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4 years ago

15

Mr. Nolan's code for his ATM card is a 4-digit number. The digits of the code are the prime factors of 84 listed from least to g

reatest. What is the code for Mr. Nolan's ATM card?

2 answers:

Nataly [62]4 years ago

8 0

The answer is 2237

Please don't forget to rate me!

Scilla [17]4 years ago

6 0

Answer:

2237

Step-by-step explanation:

We are given that Mr.Nolan's code for his ATM card is a 4- digit number.

We are given that the digits of the code are the prime factors of 84 listed from least to greatest.

We have to find the code for Mr.Nolan's ATM Card.

In order to find the code we will find the prime factors of 84.

$84=2\times 2\times 3\times 7$

The prime factors of 84 are 2,2,3 and 7.

Therefore, the code=2237

Answer:2237

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Find ∂w/∂s and ∂w/∂t using the appropriate Chain Rule.

Vesnalui [34]

Answer:

<h3>The value of $\frac{\partial w}{\partial s}$ is $e^{3t}-18e^{2s+t}$ </h3><h3>The value of $\frac{\partial w}{\partial t}$ is $3e^{3t}-9e^{2s+t}$ </h3><h3>The partial derivative at s=-5 and t=10 is $\frac{\partial w}{\partial s}$ is $e^{30}-18$ </h3><h3>The partial derivative at s=-5 and t=10 is $\frac{\partial w}{\partial t}=3(e^{30}-3)$ </h3>

Step-by-step explanation:

Given that the Function point are $w=y^3-9x^2y$

$x=e^s$ , $y=e^t$ and s = -5, t = 10

<h3>To find $\frac{\partial w}{\partial s}$ and $\frac{\partial w}{\partial t}$ using the appropriate Chain Rule : </h3>

$w=y^3-9x^2y$

Substitute the values of x and y in the above equation we get

$w=(e^t)^3-9(e^s)^2(e^t)$

$w=e^{3t}-9e^{2s}.e^t$

<h3>Now partially differentiating w with respect to s by using chain rule we have </h3>

$\frac{\partial w}{\partial ∂s}=e^{3t}-9(e^{2s}).2(e^t)$

$=e^{3t}-18e^{2s}.(e^t)$

$=e^{3t}-18e^{2s+t}$

<h3>Therefore the value of $\frac{\partial w}{\partial s}$ is $e^{3t}-18e^{2s+t}$ </h3>

$w=e^{3t}-9e^{2s}.e^t$

<h3>Now partially differentiating w with respect to t by using chain rule we have </h3>

$\frac{\partial w}{\partial t}=e^{3t}.(3)-9e^{2s}(e^t).(1)$

$=3e^{3t}-9e^{2s+t}$

<h3>Therefore the value of $\frac{\partial w}{\partial t}$ is $3e^{3t}-9e^{2s+t}$ </h3>

Now put s-5 and t=10 to evaluate each partial derivative at the given values of s and t :

$\frac{\partial w}{\partial s}=e^{3t}-18e^{2s+t}$

$=e^{3(10}-18e^{2(-5)+10}$

$=e^{30}-18e^{-10+10}$

$=e^{30}-18e^0$

$=e^{30}-18$

<h3>Therefore the partial derivative at s=-5 and t=10 is $\frac{\partial w}{\partial s}$ is $e^{30}-18$ </h3>

$\frac{\partial w}{\partial t}=3e^{3t}-9e^{2s+t}$

$=3e^{3(10)}-9e^{2(-5)+10}$

$=3e^{30}-9e{-10+10}$

$=3e^{30}-9e{0}$

$=3e^{30}-9$

$\frac{\partial w}{\partial t}=3(e^{30}-3)$

<h3>Therefore the partial derivative at s=-5 and t=10 is $\frac{\partial w}{\partial t}=3(e^{30}-3)$ </h3>

6 0

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alexandr1967 [171]

Answer:

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ArbitrLikvidat [17]

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Answer:

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Step-by-step explanation:

Given as :

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4 years ago