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AleksAgata [21]
4 years ago
14

Given quadrilateral MATH is similar to quadrilateral ROKS calculate the value of MH Picture is below

Mathematics
1 answer:
tino4ka555 [31]4 years ago
3 0
<h3>Answer:  MH = 7</h3>

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

Explanation:

The double tickmarks for quadrilateral MATH show that MA = TH. Since TH is 5 units long, this makes MA the same length as well.

For quadrilateral ROKS, we have RO = 15. For "MATH" and "ROKS" we have "MA" and "RO" as the first two letters of each four-letter sequence; meaning that MA and RO correspond together.

The ratio of the corresponding segments is RO/MA = 15/5 = 3.

The larger quadrilateral has each side length 3 times longer than the smaller quadrilateral's corresponding side lengths.

--------------

In short,

larger side = 3*(smaller side)

--------------

Using this scale factor of 3, we can find MH

larger side = 3*(smaller side)

RS = 3*(MH)

21 = 3*MH

3*MH = 21

MH = 21/3

MH = 7

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b) When t increases, both the numerator and denominator increases, but given that the grade of the polynomial of the denominator is greater than the grade of the polynomial of the numerator, then the concentration of the drug converges to zero when time diverges to the infinity. There is a monotonous decrease behavior.  

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b) When t increases, both the numerator and denominator increases, but given that the grade of the polynomial of the denominator is greater than the grade of the polynomial of the numerator, then the concentration of the drug converges to zero when time diverges to the infinity. There is a monotonous decrease behavior.  

c) From Calculus we understand that maximum concentration can be found by means of the First and Second Derivative Tests.

First Derivative Test

The first derivative of the function is:

C'(t) = \frac{(3\cdot t^{2}+5)-t\cdot (6\cdot t)}{(3\cdot t^{2}+5)^{2}}

C'(t) = \frac{1}{3\cdot t^{2}+5}-\frac{6\cdot t^{2}}{(3\cdot t^{2}+5)^{2}}

C'(t) = \frac{1}{3\cdot t^{2}+5}\cdot \left(1-\frac{6\cdot t^{2}}{3\cdot t^{2}+5} \right)

Now we equalize the expression to zero:

\frac{1}{3\cdot t^{2}+5}\cdot \left(1-\frac{6\cdot t^{2}}{3\cdot t^{2}+5} \right) = 0

1-\frac{6\cdot t^{2}}{3\cdot t^{2}+5} = 0

\frac{3\cdot t^{2}+5-6\cdot t^{2}}{3\cdot t^{2}+5} = 0

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t \approx 1.291\,h

The critical point occurs approximately at 1.291 hours after injection.

Second Derivative Test

The second derivative of the function is:

C''(t) = -\frac{6\cdot t}{(3\cdot t^{2}+5)^{2}}-\frac{(12\cdot t)\cdot (3\cdot t^{2}+5)^{2}-2\cdot (3\cdot t^{2}+5)\cdot (6\cdot t)\cdot (6\cdot t^{2})}{(3\cdot t^{2}+5)^{4}}

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C''(t) = -\frac{18\cdot t}{(3\cdot t^{2}+5)^{2}}+\frac{72\cdot t^{3}}{(3\cdot t^{2}+5)^{3}}

If we know that t \approx 1.291\,h, then the value of the second derivative is:

C''(1.291\,h) = -0.077

Which means that the critical point is an absolute maximum.

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