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AlladinOne [14]
3 years ago
15

Segment AB is parallel to segment CD. Angle 1 is congruent to angle 2 because

Mathematics
2 answers:
Klio2033 [76]3 years ago
6 0
I believe thr answer to be C.
Rom4ik [11]3 years ago
6 0

Answer:

C. Of the Corresponding Angles Postulate.

Step-by-step explanation:

We have been given a diagram, in which line segment AB is parallel to segment CD. We are asked to choose the reason that states that angle 1 is congruent to angle 2.

Let us see line BC as a transversal.

Since alternate interior angles are equal, therefore, measure of angle DCB is equal to angle ABC.

Let us see line BC as a transversal.

Since alternate interior angles are equal, therefore, measure of angle BAD is equal to angle CDA.

Since angle 1 corresponds to angle 2, therefore, angle 1 is congruent to angle 2 because of the Corresponding Angles Postulate..

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20.   The sum of two consecutive even integers is 158. Find the least of the two integers. 
Mumz [18]

Answer:

78

Step-by-step explanation:

The tricky part of this is figuring out how to assign the unknowns.  We are told that we are working with two consecutive even integers.  Consecutive means "next to" or "in order" and sum means to add.  If we use 2 and 4 as examples of our 2 consecutive even integers and assign x to 2, then in order to get from 2 to 4 we have to add 2.  So the lesser of the 2 integers is x, and the next one in order will be x + 2.  (2 and 4 are just used as examples; they mean nothing to the solving of this particular problem.  You could pick any 2 even consecutive integers and find the same rule applies.  All we are doing here with the example numbers is finding a rule for our integers.)  Now we have the 2 expressions for the integers, we will add them together and set the sum equal to 158:

x + (x + 2) = 158

The parenthesis are unnecessary since we are adding, so when we combine like terms we get

2x + 2 = 158 and

2x = 156 and

x = 78

That means that the lesser of the 2 integers in 78, and the next one in order would be 80, and 78 + 80 = 158

6 0
3 years ago
Read 2 more answers
Find the mean, meadian, mode and range for each set of data 6, 9, 2, 4, 3, 6, 5
Helga [31]

Mean: Add up the numbers and divide the sum by the number of values in the set.

6 + 9 + 2 + 4 + 3 + 6 + 5 = 35

35 / 7 = 5

Median: Sort the set from the smallest value to the largest value and select the number in the middle. If the count of the set if even, then select the two middle values and take their mean average.

2, 3, 4, 5, 6, 6, 9

            ^

So, the median average is 5.

Mode: What number appears the most frequently?

The mode of the set is 6 because it appears twice.

Range: Sort the set by ascending order and take the smallest value and subtract that from the largest value in the set.

9 - 2 = 7

The range is 7.

6 0
3 years ago
Read 2 more answers
A family on a trip budgets $1,000 for meals and gasoline. If the price of a meal for the family is $50 and if gasoline costs $3.
galben [10]

Answer:

They can buy 13meals if they buy 100 gallons of gasoline.

Step-by-step explanation:

3.50 PER gallon so 1 gallon is $3.50

if they buy 100 gallons you have to multiply 3.50 by 100 which gives you 350. you subtract 350 from 1000 so 1000-350 and get 650. now, you divide 650 by 50 because each meal is $50. And you get 13 so there you have it.

3 0
3 years ago
What is the nth term for the sequence 5 , 14 , 29 , 50 , 77
Marina86 [1]
The answer would be 3n^2 + 2.

This can be found/proven by replacing "n" with term number (1,2,3,4...), then solving to get the final number. For example 3 * 1^2 + 2. You would first do 1^2, which is 1. Next, you would multiply 1 by 3, to get 3. Finally, you'd and the 2 to get 5. 5 is the 1st term, and you can use this same equation to get the rest of the terms you need.

I hope this helps!
8 0
3 years ago
JK and LM are perpendicular diameters of a circle. They are each 12 inches long. What is the approximate length of chord LK?
MA_775_DIABLO [31]

ANSWER:

I think the approximate length og chord Lk is around 8.5 inches long

3 0
3 years ago
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