Recall the sum identity for cosine:
cos(a + b) = cos(a) cos(b) - sin(a) sin(b)
so that
cos(a + b) = 12/13 cos(a) - 8/17 sin(b)
Since both a and b terminate in the first quadrant, we know that both cos(a) and sin(b) are positive. Then using the Pythagorean identity,
cos²(a) + sin²(a) = 1 ⇒ cos(a) = √(1 - sin²(a)) = 15/17
cos²(b) + sin²(b) = 1 ⇒ sin(b) = √(1 - cos²(b)) = 5/13
Then
cos(a + b) = 12/13 • 15/17 - 8/17 • 5/13 = 140/221
You need to multiply 30 by 6 so 30times6 =your answer
Thet contradict each other, that's why both of them are incorrect.
<span>Suppose that a polynomial has four roots: s, t, u, and v. If the polynomial were evaluated at any of these values, it would have to be zero. Therefore, the polynomial can be written in this form.
p(x)(x - s)(x - t)(x - u)(x - v), where p(x) is some non-zero polynomial
This polynomial has a degree of at least 4. It therefore cannot be cubic.
Now prove Kelsey correct. We have already proved that there can be no more than three roots. To prove that a cubic polynomial with three roots is possible, all we have to do is offer a single example of that. This one will do.
(x - 1)(x - 2)(x - 3)
This is a cubic polynomial with three roots, and four or more roots are not possible for a cubic polynomial. Kelsey is correct.
Incidentally, if this is a roller coaster we are discussing, then a cubic polynomial is not such a good idea, either for a vertical curve or a horizontal curve. I hope this helps</span><span>
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You can use substitution and solve y-x=2 for y which is y= x+ 2 and plug it into the y of the other equasion and you get x=5 y=7
