Alan

I've just looked in the book myself.  The answers given are clearly wrong:

The first one balances, but the second one clearly doesn't!   It's not unusual for a book with so many equations/examples to contain a few errors.

Just type abs.  e.g. abs(-6) It will appear as |-6| = 6

$${\left|-{\mathtt{6}}\right|} = {\mathtt{6}}$$

I'll do a few for you:

I assume the first one is √(36x²) = √36*√x² = √6²*√x² = 6x

4√(32m⁷n⁹) = 4√32√m⁷√n⁹ = 4√(4²*2)*√(m⁶*m)*√(n⁸*n) = 4*4√2*m³√m*n⁴√n = 16m³n⁴√(2*m*n)

B. This is a tricky one!  5√(-5)*5*√(-2) = 5*i*√5*5*i*√2  where i is the "imaginary" number √(-1)

Since i² = -1, we have  5√(-5)*5*√(-2) = 25*i²*√5*√2 = -25√10

Like this?

Still don't get the same values as you quote!

You can use the mod function on the calculator here.

$$\left({{\mathtt{132}}}^{{\mathtt{28}}}\right) {mod} \left({\mathtt{19}}\right) = {\mathtt{1}}$$

NB You have to enter it in the form mod(132²⁸,19)

Hmm.  I'm not sure Maple agrees with me.  I based my reply above on reasoning about the two terms separately, and in this Maple agrees with me - see the first plot below (I've just used the natural log; a different base only alters the scaling, not the domain).

However, when I get Maple to plot the difference of the two terms it gives the following (which is identical to ln(x+8)):

So, Chris, it looks like the mathematical software big-hitters (Mathematica and Maple) think it's ok to combine terms before worrying about domain validity!

I'm guessing here, but perhaps this is asking for the probability density given by Student's t distribution at a value of 0.01, with 15 degrees of freedom.  If so, then it's 0.392.

In general, try to give more explanation so we know what you are asking - we're not mind readers!

Volume of one cylindrical peg = pi*r²*L  where r is the radius and L is the length.

Here we have r = 0.5 inches = 0.5*2.54 cm = 1.27 cm,  and L = 2 inches = 2*2.54 cm = 5.08 cm

The volume of one peg = pi*1.27²*5.08 cm³.

The mass of one peg is given by volume*density = pi*1.27²*5.08 cm³ * 8.05 g/cm³ = pi*1.27²*5.08*8.05 g

So the number of these you can get out of a block of mass 157.4kg is 154.7*1000/(pi*1.27²*5.08*8.05)

$${\frac{{\mathtt{157\,400}}}{\left({\mathtt{\pi}}{\mathtt{\,\times\,}}{{\mathtt{1.27}}}^{{\mathtt{2}}}{\mathtt{\,\times\,}}{\mathtt{5.08}}{\mathtt{\,\times\,}}{\mathtt{8.05}}\right)}} = {\mathtt{759.605\: \!010\: \!502\: \!234\: \!142\: \!2}}$$

Rounding down, this is 759 whole pegs (and this assumes we can reshape the block as necessary!).

Note that 157.4 kg is 157400 g, and I've converted to grams because the density is in terms of grams/cm³, not kg/cm³.

In many computer programming languages 9.0e9 can mean 9.0*10⁹ or 9 billion (often written as 9.0E9 as well). 

Indeed, this is what the calculator here thinks!