AzizHusain

17. The amplitude is 2. Each wavelength is 6. When x = 0, y = 1.

The only one that matches x = 0, y = 1 is e, f(x) = 1 + sin x since all the others don't satisfy this condition.

19. f(x) = 2x^2 + 3

f(x+h) = 2(x+h)^2 + 3 = 2x^2 + 4xh + 2h^2 + 3 -->

2x^2 + 4xh + 2h^2 + 3 - (2x^2 + 3) / h -->

4xh + 2h^2 / h -->

b, 4x + 2h.

First, let's create variables:

s = startup division

e = established division

The startup division made 80% of 1/2 of the established division. 

Let's take a look at what the above statement means: Making 80% of 1/2 means...

0.8 of 1/2 = 0.8*0.5 = 0.4 --> s = 0.4e.

Next: The startup division’s growth was 1/3 greater than the established division’s.

Seems like this is additional information is not needed.

If the divisions made 280,000 combined, how much did the startup division make?

s + e = 280,000 --> s = 280,000 - e, then we can substitude s = 0.4e for s -->

280,000 - e = 0.4e --> 1.4e = 280,000 --> e = 200,000.

This means that s = 80,000.

Hi, PlanetHunter21!

There are a total of 12 b***s of which 5 are black.

With replacement: 

$$\frac{5}{12}*\frac{5}{12}=\mathbf{\frac{25}{144}}.$$

speed = wavelength*frequency

wavelength = speed / frequency

wavelength = 3.0*10^8 m/s / 8.01*10^8 s^-1= 0.3745318352059925 = 3.75*10^-1.

± means you apply both addition and subtraction.

Near the end of the quadratic formula, you will end up with a general solution such as: (# ± #) / #.

This means you have to apply both the + and the -, which gives you two answers rather than just 1.

Example: 5 ± 3 = both 8 and 2.

A decimal such as 0.54 would be written as $$5.4*10^{-1}$$.

You have to figure out a way to remove the decimal and so you try placing it in different places. By placing the decimal in between the 5 and 4, we have to use the 10^ notation to correctly identify the original decimal place. So, we would have to move the decimal one place to the left to obtain the original decimal, 0.54. Since we moved it to the left once, we would add a -1 to 10^ giving us $$10^{-1}$$.

Depending on how many times you move the decimal and the direction, your notation of 10^ will change.

For this problem, we would NOT use $$54*10^{-2}$$ because it is incorrect. You can think of having at least 2 numbers such as 2.3, 8.6, 9.0, etc. and then *10^(#) or if you have more significant figures then 2.35, 8.67, etc. But, you notice the decimal is always after the first number, so that is an easy way to convert.

So 0.0057 would start as 5.7...

0.000013 would start as 1.3...

13.46 would start as 1.3 or 1.35...[depending on how many significant figures are required]

In any case, it is easy to remember #.##### as a general rule.

log(3,6)*log(9,6)+log(4,6)*log(18,6) =

log(3,6)*[log(3,6) + log(3,6)] + log(4,6)*[log(3,6) + log(3,6) + log(2,6)]

log(3,6) = ~0.613, log(4,6) = ~0.774, & log(2,6) = ~0.387.

So, 0.613*[0.613 + 0.613] + 0.774*[0.613 + 0.613 + 0.387]

= 0.752 + 1.248 

= 2.

u=-5i+3j, v=-7i-6j, w=-3i+12j 

These are not 3-dimensional vectors, so this must be just a dot product problem.

u dot w = (-5*(-3)) + (3*12) = 15 + 36 = 51.

v dot w = (-7*(-3)) + (-6*12) = 21 - 72 = -51.

51 + (-51) = 0.

EDIT: Forgot to add, thanks Phill. 

8.50% of 15% 

0.0850 of 0.15 

0.0850 = x/0.15 

x = 0.0850*0.15

x = 0.01275.