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3 years ago

PLZ HELP WORTH EXTRA POINTS!! 3D pythagream theroem!

Mathematics

1 answer:

Alja [10]3 years ago

6 0

Answer:

$\sqrt{69}$

Step-by-step explanation:

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What is a good word problem for 5(x+3)=170

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think of something and someone I can't really answer this question right now but if you want you can give me 1 star but I hope you get your question answered and have a WONDERFUL day

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3 years ago

Will give BRAINLEST! See image attached below.

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Answer:

See the image below for answer:)

Step-by-step explanation:

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3 years ago

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Find the directional derivative of the function at the given point in the direction of the vector v. f(x, y, z) = xe^y + ye^z +

LUCKY_DIMON [66]

Answer:

$D_{\vec{u}}f(0,0,0)=\frac{6}{\sqrt{54}}$

Step-by-step explanation:

We need to find the directional derivative of the function at the given point in the direction of the vector v.

$f(x, y, z)=xe^{y} + ye^{z} + ze^{x}$ ,point (0, 0, 0) and $v=$

By Theorem: If f is a differentiable function of x , y and z , then f has a directional derivative for any unit vector $\overrightarrow{v} =$ and

$D_{\overrightarrow{u}}f(x,y,z)=f_{x}(x,y,z)u_1+f_{y}(x,y,z)u_2+f_{z}(x,y,z)u_3$

where $\overrightarrow{u}=\frac{\overrightarrow{v}}{||v||}$

since, $v=$

then $\overrightarrow{u}=\frac{\overrightarrow{v}}{||v||}$

$\overrightarrow{u}=< \frac{6}{\sqrt{6^{2}+3^{2}+(-3)^{2}}},\frac{3}{\sqrt{6^{2}+3^{2}+(-3)^{2}}},\frac{-3}{\sqrt{6^{2}+3^{2}+(-3)^{2}}} >$

$\overrightarrow{u}=< \frac{6}{\sqrt{54}},\frac{3}{\sqrt{54}},\frac{-3}{\sqrt{54}} >$

The partial derivatives are

$f_{x}(x,y,z)=e^{y}+ze^{x}$

$f_{y}(x,y,z)=xe^{y}+e^{z}$

$f_{z}(x,y,z)=ye^{z}+e^{x}$

Then the directional derivative is

$D_{\vec{u}}f(x,y,z)=(e^{y}+ze^{x})(\frac{6}{\sqrt{54}})+(xe^{y}+e^{z})(\frac{3}{\sqrt{54}})+(ye^{z}+e^{x})(\frac{-3}{\sqrt{54}})$

so, directional derivative at point (0,0,0)

$D_{\vec{u}}f(0,0,0)=(e^{0}+0e^{0})(\frac{6}{\sqrt{54}})+(0e^{0}+e^{0})(\frac{3}{\sqrt{54}})+(0e^{0}+e^{0})(\frac{-3}{\sqrt{54}})$

$D_{\vec{u}}f(0,0,0)=\frac{6}{\sqrt{54}}+\frac{3}{\sqrt{54}}+\frac{-3}{\sqrt{54}}$

$D_{\vec{u}}f(0,0,0)=\frac{6+3-3}{\sqrt{54}}$

$D_{\vec{u}}f(0,0,0)=\frac{6}{\sqrt{54}}$

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4 years ago

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vivado [14]

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3 years ago

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Susie spent 4 1/4 hours on Monday and 3 5/8 hours on Tuesday working on a history project. About how long did she spend working

kompoz [17]

4 1\4 +3 5\8
(4+3=7)
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4 years ago

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