Will85237

Let's call the length of the side of the square "x".

We know that the area of a square is height times base. Because each side of the square has the same length, we can say the area is equal to \(x^2 \)

The perimeter is the sum of each side of the square, so it can written as \(4x\)

Because the area is equal to 5 times the perimeter, we can write the following equation and sove for x.

\(x^2=5\times4x\)

\(x^2=20x\)

\(x^2-20x=0\)

\(x(x-20)=0\)

\(x=20\)

Therefore, the length of one side is 20.

We use prefixes before units of measurement to signify that the value is multiplied by 10, or 1000, or 0.01, etc. We do this to avoid adding lots of zeroes if the number is particularly large or small.

There is a international system for this:

I'm assuming you mean the power of the engine is 25 W. "effect" is a very broad term.

Power is rate at which work can be done, shown by its formula below:

\(P=\frac{w}{t}\)

P is power, w is work, and t is time. We'd use the equation for the power and work in your question.

\(25=\frac{100}{t}\)

\(t=\frac{100}{25}\)

\(t=4s\)

The engine would take 4 seconds.

The ratio between voltage potential difference and current will be equal to resistance, assuming the circuit is ohmic.

Therefore, we can calculate current with the following equation:

\(R=\frac{V}{I}\)

Where R is resistance, V is potential difference, and I is current.

In your question's case we would use it as shown:

\(220=\frac{9}{I}\)

\(I=\frac{9}{220}\)

\(I=0.041A\)

(Please don't call it voltage. Volts are the units of potential difference. It's like calling distance "meterage" or time "secondage".)

You're going to need the formula for thermal energy:

\(Q=mc\Delta T\)

Where Q is thermal energy, m is mass, c is the specific heat capacity, and delta T is change in temperature.

SImply plug the variables you were given in question as shown (make sure they're all in the correct units!):

\(84=5\times4.18\times\Delta T\)

\(\Delta T=+4.02°C\)

Add this on to your original temperature, and you get 29°C.

First, we need to find the total number of moles in the combined solution.

Firstly, all values should be in the same units for volume. In your case, concentration is in M (AKA mol per dm^3), meaning we need to convert the volume's ml to dm^3 by dividing by 1000.

\(25 ml=0.025 mol dm^{-3}\)

\(50ml=0.05moldm^{-3}\)

Secondly, we multiply these volumes by the concentrations to find the number of moles in your combined solution.

\(0.025\times0.1=0.0025mol\)

\(0.05\times0.1=0.005mol\)

\(0.0025+0.005=0.0075mol\)

Thirdly, we find the combined volume by adding the two volumes together.

\(0.025+0.05=0.075dm^{3}\)

Finally, we divide the final number of moles by the final volume to give the final concentration.

\(0.0075/0.075=0.1M\)

In this case, because the concentrations of the initial solutions were identical, the final concetration is the same. However, this method can be used for solutions with different concentrations.

The easiest way to find vertical assymptotes is to graph the function, and look for "invisible" vertical lines which never seem to be crossed over.

If graphing isn't an option, you can do it by looking at the function. Because vertical assymptotes exist when a certain value for x is impossible, look for parts of the formula where it breaks if x is a a certain value. The most common example would be division by 0.

\(\frac{1}{x-1}\)

If x is equal to 1, the demoninator of the fraction would be 0, which makes the fraction impossible. There would therefore be an assymptote at x=1.

You'll need to use logarithms. I'm going to assume they're a new concept for you.

They're a bit like the opposite of exponents.

Logarithms can rearrange an exponent as shown below.

\(b=a^x\)

\(log_ab=x\)

This would be log base a of b equals x

You may have seen a "log" button on a calculator (the one on this website in fact!), and that's what we use to solve these equations.

So for your question, it would be.

\(0.56586=0.94^x\)

\(log_{0.94}0.56586=x\)

So x is equal to 9.2025.

Calculators can be a little funny when trying to change the log base. If there is no option for the log base, it will be log base 10.

I'm guessing you want to simplify this. You can treat the "<" as an equal sign and apply basic algebra unless you multiply by -1.

\(4a+4<12a\)

minus 4a from each side

\(4<8a\)

divide by 8

\(0.5<a\)

"a" must be greater than one half.

\(-1/2=44-x/12\)

Multiply by -12.

\(6=-528+x\)

Add 528.

\(534=x\)