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Gelneren [198K]
3 years ago
13

Please help me with this equation

Mathematics
1 answer:
klasskru [66]3 years ago
4 0

Answer: \dfrac{1}{16}x^{-5}y^{-2}z^{4}

Step-by-step explanation:

Given

Expression is \left(4xy^3z^{-2}\right)^{-2}\cdot x^{-3}y^4z^0

using the Exponent property that is

[(a)^b]^c=a^{b\cdot c}\\a^x\cdot a^y=a^{x+y}

Applying the above property

\Rightarrow \left(4xy^3z^{-2}\right)^{-2}\cdot x^{-3}y^4z^0=4^{-2}x^{-2}y^{-6}z^{4}\cdot x^{-3}y^4z^0\\\\\Rightarrow \left(4xy^3z^{-2}\right)^{-2}\cdot x^{-3}y^4z^0=4^{-2}x^{-2-3}y^{-6+4}z^{4+0}\\\Rightarrow 4^{-2}x^{-5}y^{-2}z^{4}=\dfrac{1}{16}x^{-5}y^{-2}z^{4}

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4 years ago
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6 0
3 years ago
A bucket that weighs 6 lb and a rope of negligible weight are used to draw water from a well that is 80 ft deep. The bucket is f
zaharov [31]

Answer:

3200 ft-lb

Step-by-step explanation:

To answer this question, we need to find the force applied by the rope on the bucket at time t

At t=0, the weight of the bucket is 6+36=42 \mathrm{lb}

After t seconds, the weight of the bucket is 42-0.15 t \mathrm{lb}

Since the acceleration of the bucket is the force on the bucket by the rope is equal to the weight of the bucket.

If the upward direction is positive, the displacement after t seconds is x=1.5 t

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We are now ready to calculate the work done by the rope on the bucket.

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=\left[63 t-0.2 t^{2}\right]_{0}^{36}

W=3200 \mathrm{ft} \cdot \mathrm{lb}

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Step-by-step explanation:

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