For which system of equations is the graph below a solution?

Find the value of the following expression: (2^8 • 5^-5 • 19^0)^-2 • (5^-2 over 2^3)^4 • 2^28. Write your answer in simplified f

DaniilM [7]

Recognize that "( ) over 2^3" means ( ) • 2^-3. Use the rule of exponents
.. (a^b)^c = a^(b•c)

= 2^-16•5^10•19^-2 • 5^-8*2^-12 • 2^28

Now, you can use the rule of exponents
.. (a^b)*(a^c) = a^(b+c)

= 2^(-16 -12 +28) • 5^(10 -8) • 19^-2
= 5^2 • 19^-2
= 5^2 / 19^2
= 25/361

3 0

4 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

2 years ago

Read 2 more answers

How do I solve this ?

guapka [62]

I don't like fractions, so I would turn 8/14 into a decimal.

Do 8 divided by 14 and you'll get .57

Now 8 + 0.57 = 8.57

The fraction form would be 60/7 and the mixed form would be 8 4/7

5 0

3 years ago

If f(x) = 5x + 4, which of the following is the inverse of Ax)?

Levart [38]

Answer:

$f^{-1}$ (x) = $\frac{x-4}{5}$

Step-by-step explanation:

let y = f(x) and rearrange making x the subject, that is

y = 5x + 4 ( subtract 4 from both sides )

y - 4 = 5x ( divide both sides by 5 )

$\frac{y-4}{5}$ = x

change x to $f^{-1}$ (x) and y back to x, thus

$f^{-1}$ (x) = $\frac{x-4}{5}$

6 0

4 years ago

HELPPPP<br>the last part says select 3 options

torisob [31]

Answer:

Last three options are correct answers.

Step-by-step explanation:

Hope it helps you in your learning process.

3 0

3 years ago

Read 2 more answers