The solution to that equation is p=500. What does the solution mean?

Rewrite in simplest radical form! Show your work.

vazorg [7]

$x^{\frac{1}{\frac{-3}{6} }}$

First, let's deal with the fraction in the denominator of the exponent. Multiply the top and bottom of the exponent by 6.

$x^{\frac{6}{-3} }$

Now that the fraction in the denominator is taken care of, we can reduce the denominator.

$x^{-2}$ . Some professors might accept this as simplest form, but others might ask you to get rid of the negative.

$x^{-2} = \frac{1}{x^{2} }$

7 0

2 years ago

Read 2 more answers

Lori recently quit her job and decided to start her own company, which will sell makeup to small beauty salons. What should she

skad [1K]

She should "Decide how to price her product". Which is response B.

This is because all the other options don't really make sense. You can't advertise something that you don't even know how much it costs. We've also already decided what the company sells. And we can't decide where the product will be sold unless we have made up the price for the product!

Hope this helps!

3 0

2 years ago

Rewrite each expression as a product of two or quotient of two powers

Paul [167]

Answer:

The product is $5^{40}*3^{70}$

Step-by-step explanation:

To do this we must understand how the basic operations apply to powers. If the power has the same base and we multiply them, the base remains the same while we sum the powers, if we divide them the base remains the same and we subtract the powers, if we try to make a power of a power the two powers multiply. Applying this rule we have:

$(\frac{5^8*3^7}{5^4})^{10}\\(\frac{5^8}{5^4}*3^7)^{10}\\(5^4*3^7)^{10}\\5^{40}*3^{70}$

6 0

2 years ago

Determine the domain of each piece. Domain of piece 1:[) Domain of piece 2: [) Domain of piece 3:[

zheka24 [161]

Answer:

Domain of piece 1:[) -4,-1

Domain of piece 2: [) -1,1

Domain of piece 3:[] 1,5

Step 2:

Rule for piece 1: y=-x

Rule for piece 2: y=1

Rule for piece 3: y=2-x

Step-by-step explanation:

Correct on Edgen

6 0

2 years ago

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A large tank is partially filled with 100 gallons of fluid in which 20 pounds of salt is dissolved. Brine containing 1 2 pound o

Valentin [98]

Answer:

47.25 pounds

Step-by-step explanation:

$\dfrac{dA}{dt}=R_{in}-R_{out}$

<u>First, we determine the Rate In</u>

Rate In=(concentration of salt in inflow)(input rate of brine)

$=(0.5\frac{lbs}{gal})( 6\frac{gal}{min})\\R_{in}=3\frac{lbs}{min}$

Change In Volume of the tank, $\frac{dV}{dt}=6\frac{gal}{min}-4\frac{gal}{min}=2\frac{gal}{min}$

Therefore, after t minutes, the volume of fluid in the tank will be: 100+2t

Rate Out=(concentration of salt in outflow)(output rate of brine)

$R_{out}=(\frac{A(t)}{100+2t})( 4\frac{gal}{min})\\\\R_{out}=\frac{4A(t)}{100+2t}$

Therefore:

$\dfrac{dA}{dt}=3-\dfrac{4A(t)}{100+2t}\\\\\dfrac{dA}{dt}=3-\dfrac{4A(t)}{2(50+t)}\\\\\dfrac{dA}{dt}=3-\dfrac{2A(t)}{50+t}\\\\\dfrac{dA}{dt}+\dfrac{2A(t)}{50+t}=3$

This is a linear differential equation in standard form, therefore the integrating factor:

$e^{\int \frac{2}{50+t}dt}=e^{2\ln|50+t|}=e^{\ln(50+t)^2}=(50+t)^2$

Multiplying the DE by the integrating factor, we have:

$(50+t)^2\dfrac{dA}{dt}+(50+t)^2\dfrac{2A(t)}{50+t}=3(50+t)^2\\\{(50+t)^2A(t)\}'=3(50+t)^2\\$Taking the integral of both sides\\\int \{(50+t)^2A(t)\}'= \int 3(50+t)^2\\(50+t)^2A(t)=(50+t)^3+C $ (C a constant of integration)\\Therefore:\\A(t)=(50+t)+C(50+t)^{-2}$

Initially, 20 pounds of salt was dissolved in the tank, therefore: A(0)=20

$20=(50+0)+C(50+0)^{-2}\\20-50=C(50)^{-2}\\C=-\dfrac{30}{(50)^{-2}} =-30X50^2=-75000$

Therefore, the amount of salt in the tank at any time t is:

$A(t)=(50+t)-75000(50+t)^{-2}$

After 15 minutes, the amount of salt in the tank is:

$A(15)=(50+15)-75000(50+15)^{-2}\\=47.25$ pounds$

8 0

1 year ago