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Katarina [22]
2 years ago
9

Differentiate with respect to t

Mathematics
1 answer:
True [87]2 years ago
8 0

Answer:

\displaystyle 2x\frac{dx}{dt}-3y^2\frac{dy}{dt}+4z^3\frac{dz}{dt}=0

Step-by-step explanation:

We want to differentiate the equation:

x^2-y^3+z^4=1

With respect to <em>t</em>, where <em>x, y, </em>and <em>z</em> are functions of <em>t. </em>

<em />

So:

\displaystyle \frac{d}{dt}\left[x^2-y^3+z^4\right]=\frac{d}{dt}\left[1\right]

Implicitly differentiate on the left. On the right, the derivative of a constant is simply zero. Hence:

\displaystyle 2x\frac{dx}{dt}-3y^2\frac{dy}{dt}+4z^3\frac{dz}{dt}=0

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Factor the expression completely. 6×3- 4×2 – 16x A. 0 B. 2x(3×2 – 2x – 8) C. 2x(3x + 4)(x – 2) D. 4x(2x + 1)(x – 4) E. 2x(2×2 +
Ghella [55]

Answer:

<h3>The answer is option C</h3>

Step-by-step explanation:

6x³ - 4x² - 16x

To factorize the expression first factor out the GCF out

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That's

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Next Factorize the terms in the bracket

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<u>Next factor out - 2 from the expression</u>

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We have the final answer as

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Hope this helps you

4 0
3 years ago
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tankabanditka [31]

Answer:

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Step-by-step explanation:

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Step-by-step explanation:

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How do I solve this type of problem? I'm trying to find both missing values. Please explain how to solve this as thoroughly as y
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Hi there! 

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3 0
3 years ago
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