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posledela
4 years ago
8

What is the midpoint M of that line segment?

Mathematics
1 answer:
Sladkaya [172]4 years ago
7 0

Answer: The midpoint M is located at (6,4)

==============================================

Explanation:

First focus on just the x coordinates of points S and T, which are 5 and 7 respectively. Add them up to get 5+7 = 12. Then divide that in half to get 12/2 = 6. This is the x coordinate of the midpoint.

Repeat for the y coordinates of S and T. First add up the values: 7+1 = 8. Then divide by 2 to get 8/2 = 4. This is the y coordinate of the midpoint.

The two results we get are then written as an ordered pair getting us the answer (6,4)

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If w = 5 cos (xy) − sin (xz) and x = 1/t , y = t, z = t^3 ; then find dw/dt
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In this question, we find the derivatives, using the chain's rule.

Doing this, the derivative is:

\frac{dw}{dt} = \frac{5}{t}(\sin{1} - \cos{1}) - 2t\cos{t^2}

Chain Rule:

Suppose we have a function w(x,y,z), x = x(t), y = y(t), z = z(t), and want to find it's derivative as function of t. It will be given by:

\frac{dw}{dt} = \frac{dw}{dx}\frac{dx}{dt} + \frac{dw}{dy}\frac{dy}{dt} + \frac{dw}{dz}\frac{dz}{dt}

Thus, we have to find the desired derivatives, which are:

  • w of x:

\frac{dw}{dx} = -5y\sin{(xy)} - z\cos{(xz)}

Considering x = \frac{1}{t}, y = t, z = t^3

\frac{dw}{dx} = -5t\sin{(1)} - t^3\cos{(t^2)}

  • w of y:

\frac{dw}{dy} = -5x\cos{(xy)}

Considering x = \frac{1}{t}, y = t

\frac{dw}{dy} = -\frac{5}{t}\cos{1}

  • w of z:

\frac{dw}{dz} = -x\cos{(xz)}

Considering x = \frac{1}{t}, z = t^3

\frac{dw}{dz} = -\frac{1}{t}\cos{(t^2)}

  • Derivatives of x, y and z as functions of t:

\frac{dx}{dt} = -\frac{1}{t^2}

\frac{dy}{dt} = 1

\frac{dz}{dt} = 3t^2

  • Derivative of w as function of t.

Now, we just replace what we found into the formula. So

\frac{dw}{dt} = \frac{dw}{dx}\frac{dx}{dt} + \frac{dw}{dy}\frac{dy}{dt} + \frac{dw}{dz}\frac{dz}{dt}

\frac{dw}{dt} = (-5t\sin{(1)} - t^3\cos{(t^2)})(-\frac{1}{t^2}) - (\frac{5}{t}\cos{1}) - (\frac{1}{t}\cos{(t^2)})3t^2

Applying the multiplications:

\frac{dw}{dt} = \frac{5}{t}\sin{1} + t\cos{t^2} - \frac{5}{t}\cos{1} - 3t\cos{t^2}

Applying the simplifications:

\frac{dw}{dt} = \frac{5}{t}(\sin{1} - \cos{1}) - 2t\cos{t^2}

Which is the derivative.

For more on the chain rule, you can check brainly.com/question/12795383

8 0
3 years ago
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