachother, and sides BC and EC congruent to each other, you only need that one remaining angle in between. <ACB can be proven congruent to <DCE by the Vertical Angles Theorem, and that gives you the AAS you need to prove that these two triangles are congruent
Hope this helped.
Y - y1 = m (x + x1)
Solve for m by subtracting the y's and dividing them by the difference of the 2 x's.
-32 - 1 = -33
-8 - 3 = -11
Divide the two to get 3.
Use the first point (as instructed) and plug it into the equation.
y - (-32) = 3 (x - (-8))
y + [32] = [3] (x + [5])
The brackets are the fill in the blanks.
No, she cannot use a ratio of 21 to 6. You can figure this out in a few different ways. First you can simplify down the fraction 18/3, to get 6/1. 6/1 does not equal 21/6, because 21/6 equals 7/2. However, 6/1 it does equal 36/6. You could also noticed that three doubles to make six, yet 18 does not double to make 21
Answer:
It doesn't form a right triangle
Step-by-step explanation:
According to the Pythagorean theorem, the area of the square whose side is the hypotenuse is equal to the sum of the areas of the squares on the other two sides.
a^2 + b^2 = c^2 by substituting the three values, 10, 15, and 20 you should get an equivalent value of the hypotenuse.
10^2 + 15^2 = 20^2
100 + 225 = 400
325 =/= 400
This is false.
Since this is not equivalent, it is NOT a right triangle.
Hope this helped :)
