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3 years ago

Y-3= 5(x - 2) NEED HELP

Mathematics

2 answers:

sveticcg [70]3 years ago

8 0

Answer:

y = 5x - 7

Step-by-step explanation:

y-3= 5(x-2)

y-3 = 5x - 10

y=5x -7

This is the answer if you have anymore questions please ask

DerKrebs [107]3 years ago

5 0

Answer:

Y = 5x -7

Step-by-step explanation:

That is written in slope intercept form, which I'm assuming is what you need. If not comment on my answer.

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What is the factored form of the expression. 4n^2 + 24n + 36

skad [1K]

Answer:

(2n+6)^2

Step-by-step explanation:

To solve this, you can use the given formula: x^2 + 2xy + y^2

In this case, 4n^2 is x^2, 24n is 2xy, and 36 is y^2. The next step is:

(2n)^2 + 2(2n)(6) + (6)^2

Since this equation fits into this formula ( x^2 + 2xy + y^2), we can do:

(2n+6)(2n+6) =

(2n+6)^2

Hence, the answer is (2n+6)^2

6 0

3 years ago

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) Large and small drains are opened

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

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

(ii) Only the small drain is opened

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

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

(iii) Only the big drain is opened

$\dot V_{l} = \frac{k}{t_{l}}$ (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:

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