1) if ∠J=∠E and ∠G=∠A ⇒ ΔCJG ~ Δ<span>CEA
2) </span>if ∠J=∠E and JC:JC=JG:EA ⇒ ΔCJG ~ Δ<span>CEA
3) </span>if JC:JC=JG:EA= CG:CA ⇒ ΔCJG ~ Δ<span>CEA</span>
This is a probability problem with two dependent events and conditional probability. Note that after the first donut is chosen, it is not replaced into the data set, so only 23 donuts remain. If we set A=selection of a lemon-filled, and B=selection of a custard-filled, then P(A and B) = P(A)*P(B|A), where P(B|A) means the probability of B happening given that A has already occurred.P(A) = 8/24 = 1/3 = 0.333333P(B|A) = 12/23 = 0.521739P(A and B) = 1/3(12/23) = 12/69 = 0.1739130435 or 17.4%
https://www.wyzant.com/resources/answers/296921/find_the_probability_of_selecting_a_a_lemon_filled_d...
Given a triangle (3 sides), a hexagon (6 sides) and a nonagon (9 sides).
The sides increases by 3 and the equation that represents the pattern is 3n.
Therefore, the <span>expression that represents the number of sides in the polygon for the nth member of the pattern is 3n.</span>
Answer:
F. 82
Step-by-step explanation:
To calculate the length of BC, Pythagoras theorem will be used.
It says, Hypotenuse^2 = Perpendicular^2 + Base^2
In our case, Perpendicular = AB = 18 and Base = AC = 80
therefore,
Hypotenuse^2 = 18^2 + 80^2 = 6724
Hypotenuse = BC =
= 82
Given the table showing the distance Randy drove on one day of her vacation as follows:
![\begin{tabular} {|c|c|c|c|c|c|} Time (h)&1&2&3&4&5\\[1ex] Distance (mi)&55&110&165&220&275 \end{tabular}](https://tex.z-dn.net/?f=%5Cbegin%7Btabular%7D%0A%7B%7Cc%7Cc%7Cc%7Cc%7Cc%7Cc%7C%7D%0ATime%20%28h%29%261%262%263%264%265%5C%5C%5B1ex%5D%0ADistance%20%28mi%29%2655%26110%26165%26220%26275%0A%5Cend%7Btabular%7D)
The rate at which she travels is given by

If Randy has driven for one more hour at the same rate, the number of hours she must have droven is 6 hrs and the total distance is given by
distance = 55 x 6 = 330 miles.