What is the solution to the equation? Please explain!

9. Complete the reasons in the proof below.

Arada [10]

For a), this is clearly a given as it is literally to the right of where it says “Given:”
For b), since ON bisects ∠JOH, this means that it splits it into two separate angles - JON and HON, which are similar due to that bisects mean that it splits it equally into two halves
For c), since NO is the same thing as NO, it is equal to itself
For d), since AAS (angle-angle-side) congruence states that if there are two angles that are congruent (proved in a) and b) ) as well as that a side is congruent (proved in c) ), two triangles are congruent
For e), since two triangles are congruent, every side must have one side that it matches up to in the other triangle. As the opposite side of angle H is JO and the opposite side of angle J is OH, and ∠J=∠H, those two are congruent. As JN and HN are the two sides left, they must be congruent.

Feel free to ask further questions!

4 0

4 years ago

Select two ratios that are equivalent to 2:12 A.6:40 B.12:2 C.3:24 D.1:6 E.8:48

Sever21 [200]

Answer:

E

Step-by-step explanation:

2:12 is equivalent to 8:48

7 0

3 years ago

Read 2 more answers

The chorus teacher plans to arrange the students in equal rows. Only the girls or boys will be in each row. What is the greatest

emmainna [20.7K]

Answer:

2

Step-by-step explanation:

68+48=116

116÷16=7.24 (wrong)

116÷12=9.7 (wrong)

116÷4=29 (lowest even)

116÷2=58 (highest even)

If I'm wrong, let me know

8 0

3 years ago

Which statements about the local maximums and minimums for the given function are true? Choose three options.

Nostrana [21]

Answer:

Over the interval [2, 4], the local minimum is –8.

Over the interval [3, 5], the local minimum is –8.

Over the interval [1, 4], the local maximum is 0.

Step-by-step explanation:

The true statements are:

Over the interval [2, 4], the local minimum is –8.

Over the interval [3, 5], the local minimum is –8.

Over the interval [1, 4], the local maximum is 0.

Lets discuss each option one by one:

Over the interval [1, 3], the local minimum is 0

This is a false statement. Look at the graph. The minimum point given is (3.4,-8). Therefore the local minimum is -8 not 0

Over the interval [2, 4], the local minimum is –8.

This statement is true because the given minimum point is(3.4, -8). Thus the local minimum is -8 which is true

Over the interval [3, 5], the local minimum is –8.

According to the given minimum point, the local minimum is -8 which is true

Over the interval [1, 4], the local maximum is 0.

Look at the graph. The maximum point given is (2,0). Thus this statement is true because local maximum is 0.

Over the interval [3, 5], the local maximum is 0.

This is a false statement because there is no maximum point

8 0

3 years ago

Read 2 more answers

A football team has a probability of .75 of winning when playing any of the other four teams in its conference. If the games are

Alexeev081 [22]

Answer:

0.3164 = 31.64% probability the team wins all its conference games

Step-by-step explanation:

For each conference game, there are only two possible outcomes. Either the team wins it, or they lose. The probability of winning a game is independent of any other game. This means that the binomial probability distribution is used to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

A football team has a probability of .75 of winning when playing any of the other four teams in its conference.

The probability means that $p = 0.75$ , and four games means that $n = 4$

If the games are independent, what is the probability the team wins all its conference games?

This is P(X = 4). So

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 4) = C_{4,4}.(0.75)^{4}.(0.25)^{0} = 0.3164$

0.3164 = 31.64% probability the team wins all its conference games

4 0

3 years ago