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3 years ago

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

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Answer:

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

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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

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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

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2 years ago