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3 years ago

Only do 16 pls help me i will give u brainliest if you help

Mathematics

1 answer:

USPshnik [31]3 years ago

8 0

Answer:

16) 18.5 hours

Step-by-step explanation:

40gifts = 5hrs

40/5 = 8

8gifts = 1hr

148/8 = 18.5

In 18.5 hours, 148 gifts are wrapped.

hope this helps and is right. p.s. i really need brainliest :)

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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

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