Can you prove this congruent? Using what theorem/postulate?

Mathematics

1 answer:

slavikrds [6]3 years ago

3 0

I'd say that, if the angles S and U are equal, as the SV and UV segments move towards each other, they meet at vertex V, the angles they make, angle SVT and UVT, have to be equal, because the side TV is shared by both, and has the same direction.

now, side TV is on both triangles, and is shared by both, so is the same length, so TV on one triangle is equal to TV on the other

check the picture below

so, you have one Angle, another Angle, and then a Side
A A S

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Suppose you have an experiment where you flip a coin three times. You then count the number of heads. a.)State the random variab

WITCHER [35]

Answer:

a) X=number of heads observed when flipped the coin 3 times

b) the probability distribution is

P(X=x) = 3/4 * (1/[(3-x)!*x!)])

P(X)=1/8 for x=0 and x=3 and P(X)=3/8 for x=1 and x=2

Step-by-step explanation:

the random variable will be X=number of heads observed when flipped the coin 3 times . Since the result from each flip is independent of the others , then X has a binomial probability distribution , such that

P(X=x)= n!/[(n-x)!*x!)*p^x * (1-p)^(n-x)

where

n= number of times the coin is flipped = 3

p= probability of getting heads in a flip of the coin = 1/2 (assuming that the coin is fair)

therefore

P(X=x)= 3!/[(3-x)!*x!)*(1/2)^(3-x) * (1/2)^x = 3!/[(3-x)!*x!) * (1/2)³ = 3/4 * (1/[(3-x)!*x!)])

P(X=x)= 3/4 * (1/[(3-x)!*x!)]) , for x=[0,1,2,3]

for x=0 and x=3 → P(X)=3/4 * (1/[3!*0!)]) = 1/8

for x=1 and x=2 → P(X)=3/4 * (1/[2!*1!)]) = 3/8

we can verify that is correct since the sum of all the probabilities from x=0 to x=3 is 1/8 + 3/8+ 3/8+ 1/8 = 1

8 0

3 years ago