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Papessa [141]
3 years ago
9

Y varies directly as x if x=15 when y=20, find x when y=30

Mathematics
1 answer:
mixas84 [53]3 years ago
6 0

Answer:

X=22.5

Step-by-step explanation:

Y=kX

20=15k

k=4/3

For X when Y=30

30=4/3×X

Cross multiply

4X=90

X=22.5

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What value of p makes the equation true?<br><br> -3p + 1/8 = - 1/4
allochka39001 [22]

Answer:

4

Step-by-step explanation:

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4 years ago
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Jung-Su's father gave him $144. Jung-Su bought 4 books, each of which cost $9. How much money does Jung-Su have left?
Ghella [55]

Answer:

$108

Step-by-step explanation:

9+9+9+9=36

144-36=108

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2 years ago
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The function f(x)is graphed on the coordinate plane.
professor190 [17]

Answer:

f(-2) = -2

Step-by-step explanation:

f(-2)  means find the y value when x = -2

The y value when x = -2 is-2

f(-2) = -2

5 0
2 years ago
The liquid base of an ice cream has an initial temperature of 86°C before it is placed in a freezer with a constant temperature
Karolina [17]

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

k = constant

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

T_{0} - T_{s} = 106^{o} C

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

Then, we can write that

T_{(1)}=58 = -20+106 e^{k(1)}

Then, we get

58 = -20+106 e^{k(1)}

Now, solve for k

First collect like terms

58 +20 = 106 e^{k}

78 =106 e^{k}

Then,

e^{k} = \frac{78}{106}

e^{k} = 0.735849

Now, take the natural log of both sides

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

k = -0.30673

This is the value of the constant k

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

Learn more here: brainly.com/question/11689670

6 0
2 years ago
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If (x − 2) ∶ 5 = 7 ∶ 9, then find the value of x.
WARRIOR [948]

Answer:

x=53/9

Step-by-step explanation:

(x-2):5=7:9

9x-18=35

9x=35+18

9x=53

x=53/9

4 0
2 years ago
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