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3 years ago

Solution to the equation 6(x + 4) = 20. what does x=

Mathematics

2 answers:

Sergeeva-Olga [200]3 years ago

8 0

Answer:

x= −2/3

Step-by-step explanation:

Do You Need Me To Do Step by Step?

I hope I helped you!!

Anit [1.1K]3 years ago

5 0

Answer:

x= -2/3

Step-by-step explanation:

6(x + 4)= 20

1st step: Divide each side by 6

6(x+4)/6 = 20/6

x + 4 = 3 1/3

2nd step: Subtract 4

x + 4 - 4= 3 1/3 - 4

x = -2/3

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Need to find what X is......56.4/14.4 = x/10.8

k0ka [10]

Answer:

x = 42.3.

Step-by-step explanation:

56.4 /14.4 = x / 10.8

Cross multiply:

14.4x = 56.4 * 10.8

x = (56.4 * 10.8) / 14.4

x = 42.3

6 0

4 years ago

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12) A candy company used 8 gallons of syrup to make 4 batches of candy. What is the rate of syrup per batch? Please show work

kupik [55]

It should be 8 divided by 4 which is 2:)

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3 years ago

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Alborosie

Answer:

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Step-by-step explanation:

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4 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:

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