Answer: 5, -4, 2.6, and 100
Step-by-step explanation:
Answer is <
Why? 1/4 is smaller than 1/2
1/4 can be 4 slices while 1/2 can be 2 slices
2 slices > 4 slices
Answer:
2.
Step-by-step explanation:
For #2, another way to word this question is: For which of the following trig functions is π/2 a solution? Well, go through them one by one. If you plug π/2 into sinθ, you get 1. This means that when x is π/2, y is 1. Try and visualize that. When y is 1, that means you moved off the x-axis; so y = sinθ is NOT one of those functions that cross the x-axis at θ = π/2. Go through the rest of them. For y = cos(π/2), you get 0. At θ = π/2, this function crosses the x-axis. For y = tanθ, your result is undefined, so that doesn't work. Keep going through them. You should see that y = secθ is undefined, y = cscθ returns 1, and y = cotθ returns 0. If you have a calculator that can handle trig functions, just plug π/2 into every one of them and check off the ones that give you zero. Graphically, if the y-value is 0, the function is touching/crossing the x-axis.
Think about what y = secθ really means. It's actually y = 1/(cosθ), right? So what makes a fraction undefined? A fraction is undefined when the denominator is 0 because in mathematics, you can't divide by zero. Calculators give you an error. So the real question here is, when is cosθ = 0? Again, you can use a calculator here, but a unit circle would be more helpful. cosθ = π/2, like we just saw in the previous problem, and it's zero again 180 degrees later at 3π/2. Now read the answer choices.
All multiples of pi? Well, our answer looked like π/2, so you can skip the first two choices and move to the last two. All multiples of π/2? Imagine there's a constant next to π, say Cπ/2 where C is any number. If we put an even number there, 2 will cut that number in half. Imagine C = 4. Then Cπ/2 = 2π. Our two answers were π/2 and 3π/2, so an even multiple won't work for us; we need the odd multiples only. In our answers, π/2 and 3π/2, C = 1 and C = 3. Those are both odd numbers, and that's how you know you only need the "odd multiples of π/2" for question 3.
