Evaluate z + z + z for x = 2, y = -3, z = -4. 12 -12 -9 -6 Need help!!

Ya'll fr get 50 points if you answer this. What is the rate of change of <img src="https://tex.z-dn.net/?f=f%28x%29%3D2%5Ex

Evgesh-ka [11]

Answer:

$\dfrac{dy}{dx}=2^x\ln 2$

Step-by-step explanation:

**This is a non-linear function and therefore does not have a constant rate of change. It will have a different slope depending on what points you use in the average rate of change formula: $\mathsf{average \ rate \ of \ change = \dfrac{change \ in \ y}{change \ in \ x}}$

To calculate rate of change, differentiate.

substitute y for $f(x)$ :

$\implies y=2^x$

Take natural logs of both sides:

$\implies \ln y=\ln 2^x$

Apply the log rule $\ln a^b=b \ln a$ :

$\implies \ln y=x\ln 2$

Differentiate with respect to $x$ :

$\implies \dfrac{1}{y} \frac{dy}{dx}=\ln 2$

Mulitply both sides by $y$ :

$\implies \dfrac{dy}{dx}=y\ln 2$

Replace $y$ with $y=2^x$

$\implies \dfrac{dy}{dx}=2^x\ln 2$

Therefore, rate of change of the function is :

$\dfrac{dy}{dx}=2^x\ln 2$

8 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

6 0

2 years ago

Read 2 more answers

The group of answers that satisfy an equation

eimsori [14]

solution set, monomial or term of polynomial, linear equation, relation, f(x)
i answered the same question on anoter website!!!

3 0

3 years ago

Please help!!!!!!!!!!

Sunny_sXe [5.5K]

Answer:

From the chapter Playing with numbers

we can define derek has 2×10^2 + 7 × 10^1 + 7× +0^0

as If ABC a number then ABC can be written as 10^2A+10^1B+10^0C

Hope it helps

7 0

3 years ago

What’s the dill pickle?

Fofino [41]

Answer:

Fermented cucumbers

Step-by-step explanation:

A pickled cucumber is a cucumber that has been pickled in a brine, vinegar, or other solution and left to ferment for a period of time, by either immersing the cucumbers in an acidic solution or through souring by lacto-fermentation. Pickled cucumbers are often part of mixed pickles.

6 0

3 years ago