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stealth61 [152]
4 years ago
11

Jk has endpoints j(-1,10) and k(-5,2). MN has endpoints M(9,-7) and N (1,-3).is JK congruent MN.

Mathematics
1 answer:
statuscvo [17]4 years ago
7 0
To determine if these line segments are congruent you can find the distances between the x values and the y values for each set of points.

(-1, 10) and (-5, 2). 4 on x axis and 8 on the y axis.

(9, -7) and (1, -3). 8 on the x axis and 4 on the y-axis.

This means they are congruent. Each triangle would have the same two side lengths making the hypotenuse the same.

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wea re going to the carnival on friday night . It costs $2 for each ride and $5 for snacks. Lelia has 23 total to spend. Write a
777dan777 [17]

Answer:

In total you can buy 3 snacks and ride 4 rides.

Step-by-step explanation:

$2 per ride

$5 for snacks

The equation here is $2r + $5s = $23

First I found out how many times I could multiply 5. First I though 4 but 2 doesnt go into the remaining 3, so I lowered it down to 15 which is 3.

So s = 3

Now our equation is 2r + 15 = 23

Subtract 23 - 15 = 8

Now divide

r = 8 / 2

r = 4

7 0
3 years ago
Change the improper fraction to a mixed number. 15/4
AleksAgata [21]
3 3/4 this would be your answer hope this is helpful
8 0
4 years ago
Read 2 more answers
Stephen wants to use the Corresponding Angles Theorem to determine if the street between the park and bus stop is parallel to th
Rudiy27
<span>I would say this answer should be correct since what we need is  a straight line that crosses the two streets to be able to determine ie measure the two either acute or obtuse  angles between the two lines to see if corresponding angles are the same. If they are then that shows that the two streets are parallel. d. Park, bus stop, and his friend's house</span>
3 0
3 years ago
Read 2 more answers
A fair six-sided die (with outcomes 1 through 6) is rolled twice, and two numbers are obtained. X1 = result of the first roll, a
bixtya [17]

Answer:

\dfrac{5}{12}

Step-by-step explanation:

It is given that a fair six-sided die is rolled twice.

X₁= result of the first roll

X₂=result of the second roll

Let A be the event that X₁>X₂.

A={(2,1),(3,1),(4,1),(5,1),(6,1),(3,2),(4,2),(5,2),(6,2),(4,3),(5,3),(6,3),(5,4),(6,4),(6,5)} = 15

It a fair six-sided die is rolled twice then the total number of outcomes is 36.

We need to find the probability of A.

Probability=\dfrac{\text{Favorable outcomes}}{\text{Total outcomes}}

Probability=\dfrac{15}{36}

Probability=\dfrac{5}{12}

Therefore, the probability of A is \dfrac{5}{12}.

5 0
3 years ago
Assuming boys and girls are equally​ likely, find the probability of a couple having a baby girl when their sixth child is​ born
Ivanshal [37]

Answer:  The required probability of having 6th girl is 0.5.

Step-by-step explanation:  Given that boys and girls are equally likely.

We are to find the probability of a couple having a baby girl when their sixth child is​ born, given that the first five children were all girls.

Since the events of having a boy and a girl are independent of each other, so

the probability of having 6th girl dose not depend on the birth of the first five girls.

We know that there are only two possible cases (either a boy or girl will born).

So, sample space, S = {G, B}  and the event E of having a girl is, E = {G}.

That is, n(S) = 2 and n(E) = 1.

Therefore, the probability of event E is given by

P(E)=\dfrac{n(E)}{n(S)}=\dfrac{1}{2}=0.5.

Thus, the required probability of having 6th girl is 0.5.

3 0
3 years ago
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