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3 years ago

PLEASE HELP!!!

2 answers:

7 0

<span>$100 discount loan for 7% 1) Pay back $100 in one year 2) Receive $93, today. $540 discount loan for 12.5% 1) Pay back $540.00 in 90 days. 2) Receive WHAT, today? It's less than</span>

Anastasy [175]3 years ago

5 0

16.88

523.12

at least for the question i got

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mixer [17]

Answer:the answer is c

Step-by-step explanation:

5 0

3 years ago

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

3 years ago

Solve for v.<br> 5(v+2) = -2(7v-3) +9v<br> Simplify your answer as much as possible

Andreas93 [3]

5(v+2) = -2(7v-3) + 9

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

Distribute:

5v + 10 = -14v + 6 + 9

Combine Like Terms

5v + 10 = (-14v+9v)+6

5v + 10 = -5v + 6

Subtract -5v from both sides

10v + 10 = 6

Subtract 10 from both sides

10v = -4

Reduce

5v = -2

Divide both sides by 5

v = -2/5

Simplify

v = -0.2

3 0

4 years ago

Read 2 more answers

Meghan's dog spray eats 4 ¹/4 cups of food each day. Explain how megan can determine how much food to give sparky if she needs t

Alexxandr [17]

Answer:

$\frac{17}{6}\ cups$

Step-by-step explanation:

$Total\ required\ food\ for\ sparky=4\frac{1}{4}\ cups\\\\Total\ required\ food\ for\ sparky=\frac{17}{4}\ cups\\\\Megan\ needs\ to\ feed\ him\ only\ \frac{2}{3}\ as\ much\\\\Food\ given\ by\ Megan=\frac{2}{3}\times Total\ required\ food\ for\ sparky\\\\Food\ given\ by\ Megan=\frac{2}{3}\times \frac{17}{4}\\\\Food\ given\ by\ Megan=\frac{17}{6}\ cups$

6 0

3 years ago

I will rate you brainliest!

Lapatulllka [165]

Answer:

Step-by-step explanation:

X must not be equal to -6, -1, 1 or 3

5 0

3 years ago