Mathematics, 13.01.2021 01:20 1hannacarson

Evaluate (m) (x) for x = -3. (-3) = т n

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m Sin[x] + n Cos[x] + p /. replacerule (* p + Sqrt[m^2 + n^2] Sin[x + ArcTan[m, n]] *) ... There are two or three tricks that we can use to solve for {a, b, c, ω} . First ... take the second derivative and evaluate both equatinos at x=0 and at x=π. ... Sin[b] -> σ, {ρ, σ}]; eq2 // Column Reduce[eq2, ω] (* { c n == n p, c ω^2 == p ω^2, n (-1 + ...

Step-by-step explanation:

Answer from: aparedes5

To evaluate the clinical application of the second derivative of the fingertip photoplethysmogram waveform, we performed drug administration studies (study 1) and epidemiological studies (study 2). In study 1, ascending aortic pressure was recorded simultaneously with the fingertip photoplethysmogram and its second derivative in 39 patients with a mean+/-SD age of 54+/-11 years. The augmentation index was defined as the ratio of the height of the late systolic peak to that of the early systolic peak in the pulse. The second derivative consists of an a, b, c, and d wave in systole and an e wave in diastole. Ascending aortic pressure increased after injection of 2.5 microg angiotensin from 126/74 to 160/91 mm Hg and decreased after 0.3 mg sublingual nitroglycerin to 111/73 mm Hg. The d/a, the ratio of the height of the d wave to that of the a wave, decreased after angiotensin from -0.40+/-0.13 to -0.62+/-0.19 and increased after nitroglycerin to -0.25+/-0.12 (P<0.001 and P<0.001, respectively). The negative d/a increased with increases in plethysmographic and ascending aortic augmentation indices (r=0.79, P<0.001, and r=0.80, P<0.001, respectively). The negative d/a reflects the late systolic pressure augmentation in the ascending aorta and may be useful for noninvasive evaluation of the effects of vasoactive agents. In study 2, the second derivative of the plethysmogram waveform was measured in a total of 600 subjects (50 men and 50 women in each decade from the 3rd to the 8th) in our health assessment center

Answer from: Quest

function 2 has a greater rate of change than function 1

function 2 has a greater y-intercept than function 1

step-by-step explanation:

$\newline{\text{the rate of change is change per unit time}}$ $\newline{\text{for (x,y), rate of change is }\frac{y_2-y_1}{x_2-x_1} \text{ or change in y divided by change in x}}$ $\newline{\text{for function 1, the cost goes up by 75-45=30 each time every hour}}$ $\newline{\text{that means that the slope of function 1 is 30 per hour}}$ $\newline{\text{for the 2nd function, 40 is added every hour. therefor, the slope is 40 per hour}}$ $\bigskip\newline{\text{since 40 per hour> 30 per hour, function 2 has a greater rate of change than function 1}}$ $\bigskip\newline{\text{the y intercept is the y value when x is 0}}$ $\newline{\text{for function 1, the y value when x is 0 is 45}}$ $\newline{\text{for function 2, the y value when x is 0 (or the flat rate) is 50}}$ $\newline{\text{since 50> 45, function 2 has a greater y intercept than funciton 1}}$ $\bigskip\bigskip\newline{\text{the answers are: }}$ $\newline{\text{ function 2 has a greater rate of change than function 1}}$ $\newline{\text{function 2 has a greater y-intercept than function 1.}}$

Answer from: Quest

step-by-step explanation:

draw the shape and you will see that f and k are actually at adjacent vertices (points of the hexagon that are neighboring). so find the distance between (1,2) and (4,2). that is easy since (1,2) and (4,2) is just a vertical distance to count, which is 3. it is a regular polygon which means all sides are congruent. you have 6 sides (hexa) so multiply 6 by 3 to receive 18.