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VMariaS [17]
3 years ago
5

A city has two water towers. One tower holds 7.35 x 10 with an exponent of 5, gallons of water and the other tower holds 9.78x 1

0 with an exponent of 5, gallons of water. What is the combined water capacity of the two towers?
Mathematics
2 answers:
elena55 [62]3 years ago
8 0
Sum with scientific notation. We just ignore the 10^5, because the two numbers have the same exponent and repeat it on the result of the normal sum between the numbers.

But, for example, if we had:

-> 5 x 10^4 + 5 x 10^5

We would need to equal the exponents to do the sum without having to multiply all those tens.

5 x 10^4 + 50 x 10^4
= 55 x 10^4

or 0.5 x 10^5 + 5 x 10^5
= 5.5 x 10^5

*They are the same results.

Now returning to the question:

-> 7.35 + 9.78 = 17.13
-> 17.13 x 10^5

You can prove it yourself:

-> 7.35 x 10^5 = 735,000

-> 9.78 x 10^5 = 978,000

-> 735,000 + 978,000 = 1,713,000

Or: 1713 x 10^3 or 17.13 x 10^5

Answer: The combined capacity of the two tower is 17.13 x 10^5 gallons of water.
Pavlova-9 [17]3 years ago
5 0

Answer:

the answer is actually 1.713x10^6

Step-by-step explanation:

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Kipish [7]

Answer:

So you would not be carried away by the tides

Step-by-step explanation:

hope this is what u want

3 0
3 years ago
A 1000-liter (L) tank contains 500 L of water with a salt concentration of 10 g/L. Water with a salt concentration of 50 g/L flo
djverab [1.8K]

Answer:

a) y(t)=50000-49990e^{\frac{-2t}{25}}

b) 31690.7 g/L

Step-by-step explanation:

By definition, we have that the change rate of salt in the tank is \frac{dy}{dt}=R_{i}-R_{o}, where R_{i} is the rate of salt entering and R_{o} is the rate of salt going outside.

Then we have, R_{i}=80\frac{L}{min}*50\frac{g}{L}=4000\frac{g}{min}, and

R_{o}=40\frac{L}{min}*\frac{y}{500} \frac{g}{L}=\frac{2y}{25}\frac{g}{min}

So we obtain.  \frac{dy}{dt}=4000-\frac{2y}{25}, then

\frac{dy}{dt}+\frac{2y}{25}=4000, and using the integrating factor e^{\int {\frac{2}{25}} \, dt=e^{\frac{2t}{25}, therefore  (\frac{dy }{dt}+\frac{2y}{25}}=4000)e^{\frac{2t}{25}, we get   \frac{d}{dt}(y*e^{\frac{2t}{25}})= 4000 e^{\frac{2t}{25}, after integrating both sides y*e^{\frac{2t}{25}}= 50000 e^{\frac{2t}{25}}+C, therefore y(t)= 50000 +Ce^{\frac{-2t}{25}}, to find C we know that the tank initially contains a salt concentration of 10 g/L, that means the initial conditions y(0)=10, so 10= 50000+Ce^{\frac{-0*2}{25}}

10=50000+C\\C=10-50000=-49990

Finally we can write an expression for the amount of salt in the tank at any time t, it is y(t)=50000-49990e^{\frac{-2t}{25}}

b) The tank will overflow due Rin>Rout, at a rate of 80 L/min-40L/min=40L/min, due we have 500 L to overflow \frac{500L}{40L/min} =\frac{25}{2} min=t, so we can evualuate the expression of a) y(25/2)=50000-49990e^{\frac{-2}{25}\frac{25}{2}}=50000-49990e^{-1}=31690.7, is the salt concentration when the tank overflows

4 0
3 years ago
Which of the following values are in the range of the function graphed below? check all that apply
vodomira [7]

Answer:

Option D. 1

Option E. 1/2

Step-by-step explanation:

we know that

Looking at the graph

The domain is the interval ----> [-1,1]

-1\leq x\leq 1

The domain is all real numbers greater than or equal to -1 and less than or equal to 1

The range is the interval ----> [0,1]

0\leq y\leq 1

The range is all real numbers greater than or equal to 0 and less than or equal to 1

therefore

The values that are in the range are

1 and 1/2

5 0
3 years ago
Read 2 more answers
I’m soooo sorry but can y’all plzzz help me just a couple more times!
slava [35]

Answer:

26/35

Step-by-step explanation:

Each fraction is increasing by 1/35. Therefore, 25/35 + 1/35 = 26/35.

6 0
3 years ago
Read 2 more answers
A fish was swimming LaTeX: 2\frac{1}{2}2 1 2 feet below the water's surface at 1:00 p.M. Three hours later, the fish was at a de
marin [14]

Answer:

\frac{23}{4} feet

Step-by-step explanation:

The depth of the fish to the surface of water at 1:00 pm = 2 \frac{1}{2} feet

                                                                             =   \frac{5}{2} feet

The depth of the fish to its initial position at 4:00 pm = 3 \frac{1}{4} feet

                                                                              = \frac{13}{4} feet

The position of the fish with respect to the water's surface at 4:00 pm = the depth of the fish to the surface of water at 1:00 pm + the depth of the fish to its initial position at 4:00 pm

                                        =  \frac{5}{2} + \frac{13}{4}

                                        = \frac{10 + 13}{4}

                                         = \frac{23}{4}

The position of the fish with respect to the water's surface at 4:00 pm is \frac{23}{4} feet.

8 0
3 years ago
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