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Gelneren [198K]

3 years ago

12

Mary Beth has $1500 in a savings account on January 1. She wants to have at least $800 by May 1. She estimates that she can wit

hdraw $38.50 per week, w, for food, clothing, and entertainment. Which inequality represents her estimate?

1 answer:

DerKrebs [107]3 years ago

5 0

1500-38.5x > 800
x= Number of weeks

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3 years ago

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

3 years ago