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2 years ago

8. The distance formula is d=rt,

Mathematics

2 answers:

WITCHER [35]2 years ago

7 0

252=45h
h=5.6 hours long

RoseWind [281]2 years ago

4 0

Answer:

t=5.6

Step-by-step explanation:

The given equation is d=rt.

WE are also given other information such as the distance (d) and the rate (r)

d=252 and r=45

So, we can plug this into the equation:

252=45t

Now, we divide 45 from both sides to find t.

252/45=t

t=5.6

You might be interested in

Simplify the ratio 70 pounds to 42 pounds to the lowest terms.

Ilia_Sergeevich [38]

Answer:
5/3

Explanation:
The given is:
70/42
We can start by dividing both the numerator and denominator by 7 as follows:
(70/7) / (42/7)
10 / 6
The result can be further simplified by dividing both the numerator and the denominator by 2 as follows:
(10/2) / (6/2)
This will give us:
5/3
The reached result cannot be further simplified.
Therefore:
70 pounds to 42 pounds is equivalent t 5 pounds to 3 pounds in the simplest form.

Hope this helps :)

7 0

2 years ago

Read 2 more answers

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:

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