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astraxan [27]
3 years ago
14

Please help I’m really confused and need to learn this

Mathematics
1 answer:
egoroff_w [7]3 years ago
6 0

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.

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Find two consecutive integers whose product is 50
GarryVolchara [31]

n, n+1 - two consecutive integers

n(n + 1) = 50     <em>use distributive property</em>

n² + n = 50     <em>subtract 50 from both sides</em>

n² + n - 50 = 0

-----------------------------------------------------

ax² + bx + c =0

if b² - 4ac > 0 then we have two solutions:

[-b - √(b² - 4ac)]/2a and [-b - √(b² + 4ac)]/2a

if b² - 4ac = 0 then we have one solution -b/2a

if  b² - 4ac < 0 then no real solution

----------------------------------------------------------

n² + n - 50 = 0

a = 1, b = 1, c = -50

b² - 4ac = 1² - 4(1)(-50) = 1 + 200 = 201 > 0 → two solutions

√(b² - 4ac) = √(201) - it's the irrational number

Answer: There are no two consecutive integers whose product is 50.

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3 years ago
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We know that
see the picture to better understand the problem
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