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Semmy [17]
3 years ago
12

Which of the following points is a solution of y > |x| + 5? A) (1,7) B) (0,5) C) (7,1)

Mathematics
1 answer:
bearhunter [10]3 years ago
3 0

Answer:

A

Step-by-step explanation:

We can plug in all of the x and y values to check if the point is a solution to the inequality.

Point A: x = 1, y = 7

7 > |1| + 5

7 > 1 + 5

7 > 6

This is a true statement, which means that this point is a solution to the inequality. We don't have to check any more points since we have found our answer.

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How many different 5​-letter radio station call letters can be made a. if the first letter must be Upper C comma Upper X comma U
Lady_Fox [76]

Answer:

a) 1,518,000

b) 2,284,880

c) 60,720

Step-by-step explanation:

a) a. if the first letter must be Upper C comma Upper X comma Upper T comma or Upper M and no letter may be​ repeated?

We draw 5 boxes, and based on that we will see the total possible cases. There are 26 alphabets

The first box should have C or X or T or M .No letter may be repeated.

Any                    Any             Any                 Any                    Any

5 alphabets    of the            of the              of the                 of the

C,X , T , M      remaining     remaining       remaining          remaining

                   25 alphabets   24 alphabets   23 alphabets   22 alphabets

Therefore; total possible call letters = 5 × 25 × 24 × 23 × 22 = 1,518,000

b)

The first box should have C or X or T or M Repeats as allowed

Any                    Any             Any                 Any                    Any

5 alphabets    of the            of the              of the                 of the

C,X , T , M      remaining     remaining       remaining          remaining

                   26 alphabets   26 alphabets   26 alphabets   26 alphabets

Therefore Total possible call letters = 5 × 26 × 26 × 26 × 26 = 2,284,880

c)   The first box should have   C,X , T , M  and end with S

So the last place if fixed, and we now have 25 alphabets. The first box can go in 5 ways. The next box then will have only 24 letters to choose from, as the first box has taken a letter and the last box already has S in it. Repetition not allowed

Any                    Any             Any                 Any                       S

5 alphabets    of the            of the              of the                  is fixed

C,X , T , M      remaining     remaining       remaining           here

                   24 alphabets   23 alphabets   22 alphabets  

Therefore Total possible call letters = 5 × 24 × 23 × 22  × 1 =  60,720

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Step-by-step explanation:

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