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IgorC [24]
2 years ago
10

Geometry problem in the picture: Find the measure of the angle to the nearest degree

Mathematics
2 answers:
Anit [1.1K]2 years ago
5 0

Answer:

? Is 41 degrees

Step-by-step explanation:

SOH CAH TOA

Hypotenuse is 60

Adjacent is 45

Hypotenuse and adjacent - Cos

(Cos inverse to find the angle)

Cos(?) = a/h

Cos(?) = 45/60

Cos-(45/60) =?

? = 41.40962211

Hope this helps!

nalin [4]2 years ago
4 0

Answer:

?=41.4

Step-by-step explanation:

In the photo is explained

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What is the domain and range of y=-1/4(x)
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Domain: (-\infty, \infty)

Range: (-\infty, \infty)

Explanation:

The function is y=-\frac{1}{4} x

The domain of a function is the set of all input values for which the function is well defined. Generally, domain consists of all x-values of the function. Hence, the function y=-\frac{1}{4} x is defined in the interval (-\infty, \infty).

The range of a function is the set of output values obtained by substituting the value of x in the function. Hence, the function y=-\frac{1}{4} x is defined in the interval (-\infty, \infty).

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Factorise the following completely 3ax+2ay+6bx+4by​
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Answer:

(a+2b)(3x+2y)

Step-by-step explanation:

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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}

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Then,

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

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Now, take the natural log of both sides

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

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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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