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antoniya [11.8K]
2 years ago
8

G(x)= (x^2 +4x-12)(x-3) find the roots

Mathematics
1 answer:
KengaRu [80]2 years ago
8 0

Answer:

x = 3 or x = 2 or x = -6

Step-by-step explanation:

Solve for x:

(x - 3) (x^2 + 4 x - 12) = 0

Split into two equations:

x - 3 = 0 or x^2 + 4 x - 12 = 0

Add 3 to both sides:

x = 3 or x^2 + 4 x - 12 = 0

The left hand side factors into a product with two terms:

x = 3 or (x - 2) (x + 6) = 0

Split into two equations:

x = 3 or x - 2 = 0 or x + 6 = 0

Add 2 to both sides:

x = 3 or x = 2 or x + 6 = 0

Subtract 6 from both sides:

Answer: x = 3 or x = 2 or x = -6

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3 0
3 years ago
Select the functions that have a value of -1. sin180° cos180° tan180° csc180° sec180° cot180°
kupik [55]

We have to break each degree in terms of 90

A) sin180^\circ=sin(90\times2+0)

Which is in third quadrant, therefore sine is negative hence

sin(90\times2+0)= -sin0 ^\circ = 0


B) cos180^\circ =cos(90\times2+0)

Which is in third quadrant, therefore cosine is negative hence

cos(90\times2+0)= -cos0^\circ  = -1


C) tan180^\circ=tan(90\times2+0)

Which is in third quadrant, therefore tangent is positive hence

tan(90\times2+0)= tan0^\circ  = 0


D) csc180^\circ=csc(90\times2+0)

Which is in third quadrant, therefore cosec is negative hence

cosec(90\times2+0)= -csc0^\circ  =not defined


E)sec180^\circ=sec(90\times2+0)

Which is in third quadrant, therefore secant is negative hence

sec(90\times2+0)= -sec0^\circ  = -1


F) cot180^\circ=cot(90\times2+0)

Which is in third quadrant, therefore tangent is positive hence

cot(90\times2+0)= cot0^\circ = not defined


Hence only cos 180^\circ and

sec180^\circ have value -1

Hope this will help

7 0
3 years ago
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Answer:

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Step-by-step explanation:

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7 0
2 years ago
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Aleksandr-060686 [28]
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