Subtract. Write your answer as a fraction in simplest form. 1/7

2.5 = n 3.5 2 What is the variable?

attashe74 [19]

N is the variable I believe.

8 0

4 years ago

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STUDY 1. Simplify the expression below 5p + 14p 4(x - 2) + 6x 2. Evaluate

melisa1 [442]

Answer:

(11)(B) Simplify numeric and algebraic expressions using the laws of exponents, including integral ... Evaluate the expression for d = -2, d = 0, and d = 1.

Step-by-step explanation:

3 0

3 years ago

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Why are halves a good choice for benchmark fractions for 1 1/3

Natalka [10]

Answer and explanation:

Benchmark fractions are fractions that are used as references in measuring other fractions. They are easily estimated and so can be used in measuring more "specific" fractions such as 1/5, 7/9, 3/7, 1/3 etc. If I wanted to measure 1 1/3cm for instance using a calibrated ruler, having centimeter measurements, I would first find 1cm on the ruler and then find half of one centimeter. Seeing that half is bigger than 1/3 but close, I could then estimate 1/3 to be somewhere less than 1/2 but a bit close to it

3 0

3 years ago

Solve each equation 19. <img src="https://tex.z-dn.net/?f=%20%7B2%7D%5E%7Bx%20%3D%2016%7D%20" id="TexFormula1" title=" {

Natasha2012 [34]

How many 2 s are in 16.
4

X=4

8 0

3 years ago

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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

Subtract. Write your answer as a fraction in simplest form. 1/7 - (-4/7)= ?