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3 years ago

What is 1189^2 x 1809

Mathematics

2 answers:

Trava [24]3 years ago

8 0

Answer:

the answer is : 2557421289

blondinia [14]3 years ago

7 0

Answer:

the answer is 2,544,697.8

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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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2 years ago

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How to slove - Tan x°= 1.11 and what is the answer?

jarptica [38.1K]

$\bf tan(x^o)=1.11\impliedby \textit{taking }tan^{-1}\textit{ to both sides}
\\\\\\
tan^{-1}[tan(x^o)]=tan^{-1}(1.11)\implies \measuredangle x=tan^{-1}(1.11)$

plug that in your calculator, make sure the calculator is in Degree mode

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3 years ago

Choose the equation in point slope form that passes through the point (3,-4) and has a slope of 4

netineya [11]

Answer:

y+4=4(x-3)

Step-by-step explanation:

y-y1=m(x-x1)

y-(-4)=4(x-3)

y+4=4(x-3)

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3 years ago

Can someone help me out with this my grade are bad

Tju [1.3M]

Answer:

2 2/5

Step-by-step explanation:

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2 years ago

Type the correct answer in the box. Use numerals instead of words. If necessary, use / for the fraction bar. One of the x-interc

Marina CMI [18]

3x^2 + 6x - 10 = 0
3(x^2 + 2x) - 10 = 0
3[ (x + 1)^2 - 1 ] - 10 = 0
3(x+1)^2 - 13 = 0

so the vertex is at (-1,-13)
the roots will be same distance from x = -1
that is a distance 1.08 --1 = 2.08

so other root is approximately -1 -2.08 = -3.08

the other intercept is at (-3.08,0)

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3 years ago

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