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4 years ago

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The variables x and y have a quadratic relationship. Tripling the value of x causes the value of y to increase by a factor of ni

ne. When x has a value of 3, y has a value of 2. What will be the value of y when x has a value of 9?

1 answer:

Lady bird [3.3K]4 years ago

3 0

Answer:

<em>The value of y when x has a value of 9 is 18</em>

Step-by-step explanation:

<u>Functions</u>

The variables x and y are said to have a quadratic relationship. That information is not enough to set up a general equation for both variables.

Furthermore, we know that tripling the value of x causes the value of y to be multiplied by 9. That leads us to establish a proportional equation between them, as follows:

$y=kx^2$

Where k is an unknown constant.

Let's use the given point (3,2) to find the value of k:

$2=k(3)^2$

$2=9k$

Solving for k:

$\displaystyle k=\frac{2}{9}$

The equation is now:

$\displaystyle y=\frac{2}{9}x^2$

Now find y when x=9:

$\displaystyle y=\frac{2}{9}(9)^2$

$\displaystyle y=\frac{2}{9}\cdot 81=18$

The value of y when x has a value of 9 is 18

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Shanking draws a line that is 25cm long. Kathi draws a line that is 30cm longer than shaniks line. How long is kathis line

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Answer:

55

Step-by-step explanation:

If Kathi´s line is 30cms longer than Shanking´s, you would just add 30 and 25 to get the length of Kathi´s line.

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4 years ago

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A large fish tank at an aquarium needs to be emptied so that it can be cleaned. When its

VikaD [51]

Answer:

The draining time when only the big drain is opened is 2.303 hours.

The draining time when only the small drain is opened is 5.303 hours.

Step-by-step explanation:

From Physics, we know that volume flow rate ( $\dot V$ ), measured in liters per hour, is directly proportional to draining time ( $t$ ), measured in hours. That is:

$\dot V \propto \frac{1}{t}$

$\dot V = \frac{k}{t}$ (Eq. 1)

Where $k$ is the proportionality constant, measured in liters.

From statement, we have the following three expressions:

(i) <em>Large and small drains are opened</em>

$\dot V_{s}+\dot V_{l} = \frac{k}{2}$ (Eq. 2)

$\frac{\dot V_{s}+\dot V_{l}}{k} = \frac{1}{2}$

(ii) <em>Only the small drain is opened</em>

$\dot V_{s} = \frac{k}{t_{l}+3}$ (Eq. 3)

$\frac{\dot V_{s}}{k} = \frac{1}{t_{l}+3}$

(iii) <em>Only the big drain is opened</em>

$\dot V_{l} = \frac{k}{t_{l}}$ (Eq. 4)

$\frac{\dot V_{l}}{k} = \frac{1}{t_{l}}$

By applying (Eqs. 3, 4) in (Eq. 2) and making some algebraic handling, we find that:

$\frac{1}{t_{l}+3}+\frac{1}{t_{l}} = \frac{1}{2}$

$\frac{t_{l}+t_{l}+3}{t_{l}\cdot (t_{l}+3)} = \frac{1}{2}$

$2\cdot t_{l}+3 = t_{l}^{2}+3\cdot t_{l}$

$t_{l}^{2}-t_{l}-3 = 0$ (Eq. 5)

Whose roots are determined by the Quadratic Formula:

$t_{l,1}\approx 2.303\,h$ and $t_{l,2} \approx -1.302\,h$

Only the first roots offers a solution that is physically reasonable. Hence, the draining time when only the big drain is opened is 2.303 hours. And the time needed for the small drain is calculated by the following formula:

$t_{s} = 2.303\,h+3\,h$

$t_{s} = 5.303\,h$

The draining time when only the small drain is opened is 5.303 hours.

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