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3 years ago

5/17 multiplied by 3/8

Mathematics

2 answers:

tino4ka555 [31]3 years ago

6 0

5/17 × 3/8 = 15/ = 15/136

wariber [46]3 years ago

4 0

15/136 is the answer

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PLEASE HELP (100 POINTS)

Artyom0805 [142]

$\\ \rm\Rrightarrow f(x)=-5cos\dfrac{x}{2}+5$

Compare to standard equation of SHM

$\\ \rm\Rrightarrow y=Acos(\omega x+\phi)$

Here

Amplitude=A=-5

$\\ \rm\Rrightarrow \omega x=\dfrac{x}{2}$

$\\ \rm\Rrightarrow \omega=1/2$

Well the above one are for extra knowledge .

Solution:-

$\\ \rm\Rrightarrow 10=-5cos\dfrac{x}{2}+5$

$\\ \rm\Rrightarrow 10-5=-5cos\dfrac{x}{2}$

$\\ \rm\Rrightarrow 5=-5cos\dfrac{x}{2}$

$\\ \rm\Rrightarrow cos\dfrac{x}{2}=-1$

$\\ \rm\Rrightarrow \dfrac{x}{2}=cos^{-1}(-1)$

$\\ \rm\Rrightarrow \dfrac{x}{2}=\pi$

$\\ \rm\Rrightarrow x=2\pi$

So.

$\\ \rm\Rrightarrow x=2\pi+2\pi n$

3 0

2 years ago

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Need help- Find the measure of the angle.<br><br> THANK YOU

snow_tiger [21]

Answer:

Step-by-step explanation:

Remember triangle always equal to 180

180=70+4x-5+6x-15

x=13

4(13)-5

Angle A is 47

6 0

2 years ago

Please help me. Somebody! I need help.

kogti [31]

Answer:

https://www.imsdb.com/scripts/Shrek.html

Step-by-step explanation: Follow this link. DO it no ballz

5 0

3 years ago

What is 25%of395 thanks

tia_tia [17]

98.75
Just do 0.25 (25%) times 395
its the same as doing 395 divided by 25

6 0

3 years ago

Read 2 more answers

The concentration C of certain drug in a patient's bloodstream t hours after injection is given by

frozen [14]

Answer:

a) The horizontal asymptote of $C(t)$ is $c = 0$ .

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) The time at which the concentration is highest is approximately 1.291 hours after injection.

Step-by-step explanation:

a) The horizontal asymptote of $C(t)$ is the horizontal line, to which the function converges when $t$ diverges to the infinity. That is:

$c = \lim _{t\to +\infty} \frac{t}{3\cdot t^{2}+5}$ (1)

$c = \lim_{t\to +\infty}\left(\frac{t}{3\cdot t^{2}+5} \right)\cdot \left(\frac{t^{2}}{t^{2}} \right)$

$c = \lim_{t\to +\infty}\frac{\frac{t}{t^{2}} }{\frac{3\cdot t^{2}+5}{t^{2}} }$

$c = \lim_{t\to +\infty} \frac{\frac{1}{t} }{3+\frac{5}{t^{2}} }$

$c = \frac{\lim_{t\to +\infty}\frac{1}{t} }{\lim_{t\to +\infty}3+\lim_{t\to +\infty}\frac{5}{t^{2}} }$

$c = \frac{0}{3+0}$

$c = 0$

The horizontal asymptote of $C(t)$ is $c = 0$ .

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:

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

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

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

$5-3\cdot t^{2} = 0$

$t = \sqrt{\frac{5}{3} }\,h$

$t \approx 1.291\,h$

The critical point occurs approximately at 1.291 hours after injection.

Second Derivative Test

The second derivative of the function is:

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

$C''(t) = -\frac{6\cdot t}{(3\cdot t^{2}+5)^{2}}- \frac{12\cdot t}{(3\cdot t^{2}+5)^{2}}+\frac{72\cdot t^{3}}{(3\cdot t^{2}+5)^{3}}$

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

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.

The time at which the concentration is highest is approximately 1.291 hours after injection.

5 0

2 years ago