Answer:
I believe D is the answer. (Triangles BPA and DPC are congruent.)
Angles BAD and ADC are not congruent.
Sides CD and DA are also not congruent
Answer:
- 3.2
Step-by-step explanation:
- 5x + 4 = 20
- 5x = 20 - 4
- 5x = 16
x = 16 / - 5
x = - 3.2
Answer:
II
Step-by-step explanation:
All the other options create a number that is smaller than 7, while the inequality suggests otherwise.
While, II create the number 9 - which is indeed bigger then 7
Glad to help ÷)
Since you're working with the ASA postulate, you're looking to show congruence of the angles at either end of a side. You're given side AC and angle A as congruent with their counterparts. Obviously, you also need to show congruence of angle C with its counterpart, angle Z.
selection B is appropriate
Simplify the following:
((x^2 - 11 x + 30) (x^2 + 6 x + 5))/((x^2 - 25) (x - 5 x - 6))
The factors of 5 that sum to 6 are 5 and 1. So, x^2 + 6 x + 5 = (x + 5) (x + 1):
((x + 5) (x + 1) (x^2 - 11 x + 30))/((x^2 - 25) (x - 5 x - 6))
The factors of 30 that sum to -11 are -5 and -6. So, x^2 - 11 x + 30 = (x - 5) (x - 6):
((x - 5) (x - 6) (x + 5) (x + 1))/((x^2 - 25) (x - 5 x - 6))
x - 5 x = -4 x:
((x - 5) (x - 6) (x + 5) (x + 1))/((x^2 - 25) (-4 x - 6))
Factor -2 out of -4 x - 6:
((x - 5) (x - 6) (x + 5) (x + 1))/(-2 (2 x + 3) (x^2 - 25))
x^2 - 25 = x^2 - 5^2:
((x - 5) (x - 6) (x + 5) (x + 1))/(-2 (x^2 - 5^2) (2 x + 3))
Factor the difference of two squares. x^2 - 5^2 = (x - 5) (x + 5):
((x - 5) (x - 6) (x + 5) (x + 1))/(-2(x - 5) (x + 5) (2 x + 3))
((x - 5) (x - 6) (x + 5) (x + 1))/((x - 5) (x + 5) (-2) (2 x + 3)) = ((x - 5) (x + 5))/((x - 5) (x + 5))×((x - 6) (x + 1))/(-2 (2 x + 3)) = ((x - 6) (x + 1))/(-2 (2 x + 3)):
((x - 6) (x + 1))/(-2 (2 x + 3))
Multiply numerator and denominator of ((x - 6) (x + 1))/(-2 (2 x + 3)) by -1:
Answer: (-(x - 6) (x + 1))/(2 (2 x + 3))
