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densk [106]
3 years ago
13

Suppose x, y, and z are positive integers such that xy + yz = 29 and xz + yz = 81. Which of the following variables has exactly

one unique solution?
(i) x
(ii) y
(iii) z
A. none
B. ii only
C. iii only
D. i and ii only
E. ii and iii only
Mathematics
1 answer:
ss7ja [257]3 years ago
7 0

Answer:

Step-by-step explanation:

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Miranda le debe $120.00 a max .Le paga 30% de lo que le debe ¿cuanto le falta pagarle?
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Answer:

84

Step-by-step explanation:

el 10% de 120= 12. 120-12= 108

el 20% de 120=24. 120-24=96

el 30% de 120=36. 120-36=84

el 40% de 120=48. 120-48=72

el 50% de 120=60. 120-60=60

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Find the value of A if the 4-digit number 3A5A is divisible by both 3 and 5.
Arturiano [62]

Answer:

The value of A is 5

Step-by-step explanation:

- The number is divisible by 3 if the sum of its digits is a number

  divisible by 3

- Ex: 126 is divisible by 3 because the sum of its digits = 1 + 2 + 3 = 6

  and 6 is divisible by 3

- The number is divisible by 5 if its ones digit is zero or 5

- Ex: 675 is divisible by 5 because its ones digit is 5

        890 is divisible by 5 because its ones digit is 0

- We are looking for the value of A in the 4-digit number 3A5A which

  makes the number divisible by both 3 and 5

∵ A is in the ones position

∴ A must be zero or 5

- Let us try A = 0

∵ A = 0

∴ The number is 3050

∵ The sum of the digits of the number = 3 + 0 + 5 + 0 = 8

∵ 8 is not divisible by 3

∴ 3050 is not divisible by both 3 and 5

∴ A can not be zero

- Let us try A = 5

∵ A = 5

∴ The number is 3555

∵ The sum of the digits of the number = 3 + 5 + 5 + 5 = 18

∵ 18 is divisible by 3

∴ 3555 is divisible by both 3 and 5

∴ A must be equal 5

* <em>The value of A is 5</em>

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3 years ago
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