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2 years ago

Find all solutions to the equation in the interval [0, 2īt). 4) sin 2x = -sin x

Mathematics

1 answer:

Sonbull [250]2 years ago

8 0

Answer:

x={0, 2pi/3, pi, 4pi/3}

Step-by-step explanation:

First, move sin x by adding sin x to both sides. sin 2x+ sin x = 0. Next using the double angle identity, sin 2x=2(sin x)(cos x), so 2(sin x)(cos x)+sinx=0. Factoring, sin x(2cos x +1)=0. Solving, sin x=0 and 2cos x +1 = 0, or cos x = -1/2.

sin x = 0. Using the unit circle, sin x=0 when x=0 and x=pi.

cosx=-1/2. Using the unit circle, cos x= -1/2 when x=2pi/3 and x=4pi/3.

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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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1 year ago

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Given f(x) = 6x^4 – 10x^3 + 40x – 50, find f(2)

ASHA 777 [7]

F(x)=6x^4-10x^3+40x-50, plug 2 in for x
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f(2)=12^4-30^3+80-50
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2 years ago

Find the missing denominator or numerator of 1/2 = /4

Jet001 [13]

Simply cross-multiply about,

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2 years ago

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Write an equation that is non-proportional.

elena-s [515]

Answer:

y = mx + b. When b ≠ 0, the relationship between x and y is nonproportional.

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1 year ago

3 On a coordinate grid, the location of a lighthouse is at L, and the location of a buoy is at B. At noon, a ship is at the midp

Zarrin [17]

Answer:

$(\frac{1}{2}, 2)$

Step-by-step explanation:

Using the midpoint formula, $M(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2})$ , we can find the coordinates that represents the position of the ship at noon.