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Yakvenalex [24]

3 years ago

5

Need help on this please

1 answer:

scoundrel [369]3 years ago

8 0

Answer:

It's D

Step-by-step explanation:

2t - 30( $\frac{1}{5}$ ) = 20

2t - $\frac{30}{1}$ ( $\frac{1}{5}$ ) = 20

2t - $\frac{30}{5}$ = 20

2t - 6 = 20

2t - 6 + 6 = 20 +6

2t = 26

$\frac{2t}{2}$ = $\frac{26}{2}$

t = 13

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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>

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3 years ago

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alex41 [277]

Answer:

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Step-by-step explanation:

I hope this helps. Best of luck.

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3 years ago

Read 2 more answers

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AysviL [449]

Answer:

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Step-by-step explanation:

P(a/3 , 2) is mid-point of Q (-5 , 4) and R (-1 , 0)

a/3 = (-5 + (-1))/2 = -6/2 = -3

a = -9

3 0

3 years ago