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3 years ago

In slope intercept form, write the equation for the line that passes through the point (2,3) and the point (7,13).

1 answer:

8 0

First, Use the slope equation to find the slope of the line passing thru these two points:

m=rise/run

Here, the rise is 13-3, or 10, and the run is 7-2, or 5. Thus, the slope, m, is 10/5, or 2: m=2.

We want the slope-intercept form, so let's begin with its general form:

y=mx+b. Substitute the slope 2 for m: y=2x+b. Now choose either of the given points. Arbitrarily I am choosing (2,3). Then x=2 and y=3.

Substituting these values into y=2x+b: 3 = 2(2) + b, or b= 3 -4, or b = -1.

Then the equation of this line, in slope-intercept form, is y = 2x - 1.

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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

8% of what is 2.8??

Ilia_Sergeevich [38]

Write that as an equation and solve. (In math, "of" often means "times".)
8% × what = 2.8
0.08 × what = 2.8
what = 2.8/0.08
what = 35

8% of 35 is 2.8.

6 0

3 years ago

Highlight very briefly why/how when you subtract a negative number, that it yields a positive number. Will give brainliest

lbvjy [14]

When we build integers from natural numbers, we're looking for additive inverse of natural numbers?

What's an additive inverse? Well, for example, the additive inverse of 2 is a numbers $x$ such that

$2+x = 0$

We call this number -2. So, the real meaning behind the negative sign is "if you add me and my positive counterpart, the result is zero".

So, -5 is the additive inverse of 5, -16 is the additive inverse of 16, and so on, because

$5-5=0,\quad 16-16=0,\ldots$

Note that this is a symmetrical relation: if -5 is the inverse of 5, it is also true that 5 is the inverse of -5.

So, when you write something like

$5-(-4)$

it means that you want to add 5 and the inverse of the inverse of 4. But given what we just said, the inverse of the inverse of a number is the number itself, which is why subtracting a negative number is the same as adding it.

3 0

3 years ago

Emily measured the height of a dresser as 4.27 feet, but the actual height was 4 feet. What is the percentage of error in Emily’

Assoli18 [71]

We need to find the percentage of the extra feet that she measured, so first we have to find out how much that is. We can do this by simply subtracting:

4.27 - 4 = 0.27

Now we can set up a proportion to find the percentage.

$\frac{0.27}{4} \frac{x}{100}$

Solve for x

0.27 x 100 = 27
27/4 = 6.75

The answer is 6.75% :)

Hopefully this helps! If you have any more questions or don't understand, feel free to DM me, and I'll get back to you ASAP! :)

6 0

3 years ago

A bag contains 52 balls lettered from a to z and A to Z (i.e. uppercase or lowercase letters). One ball is withdrawn. What is th

never [62]

the probability that the ball will have a letter that is lower case or earlier in the alphabet than d (or D) is 0.557

<u>Step-by-step explanation:</u>

Here we have ,A bag contains 52 balls lettered from a to z and A to Z (i.e. uppercase or lowercase letters). One ball is withdrawn. We need to find What is the probability that the ball will have a letter that is lower case or earlier in the alphabet than d (or D)? Let's find out:

There are 26 lowercase letters and 26 uppercase letters ! According to question we need to choose a ball which is either a lowercase ( i.e. from 26 letters ) or earlier in the alphabet than d ( or D ) which is A or B or , C .

Probability = (Favorable outcome)/(Total outcome)

Favorable outcome = 26+3=29

Total Outcome = 52

So ,

⇒ $Probability = \frac{29}{52}$

⇒ $Probability = 0.557$

Therefore , the probability that the ball will have a letter that is lower case or earlier in the alphabet than d (or D) is 0.557

7 0

3 years ago