Alan

If the domain is from x=0 to x = 4 then the range of y = x² is from y = 0 to y = 16.  It includes all real numbers not just integers.

It often helps to draw a graph:

KOH --> K⁺ + OH^-

pH + pOH = 14

pOH = -log₁₀([OH^-]) = -log₁₀(7.4*10^-2)

pH = 14 - (-log₁₀(7.4*10^-2))

$${\mathtt{pH}} = {\mathtt{14}}{\mathtt{\,\small\textbf+\,}}{log}_{10}\left({\mathtt{7.4}}{\mathtt{\,\times\,}}{{\mathtt{10}}}^{-{\mathtt{2}}}\right) \Rightarrow {\mathtt{pH}} = {\mathtt{12.869\: \!231\: \!719\: \!730\: \!976\: \!2}}$$

or pH ≈ 12.9

Depends what information you are given, but the equation of a straight line can be written as y = m*x + c where m is the slope and c is the y-intercept (i.e. the value of y when x = 0).

I assume you have a function f(x) = 2x² - x and you want to find it's value when x = -3.  You could just do this in the calculator here.

(-3)² = 9 and -(-3) = +3, so f(-3) = 2*9 + 3 = 21

My view is as follows (though the question isn't really about mathematics at all - it's more about presentation!):

Which of the following best describes the graph?

1.  It will not be spread out horizontally across the entire coordinate plane because in Step 4, Nancy selected an incorrect scale on the axis.  This is true (though I wouldn't have used the word "incorrect") but the "incorrectness" is relatively small.

2.  It will not be spread out vertically across the entire coordinate plane because in Step 5, Nancy selected an incorrect scale on the y-axis. This is true and the "incorrectness" is relatively large. 

3.  It will not be spread out horizontally across the entire coordinate plane because in Step 2, Nancy plotted time instead of temperature on the axis.  This is plain nonsense!

4.  It will not be spread out vertically across the entire coordinate plane because in Step 3, Nancy plotted temperature instead of time on the y-axis.  This is also plain nonsense!

So, of the options on offer, I think 2. best describes the graph (but it's a really silly question!!).

sin x = -√(1-(2/9)²)  It has to be negative because cos x is positive and tan x is negative only in the fourth quadrant, where sin x is negative.

$${\mathtt{sinx}} = {\mathtt{\,-\,}}{\sqrt{{\mathtt{1}}{\mathtt{\,-\,}}{\left({\frac{{\mathtt{2}}}{{\mathtt{9}}}}\right)}^{{\mathtt{2}}}}} \Rightarrow {\mathtt{sinx}} = -{\mathtt{0.974\: \!996\: \!043\: \!043\: \!569\: \!1}}$$

Here's the integral:

Haven't a clue how it's obtained though!  

(Because it's an indefinite integral you should add a constant, not shown above.)

Here's a graph over a section of the domain and range.

Can you figure it out using this?

Don’t be so amazed, Melody. That happens every once in a while. LOL   Ditto!

...the M1 and F1 should still switch—shouldn’t they?   M1 must always stay in the same place, otherwise the result is the equivalent of a rotation of one of the other arrangements.

One thing about hyper-geometric distributions, they hyper-stretch one's brain.  Hear hear!  That's why I like to draw pictures where possible!