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2 years ago

How do you write the standard number 0.00009 in scientific notation?

Mathematics

1 answer:

tekilochka [14]2 years ago

7 0

$0.\underbrace{00009}\\~~~~~~~5\\\\\\0.00009 =\boxed{\bf{ 9 \times 10^{-5}}}$

Your answer is A.

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Given cos theta = - 2/5 and sin theta < 0, find the six trigonometric values. I need help with this

Lena [83]

Answer:

$\sin\,\theta =-\frac{\sqrt{21} }{5}$

$\tan\,\theta =\frac{\sqrt{21} }{2}$

$\sec\,\theta = \frac{-5}{2}$

$cosec\,\theta =\frac{-5}{\sqrt{21} }$

$\cot\,\theta =\frac{2}{\sqrt{21} }$

Step-by-step explanation:

$\cos\theta =\frac{-2}{5}$

As both $sin\,\theta$ ,

$\theta$ lies in the third quadrant.

In the third quadrant,

$\sin\theta$

$\sin\,\theta =-\sqrt{1-\cos^2\,\theta} \\=-\sqrt{1-(\frac{-2}{5})^2 } \\\\=-\sqrt{1-\frac{4}{25} }\\\\=-\sqrt{\frac{25-4}{25} }\\\\=-\frac{\sqrt{21} }{5}$

$\tan\,\theta = \frac{\sin\,\theta}{\cos\,\theta }\\\\=\frac{\frac{-\sqrt{21} }{5} }{\frac{-2}{5} }\\\\=\frac{\sqrt{21} }{2}$

$\sec\,\theta =\frac{1}{\cos\,\theta }\\\\=\frac{1}{\frac{-2}{5} }\\\\=\frac{-5}{2}$

$\ cosec \,\theta = \frac{1}{sin\,\theta }\\\\=\frac{1}{\frac{-\sqrt{21} }{5} }\\\\=\frac{-5}{\sqrt{21} }$

$\cot\,\theta =\frac{1}{\tan\,\theta}\\\\=\frac{1}{\frac{\sqrt{21} }{2} }\\\\=\frac{2}{\sqrt{21} }$

3 0

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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Jonathan is rolling to dice and needs to roll and 11 to win the game he is playing what is the probability that Jonathan wins th

hjlf

Answer:

1 / 18

Step-by-step explanation:

In a roll of two dice :

Number of faces on a dice = 6

Total sample space for 2 6-sided dice = (number of faces)^Number of dice = 6^2 = 36

Total possible outcomes = 36

Required outcome = sum of 11

11 = {(5,6) ; (6, 5)}) = 2 possibilities

Probability = required outcome / Total possible outcomes

P(obtaining a sum of 11) = 2 / 36 = 1/18

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2 years ago

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Evgesh-ka [11]

Answer: c

Step-by-step explanation:

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2 years ago

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Margaret [11]

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3 years ago

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