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3 years ago

BRAINLIEST !!! Can someone add me as a friend plz need to look someone up but cant till someone adds me

Mathematics

2 answers:

marishachu [46]3 years ago

8 0

Answer:

no hahahahahahahahahahahahahahahahahahahah

eimsori [14]3 years ago

4 0

Answer:

I have done what you want

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Help much appreciated ):

sweet-ann [11.9K]

Answer

Step-by-step explanation:

help much appreciated ):help much appreciated ):help much appreciated ):help much appreciated ):help much appreciated ):help much appreciated ):help much appreciated ):help much appreciated ):help much appreciated ):help much appreciated ):help much appreciated ):help much appreciated ):help much appreciated ):help much appreciated ):help much appreciated ):help much appreciated ):help much appreciated ):help much appreciated ):help much appreciated ):help much appreciated ):help much appreciated ):help much appreciated ):help much appreciated ):help much appreciated ):help much appreciated ):help much appreciated ):help much appreciated ):help much appreciated ): hope this helps leave a heart c:

7 0

3 years ago

Necesito las respuestas

Stels [109]

Answer:

I need it in English please so I can helps

3 0

2 years ago

Point D is the centroid of Triangle ABC. Find CD and CE.<br> DE = 9

sukhopar [10]

Given:

Point D is the centroid of Triangle ABC and DE = 9.

To find:

The measures of CD and CE.

Solution:

We know that, centroid is the intersection of medians and it divides each median in 2:1.

In triangle ABC, CE is a meaning and centroid D divided CE in 2:1. So,

Let the measures of CD and DE are 2x and x respectively.

DE = 9 (Given)

$x=9$

Now,

$CD=2x$

$CD=2(9)$

$CD=18$

And,

$CE=CD+DE$

$CE=18+9$

$CE=27$

Therefore, the measure of CD is 18 units and the measure of CE is 27 units.

6 0

3 years ago

PLS HELP ME ON THIS QUESTION I WILL MARK YOU AS BRAINLIEST IF YOU KNOW THE ANSWER!!

Taya2010 [7]

Answer:

1071.79 yd³

Step-by-step explanation:

radius r = 16/2 = 8 yd

height = 16 yd

volume = (1/3)×πr²h

= (1/3)×3.14×8²×16

= 1071.79 yd³ (rounded to the nearest tenth)

3 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