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salantis [7]
2 years ago
8

Write 25×10^6 in standard form. Please can someone help me

Mathematics
2 answers:
Dovator [93]2 years ago
8 0

Answer:

2.5*10^7

Step-by-step explanation:

Standard form is always a number between 1 and 10 (but never exactly 10) multiplied by 10 to the power of something. Just divide the 25 by 10, so it can be used in standard form, and increase the power by 1. Hope this helped! :^)

Lemur [1.5K]2 years ago
4 0
Answer:
2.5 • 10^6

Explanation:
In order for 25 • 10^6 to be written in standard form, the ‘25’ needs to be less than 10. To do that we will move the decimal one place to the left so that 25 is now 2.5; it is now a number less than 10. It should now look like 2.5 • 10^6. It is now in standard form.
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A swimming pool with a volume of 30,000 liters originally contains water that is 0.01% chlorine (i.e. it contains 0.1 mL of chlo
SpyIntel [72]

Answer:

R_{in}=0.2\dfrac{mL}{min}

C(t)=\dfrac{A(t)}{30000}

R_{out}= \dfrac{A(t)}{1500} \dfrac{mL}{min}

A(t)=300+2700e^{-\dfrac{t}{1500}},$  A(0)=3000

Step-by-step explanation:

The volume of the swimming pool = 30,000 liters

(a) Amount of chlorine initially in the tank.

It originally contains water that is 0.01% chlorine.

0.01% of 30000=3000 mL of chlorine per liter

A(0)= 3000 mL of chlorine per liter

(b) Rate at which the chlorine is entering the pool.

City water containing 0.001%(0.01 mL of chlorine per liter) chlorine is pumped into the pool at a rate of 20 liters/min.

R_{in}=(concentration of chlorine in inflow)(input rate of the water)

=(0.01\dfrac{mL}{liter}) (20\dfrac{liter}{min})\\R_{in}=0.2\dfrac{mL}{min}

(c) Concentration of chlorine in the pool at time t

Volume of the pool =30,000 Liter

Concentration, C(t)= \dfrac{Amount}{Volume}\\C(t)=\dfrac{A(t)}{30000}

(d) Rate at which the chlorine is leaving the pool

R_{out}=(concentration of chlorine in outflow)(output rate of the water)

= (\dfrac{A(t)}{30000})(20\dfrac{liter}{min})\\R_{out}= \dfrac{A(t)}{1500} \dfrac{mL}{min}

(e) Differential equation representing the rate at which the amount of sugar in the tank is changing at time t.

\dfrac{dA}{dt}=R_{in}-R_{out}\\\dfrac{dA}{dt}=0.2- \dfrac{A(t)}{1500}

We then solve the resulting differential equation by separation of variables.

\dfrac{dA}{dt}+\dfrac{A}{1500}=0.2\\$The integrating factor: e^{\int \frac{1}{1500}dt} =e^{\frac{t}{1500}}\\$Multiplying by the integrating factor all through\\\dfrac{dA}{dt}e^{\frac{t}{1500}}+\dfrac{A}{1500}e^{\frac{t}{1500}}=0.2e^{\frac{t}{1500}}\\(Ae^{\frac{t}{1500}})'=0.2e^{\frac{t}{1500}}

Taking the integral of both sides

\int(Ae^{\frac{t}{1500}})'=\int 0.2e^{\frac{t}{1500}} dt\\Ae^{\frac{t}{1500}}=0.2*1500e^{\frac{t}{1500}}+C, $(C a constant of integration)\\Ae^{\frac{t}{1500}}=300e^{\frac{t}{1500}}+C\\$Divide all through by e^{\frac{t}{1500}}\\A(t)=300+Ce^{-\frac{t}{1500}}

Recall that when t=0, A(t)=3000 (our initial condition)

3000=300+Ce^{0}\\C=2700\\$Therefore:\\A(t)=300+2700e^{-\dfrac{t}{1500}}

3 0
3 years ago
The number -7/5 is ___ than -1 1/5 because -7/5 lies ____ -1 1/5 on a vertical number line
Airida [17]

greater, above I hope this helps out!

7 0
2 years ago
Read 2 more answers
Which equation is true?
nikitadnepr [17]

Answer:

i think its the 3rd option

5 0
3 years ago
25 POINTS!!!!! REAL ANSWERS ONLY!!!!!!!! I WILL REPORT LINKS AND FAKE/WRONG ANSWERS!!!!!!!​
Nonamiya [84]

Answer:

7.

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curved surface area=2[area oftraingle]=2×1/2×4×3

=12cm²

now

total surface area=72cm²+12cm²=84cm²

8.

volume of traingular prism=area of traingle ×length

=1/2×12×3.4×4=81.6cm³

3 0
2 years ago
Please help me please
alexandr1967 [171]
I believe it is the second one
8 0
3 years ago
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