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2 years ago

Find The value of x =

Mathematics

1 answer:

Levart [38]2 years ago

4 0

Answer:

x = 10

Step-by-step explanation:

5 / sin 30° = x / sin 90°

5 / (1/2) = x / 1

5 . 1 = 1/2 . x

5 = 1/2x

x = 5/1/2

x = 10/1

x = 10

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First 4 sets of (x, y) show that when x increases 2, y increases 3. It seems the relationship is linear. But last 2 set show that x increases 1, y increases 3. So the relation is NOT linear

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The box plot shows the time it takes you and your classmates to get to school. What is the median time spent traveling to school

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2 years ago

C С 154 -Y Х 2 3 5 w 4 6 7 Z D

Sergio [31]

Answer:

m<1 = 26°

m<2 = 154°

m<3 = 26°

m<4 = 26°

m<5 = 154°

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m<7 = 26°

Step-by-step explanation:

What is required was not stated, however, let's find the value of every angle labelled in this diagram.

✔️m<1 = 180° - 154° (linear pair theorem)

m<1 = 26°

✔️m<2 = 154° (vertical angles theorem)

m<2 = 154°

✔️m<3 = m<1 (vertical angles theorem)

m<3 = 26° (substitution)

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m<4 = 26° (substitution)

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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

6 0

1 year ago

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