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3 years ago

5

Which of the following points would fall on the line produced by the point-slope form equation y - 8 = 4/5(x - 1) when graphed

1 answer:

HACTEHA [7]3 years ago

6 0

This says that x will equal -9

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Given the parent function of f(x) = x³, what is the value of k in the translated graph of f(x - h) + k?

jekas [21]

The value of k in the translated graph of f(x - h) + k is; k = 2

<h3>How to carry out Translations on Graphs?</h3>

We are given the parent function of f(x) = x³.

Now, the point (0, 0) in the graph of f(x) is the point (3, 2) in the translated graph. Thus;

The rule of translation is;

(x, y) → (x + 3, y + 2)

That means;

The translation is 3 units to the right and 2 units up

The equation of the translated graph is equal to;

f(x) = (x - 3)³ + 2

Thus;

The value of k is equal to 2.

Read more about Graph Translations at; brainly.com/question/4025726

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1 year ago

What is the value of 6 divided by 3/8

V125BC [204]

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2 years ago

Amy mixed 4 cups of red paint with 1 cup of white paint to make pink paint.

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4 cups of white paint since there was 4 cups of red paint

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2 years ago

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Which set of numbers could be the lengths of the sides of a triangle?

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3 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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