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Arturiano [62]
3 years ago
15

Derivative of the (sqrt of t/4t-3)

Mathematics
1 answer:
Soloha48 [4]3 years ago
3 0
Next time, please enclose the function in the denominator inside parentheses.

Derivative of the (sqrt of t/4t-3) should be 
<span>
Derivative of the (sqrt of t/[4t-3])

then f(t) = sqrt( t / [4t-3] ), or    f(t) = ( t/ [4t-3])^(1/2)

Use the power rule here.  f '(t) = (1/2) [ t/{4t-3} ]^(-1/2) times the derivative of [ t / (4t-3) ].  Can you finish this differentiation?


</span>
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Step-by-step explanation:

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2 years ago
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Need help on number 25 and 26 can you please explain how you got the answer thanks!:)
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In both problems, the sum of side lengths is the perimeter. Opposite sides of a parallelogram (or rectangle) are equal in length, so you can find the perimeter by doubling the sum of adjacent sides.

25. 2(x +(x +15)) = (x +45) +(x +40) +(x +25)
.. 4x +30 = 3x +110 . . . . . . . . . . . . . . . . . . . . . . simplify
.. x = 80 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . subtract 3x+30
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26. 2(x +(x +2)) = (x) +(x +6) +(x +4)
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Can someone please help
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You can find the answer by substituting the x-values provided in the question and seeing if you get what the question says.

a satisfies all of the parameters.
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3 years ago
What is the ratio of the area of the inner square to the area of the outer square?
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Answer:

\frac{(a-b)^2+b^2}{a^2}

Step-by-step explanation:

Since, By the given diagram,

The side of the inner square = Distance between the points (0,b) and (a-b,0)

=\sqrt{(a-b-0)^2+(0-b)^2}

=\sqrt{(a-b)^2+b^2}

Thus the area of the inner square = (side)²

=(\sqrt{(a-b)^2+b^2})^2

=(a-b)^2+b^2\text{ square cm}

Now, the side of the outer square = Distance between the points (0,0) and (a,0),

=\sqrt{(a-0)^2+0^2}

=\sqrt{a^2}=a

Thus, the area of the outer square = (side)²

=a^2\text{ square cm}

Hence, the ratio of the area of the inner square to the area of the outer square

=\frac{(a-b)^2+b^2}{a^2}

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