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Ne4ueva [31]
3 years ago
6

In a certain deck of cards, each card has a positive integer written on it. In a multiplication game, a child draws a card and m

ultiplies the integer on the card by the next larger integer. If each possible product is between 15 and 200, then the least and greatest integers on the cards could be

Mathematics
1 answer:
ASHA 777 [7]3 years ago
8 0

Answer:

  4 and 13

Step-by-step explanation:

You want integer solutions to ...

  15 ≤ n(n+1) ≤ 200

If we let the limits be represented by "a", then the equality is represented by ...

  n² +n -a = 0

  (n² +n +1/4) -a -1/4 = 0

  (n +1/2)^2 = (a +1/4)

  n = -1/2 + √(a +1/4)

For a=15, we have

  n ≥ -1/2 + √15.25 ≈ 3.4 . . . . . minimum n is 4

For a=200, we have

  n ≤ -1/2 + √200.25 ≈ 13.7 . . . maximum n is 13

The least and greatest integers on the cards are 4 and 13.

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Answer:

The greatest number of bunches of bananas that Andrea can buy is 15 and the greatest number of boxes of raspberries that Andrea can buy is 21 to maintain the ratio 7:5

Step-by-step explanation:

Let

x ---->the greatest number of boxes of raspberries that Andrea can buy

y ---->the greatest number of bunches of bananas that Andrea can buy

we know that

100\%+8\%=106\%=108/100=1.08 ---> sales tax

\frac{x}{y} =\frac{7}{5}

x=\frac{7}{5}y ----> equation A

1.08[2.50x+3.00y]\leq 135 ----> inequality B

substitute equation A in the inequality B

1.08[2.50(\frac{7}{5}y)+3.00y]\leq 135

solve for y

Multiply by 5 both sides to remove the fraction

18.9y+16.2y\leq 675

35.1y\leq 675

y\leq 19.23

so

The maximum value of y is 19

Find the value of x

replace the value of y in equation A until you get an integer value that satisfies the ratio 7:5  

For y=19

x=\frac{7}{5}(19)

x=26.6

For y=18

x=\frac{7}{5}(18)

x=25.2

For y=17

x=\frac{7}{5}(17)

x=23.8

For y=16

x=\frac{7}{5}(16)

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For y=15

x=\frac{7}{5}(15)

x=21

therefore

The greatest number of of bunches of bananas that Andrea can buy is 15 and the greatest number of boxes of raspberries that Andrea can buy is 21 to maintain the ratio 7:5

<u><em>Verify the inequality B</em></u>

1.08[2.50(21)+3.00(15)]\leq 135

1.08[52.5+45]\leq 135

105.3\leq 135 ----> is true

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Step-by-step explanation:

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\left( \cfrac{x^2}{3y^{- \frac{1}{2} }} \right)^{-6}=\left( \cfrac{3y^{- \frac{1}{2} }}{x^2} \right)^{6}=\left( \cfrac{3}{x^2y^{ \frac{1}{2} }} \right)^{6}= \cfrac{3^6}{x^{2*6}y^{ \frac{1}{2}*6 }} = \cfrac{729}{x^{12}y^3}
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