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marshall27 [118]
3 years ago
13

What is sec 600 degrees

Mathematics
2 answers:
Flauer [41]3 years ago
8 0

Answer:

-1

Step-by-step explanation:

finlep [7]3 years ago
8 0

Answer:

-2

Step-by-step explanation:

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Find an nth degree polynomial function with real coefficients satisfying the given conditions..
gregori [183]

we know that if we have a polynomial with real coefients and one root is a+bi, another root is a-bi


so

some roots are

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teh factors are

(x-r1)(x-r2)

so

we can put them in and

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to keep the same coefients, we multiply whole thing by c

c(x^4-11x^3+37x^2+205x-732)

we don't know what c is

we know that

f(1)=250

f(1)=c(x^4-11x^3+37x^2+205x-732)

find c

f(1)=c(1^4-11(1^3)+37(1^2)+205(1)-732)=250

f(1)=c(-500)=250


-500c=250

divide both sides by -500

c=-1/2


the polynomial (factored with real coefients) is

f(x)=-0.5(x-3)(x+4)(2x^2-12x+61)

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So your equation is in the form (y-k)^2 = 4p(x-h), where focus is at (h+p,k) and vertex is at (h,k) If h = 4 from the vertex and h + p = 3 from the focus, p = -1 in your parabola. So your equation looks like (y--7)^2 = -4(x - 4). Clean that up and you're all set.
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