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levacccp [35]
4 years ago
5

Given f(1) = 2. 5 and g(2) = 8.1 + 2then what is f(g(-9)) ?​

Mathematics
1 answer:
sasho [114]4 years ago
8 0

Answer:

f(g(-9)) = 3.6

Step-by-step explanation:

-9 + 8.1 + 2 = 1.1

2.5  + 1.1 = 3.6

f(g(-9)) = 3.6

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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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