A diagram would be nice. The quick answer is C.
The first answer makes AE 8 which is twice as long as AC. I don't think the intent of the question is to use an external point.
The second one is also incorrect. This time E is in a place such that it makes AC and AE both equal and that too isn't possible.
The third one is the answer because D and E are both 1/2 the distance between the end point of the line containing them.
D is wrong because E is the midpoint of AC but D is not the midpoint of AB.
Answer:
First choice
Step-by-step explanation:
The discontinuity is removable if by reducing the fraction that discontinuity doesn't continue to exist as a discontinuity.
Example of, (x-1)/(x-1) has a a discontinuity at x=1 and it's removable because the fraction reduces to 1 which doesn't have a discontinuity at x=1.
Example not of, (x-1)/(x-2) has a discontinuity at x=2 and it is not removable because we can't get rid of the x-2 factor in the denominator.
The first choice has a discontinuity at x=-1 and it is removable because x^2-x-2=(x-2)(x+1) and the x+1's will cancel on top and bottom making the point at x=-1 a removable discontinuity.
Answer:
i'm guessing b or a sorry if i'm wrong TwT
Step-by-step explanation:
sorry
Answer:
x = 3
y = 15
Step-by-step explanation:
If △XPS ≅△DNF, their corresponding sides would be congruent. This implies that:
XP ≅ DN
PS ≅ NF
XS ≅ DF
Given that:
XP = 4y - 3
DN = 57
NF = 51
XS = 17x + 3
DF = 54
Therefore:
XP = DN
4y - 3 = 57 (Substitution)
Add 3 to both sides
4y = 57 + 3
4y = 60
Divide both sides by 4
y = 60/4
y = 15
Also,
XS = DF
17x + 3 = 54 (substitution)
Subtract 3 from each side
17x = 54 - 3
17x = 51
Divide both sides by 17
x = 51/17
x = 3
The answer is 12.6 that’s the answer right there
