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3 years ago

ANSWER QUICKLY FOR THANKS AND BRAINLIEST

Mathematics

2 answers:

DerKrebs [107]3 years ago

6 0

Answer:

Hypothesis: If the temperature gets below normal the pool will be closed.

Conclusion: As a result of the temperature dropping below the regular or room temperature needed in order to be used the pool will be closed

Step-by-step explanation:

Firlakuza [10]3 years ago

3 0

Answer:

Hypothesis: If the temperature gets below normal the pool will be closed.

Conclusion: As a result of the temperature dropping below the regular or room temperature needed in order to be used the pool will be closed.

Step-by-step explanation:

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3 years ago

Three bouquets of flowers are ordered at a florist. 3 roses, 2 carnations, and 1 tulip cost $14. 6 roses, 2 carnations, and 6 tu

STALIN [3.7K]

cost of 1 rose = $ 3

cost of 1 carnation = $ 1

cost of 1 tulip = $ 3

<em><u>Solution:</u></em>

Let "r" be the cost of 1 rose

Let "c" be the cost of 1 carnation

Let "t" be the cost of 1 tulip

<em><u>3 roses, 2 carnations, and 1 tulip cost $14</u></em>

So we can frame a equation as:

3 roses x cost of 1 rose + 2 carnations x cost of 1 carnation + 1 tulip x cost of 1 tulip = $ 14

$3 \times r + 2 \times c + 1 \times t = 14$

3r + 2c + 1t = 14 ----- eqn 1

<em><u>6 roses, 2 carnations, and 6 tulips cost $38</u></em>

So we can frame a equation as:

6 roses x cost of 1 rose + 2 carnations x cost of 1 carnation + 6 tulip x cost of 1 tulip = $ 38

$6 \times r + 2 \times c + 6 \times t = 38$

6r + 2c + 6t = 38 ------ eqn 2

<u><em>1 rose, 12 carnations, and 1 tulip cost $18</em></u>

So we can frame a equation as:

1 rose x cost of 1 rose + 12 carnations x cost of 1 carnation + 1 tulip x cost of 1 tulip = $ 18

$1 \times r + 12 \times c + 1 \times t = 18$

r + 12c + t = 18 ----- eqn 3

<em><u>Let us solve eqn 1 and eqn 2 and eqn 3 to find values of "r" "c" "t"</u></em>

3r + 2c + 1t = 14 ----- eqn 1

6r + 2c + 6t = 38 ------ eqn 2

r + 12c + t = 18 ----- eqn 3

From eqn 1,

3r = 14 - 2c - t

$r = \frac{14 - 2c - t}{3}$

Substitute the above value of r in eqn 2

$6(\frac{14 - 2c - t}{3})+ 2c + 6t = 38\\\\28 - 4c - 2t + 2c + 6t = 38\\\\-2c +4t = 10\\\\-2c = 10 - 4t\\\\2c = 4t - 10\\\\c = 2t - 5$

Substitute c = 2t - 5 and $r = \frac{14 - 2c - t}{3}$ in eqn 3

$12(2t - 5) + \frac{14 - 2c - t}{3} + t = 18\\\\24t - 60 + \frac{14-2(2t - 5) - t}{3} + t = 18\\\\72t - 180 + 14 - 4t +10 - t + 3t = 54\\\\70t = 54 + 180 - 14 -10\\\\70t = 210\\\\t = 3$

Substitute t = 3 in c = 2t - 5

c = 2(3) - 5

Substitute t = 3 and c = 1 in $r = \frac{14 - 2c - t}{3}$

$r = \frac{14 - 2(1) - 3}{3}\\\\r = \frac{14 - 2 - 3}{3}\\\\r = \frac{9}{3} = 3$

<em><u>Summarizing the results:</u></em>

cost of 1 rose = $ 3

cost of 1 carnation = $ 1

cost of 1 tulip = $ 3

6 0

3 years ago