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astraxan [27]
3 years ago
7

Pls Help the questions are on the picture

Mathematics
1 answer:
natima [27]3 years ago
5 0

coronanananananannana cornananananan cornananana coronananannanan ornanananan cornananan ebola ebola ebolalalalla

Step-by-step explanation:

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Use chain rule to find dw/dt w=ln(x^2 + y^2 + z^2)^(1/2) , x=sint, y=cost, z=tant
son4ous [18]
The chain rule states that

\dfrac{\mathrm dw}{\mathrm dt}=\dfrac{\partial w}{\partial x}\dfrac{\mathrm dx}{\mathrm dt}+\dfrac{\partial w}{\partial y}\dfrac{\mathrm dy}{\mathrm dt}+\dfrac{\partial w}{\partial z}\dfrac{\mathrm dz}{\mathrm dt}

Since \ln(x^2+y^2+z^2)^{1/2}=\dfrac12\ln(x^2+y^2+z^2), you have

\dfrac{\partial w}{\partial x}=\dfrac x{x^2+y^2+z^2}
\dfrac{\partial w}{\partial x}=\dfrac y{x^2+y^2+z^2}
\dfrac{\partial w}{\partial z}=\dfrac z{x^2+y^2+z^2}

and

\dfrac{\mathrm dx}{\mathrm dt}=\cos t
\dfrac{\mathrm dy}{\mathrm dt}=-\sin t
\dfrac{\mathrm dz}{\mathrm dt}=\sec^2t

Also, since x^2+y^2+z^2=\sin^2t+\cos^2t+\tan^2t=1+\tan^2t=\sec^2t, the derivative is

\dfrac{\mathrm dw}{\mathrm dt}=\dfrac{\sin t\cos t}{\sec^2t}-\dfrac{\sin t\cos t}{\sec^2t}+\dfrac{\tan t\sec^2t}{\sec^2t}
\dfrac{\mathrm dw}{\mathrm dt}=\tan t
4 0
3 years ago
Read and choose the correct words for the blank that match the image. (1 point) A cook using pepper on food cooking in a pan Me
spayn [35]

Answer:

C) la pimienta

Step-by-step explanation:

8 0
2 years ago
Members of a community center are planning a field trip to the zoo. More than 50 people will go on the trip. Ten people are prov
slava [35]
The answer is 5(k+10)>50

6 0
3 years ago
Read 2 more answers
A car starts moving at 9:00 am and moves at a constant speed. At what time will the car be at the distance 200 miles from the st
andriy [413]

. i got the answer of 2:00

Step-by-step explanation:

yea

6 0
3 years ago
Read 2 more answers
Help!! 50 points and brainliest!
Viktor [21]

Answer:

Second choice:

x=2t

y=4t^2+4t-3

Fifth choice:

x=t+1

y=t^2+4t

Step-by-step explanation:

Let's look at choice 1.

x=t+1

y=t^2+2t

I'm going to subtract 1 on both sides for the first equation giving me x-1=t. I will replace the t in the second equation with this substitution from equation 1.

y=(x-1)^2+2(x-1)

Expand using the distributive property and the identity (u+v)^2=u^2+2uv+v^2:

y=(x^2-2x+1)+(2x-2)

y=x^2+(-2x+2x)+(1-2)

y=x^2+0+-1

y=x^2

So this not the desired result.

Let's look at choice 2.

x=2t

y=4t^2+4t-3

Solve the first equation for t by dividing both sides by 2:

t=\frac{x}{2}.

Let's plug this into equation 2:

y=4(\frac{x}{2})^2+4(\frac{x}{2})-3

y=4(\frac{x^2}{4})+2x-3

y=x^2+2x-3

This is the desired result.

Choice 3:

x=t-3

y=t^2+2t

Solve the first equation for t by adding 3 on both sides:

x+3=t.

Plug into second equation:

y=(x+3)^2+2(x+3)

Expanding using the distributive property and the earlier identity mentioned to expand the binomial square:

y=(x^2+6x+9)+(2x+6)

y=(x^2)+(6x+2x)+(9+6)

y=x^2+8x+15

Not the desired result.

Choice 4:

x=t^2

y=2t-3

I'm going to solve the bottom equation for t since I don't want to deal with square roots.

Add 3 on both sides:

y+3=2t

Divide both sides by 2:

\frac{y+3}{2}=t

Plug into equation 1:

x=(\frac{y+3}{2})^2

This is not the desired result because the y variable will be squared now instead of the x variable.

Choice 5:

x=t+1

y=t^2+4t

Solve the first equation for t by subtracting 1 on both sides:

x-1=t.

Plug into equation 2:

y=(x-1)^2+4(x-1)

Distribute and use the binomial square identity used earlier:

y=(x^2-2x+1)+(4x-4)

y=(x^2)+(-2x+4x)+(1-4)

y=x^2+2x+-3

y=x^2+2x-3.

This is the desired result.

3 0
3 years ago
Read 2 more answers
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