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Law Incorporation [45]
2 years ago
8

Which shows solutions to X+4y=9 if X and y must be whole numbers? A. {(0,9)(1,5)(2,1)} B.{(2,1)(1,6)(0,9)} C.{(1,2)(5,1)(9,0) D.

{(0,9)(1,4)(2,1)
Mathematics
2 answers:
Jlenok [28]2 years ago
5 0

Answer with Step-by-step explanation:

We find the value of y when x=0,1,2,5 and 9

Since, all the options have x values to be amongst these only

Now, when x=0, then 4y=9

i.e. y=9/4

so, (0,9/4) is a solution

So, option A and D are incorrect

when x = 1, then 1 + 4y = 9

4y = 9 - 1

4y = 8

y = 2

So, (1,2) is a solution

So, option B is incorrect

when x = 5, then 5 + 4y = 9

4y = 9 - 5

4y = 4

y = 1

So, (5,1) is the solution

Lastly, if x = 9, then 9 + 4y = 9

4y = 9 - 9

y = 0/4

y = 0

So, (9,0) is the solution

Hence, option C is correct

Dafna11 [192]2 years ago
4 0
The answer here is C. Let's proof.

Since we are dealing with whole numbers, select a constant for x to satisfy that y will result a whole number.
If x = 1, then the function would be 1 + 4y = 9. Solving for y,
4y = 9 - 1
4y = 8
y = 2
In ordered pair, that is (1,2)

Next, if x = 5, then 5 + 4y = 9. Solving for y,
4y = 9 - 5
4y = 4
y = 1
In ordered pair, that is (5,1).

Lastly, if x = 9, then 9 + 4y = 9. Solving for y,
4y = 9 - 9
y = 0/4
y = 0
In order pair, that is (9,0). 
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The temperature of the ice cream 2 hours after it was placed in the freezer is 37.40 °C

From Newton's law of cooling, we have that

T_{(t)}= T_{s}+(T_{0} - T_{s})e^{kt}

Where

(t) = \ time

T_{(t)} = \ the \ temperature \ of \ the \ body \ at \ time \ (t)

T_{s} = Surrounding \ temperature

T_{0} = Initial \ temperature \ of \ the \ body

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From the question,

T_{0} = 86 ^{o}C

T_{s} = -20 ^{o}C

∴ T_{0} - T_{s} = 86^{o}C - -20^{o}C = 86^{o}C +20^{o}C

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Therefore, the equation T_{(t)}= T_{s}+(T_{0} - T_{s})e^{kt} becomes

T_{(t)}=-20+106 e^{kt}

Also, from the question

After 1 hour, the temperature of the ice-cream base has decreased to 58°C.

That is,

At time t = 1 \ hour, T_{(t)} = 58^{o}C

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Then, we get

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ln(e^{k}) =ln( 0.735849)

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Now, for the temperature of the ice cream 2 hours after it was placed in the freezer, that is, at t = 2 \ hours

From

T_{(t)}=-20+106 e^{kt}

Then

T_{(2)}=-20+106 e^{(-0.30673 \times 2)}

T_{(2)}=-20+106 e^{-0.61346}

T_{(2)}=-20+106\times 0.5414741237

T_{(2)}=-20+57.396257

T_{(2)}=37.396257 \ ^{o}C

T_{(2)} \approxeq  37.40 \ ^{o}C

Hence, the temperature of the ice cream 2 hours after it was placed in the freezer is 37.40 °C

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