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3 years ago

Which ordered pair would form a proportional relationship with the point graphed below?

Mathematics

1 answer:

kvasek [131]3 years ago

4 0

Answer:

$(x,y) = (-30,10)$

Step-by-step explanation:

Given

$(x,y) = (60,-20)$

Required:

Determine a pair with a proportional relationship

A proportional relationship is:

$k = \frac{y}{x}$

Where k is the constant of proportionality.

$(x,y) = (60,-20)$

So, we have:

$k = \frac{-20}{60}$

$k = -\frac{1}{3}$

This means that for a pair to have a proportional relation, k must be -1/3.

This is true for (c).

Where:

$(x,y) = (-30,10)$

So, we have:

$k = \frac{10}{-30}$

$k = -\frac{1}{3}$

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What is the value of 35 to the second power

valina [46]

35 to the second power would be the same as multiplying 35 by itself so.....
Ex. 35 to the second power would be

35x35

35 to the third power would be

35x35x35

35 to the fourth power would be

35x35x35x35

And etc.

35x35=1,225

7 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 the equation:<br> y^2=-11y

bazaltina [42]

Answer:

-5.5

Step-by-step explanation:

divide both sides by y which deleted y on the right side. then the left side turns into y instead of y squared. then you have y equals -5.5

8 0

3 years ago

This question difficult and i need some help would anyone please help me

wel

Answer:

x = 30

F = 130

G = 50

Step-by-step explanation:

f and g are supplementary which means they add to 180

5x-20 + 3x - 40 = 180

Combine like terms

8x - 60 = 180

Add 60 to each side

8x-60+60 = 180+60

8x = 240

Divide by 8

8x/8 = 240/8

x = 30

F = 5x -20 = 5*30 -20 = 150 -20 = 130

G = 3x-40 = 3*30 -40 = 90-40 = 50

5 0

3 years ago

Read 2 more answers

Elimination Method<br> 2x-4y=12 -x+2y=-6

Readme [11.4K]

Answer:

The given system has INFINITE NUMBER OF SOLUTIONS.

Step-by-step explanation:

Here, the given system of equation is given as:

2 x - 4 y = 12 ... (1)

- x + 2 y = -6 ...... (2)

Now, to use the ELIMINATION METHOD we either need to cancel out x coefficients or the y coefficient.

Now, to cancel out the x- coefficient in both the equations:

Multiply equation (2) with 2 and add with (1), we get:

2 x - 2x + 4 y - 4 y = 12 - 12

or, 0 = 0

Hence, the given system has INFINITE NUMBER OF SOLUTIONS.

5 0

3 years ago