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3 years ago

10

What is the range of f(x) = A cos(x) +b or g(x) = A sin(x)+b?

1 answer:

vagabundo [1.1K]3 years ago

3 0

Answer:

Step-by-step explanation:

The range of f(x) = cos(x) is [-1, 1].

That of f(x) = A cos(x) is [-A, A], and

That of f(x) = A cos(x) + 2 is [ b-A, b+A ].

The function g(x) = A sin(x)+b has the same range.

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Which of the following functions best fits the data shown in the scatterplot below? y=100x+2y is equal to 100 x plus 2 y=20x2+10

KengaRu [80]

The equation of the scatter plot is (a) y = 100x + 2

<h3>How to determine the equation of the scatterplot?</h3>

The line of best fit of the scatter plot passes through the points

(x,y) = (0,2) and (1,102)

Start by calculating the slope using:

$m = \frac{y_2 -y_1}{x_2 -x_1}$

So, we have:

$m = \frac{102 - 2}{1 - 0}$

m = 100

The equation is then calculated as:

y = m(x - x1) + y1

This gives

y = 100(x -0) + 2

Evaluate

y = 100x + 2

Hence, the equation of the scatter plot is (a) y = 100x + 2

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1 year ago

Which is a rational number?<br> 3.<br> 4.<br> 5/5<br> 6.12

k0ka [10]

Answer:

i pretty sure one of em is 6.12

4 0

2 years ago

Each month, Anna pays $630 in rent. How much rent does she pay over the course of 18 months?

Ilia_Sergeevich [38]

Answer:

$11,340

Step-by-step explanation:

1 month: 630

18 months: 18×630=11,340

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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

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2 years ago

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Hi, can someone please answer this question? Thanks!!!

cricket20 [7]

Answer:

sorry im not so sure abut the answer

Step-by-step explanation:

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8 0

3 years ago