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disa [49]
3 years ago
12

Multiply the polynomials: (x + 3)(x2 – 5x + 12)

Mathematics
2 answers:
LekaFEV [45]3 years ago
3 0

Answer:

x³ - 2x² - 3x + 36

Step-by-step explanation:

(x + 3)(x² - 5x + 12)

Each term in the second factor is multiplied by each term in the first factor , that is

x(x² - 5x + 12) + 3(x² - 5x + 12) ← distribute parenthesis

= x³ - 5x² + 12x + 3x² - 15x + 36 ← collect like terms

= x³ - 2x² - 3x + 36

yKpoI14uk [10]3 years ago
3 0
I hope you understood!! :)

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Find the equation of a sphere if one of its diameters has endpoints: (-9, -12, -6) and (11, 8, 14).
baherus [9]

Answer:

Hence, the equation of a sphere with one of its diameters with endpoints (-9, -12, -6) and (11, 8, 14) is (x-1)^{2}+(y+2)^{2}+(z-4)^{2} = 30.

Step-by-step explanation:

There are two kew parameters for a sphere: Center (h, k, s) and Radius (r). The radius is the midpoint of the line segment between endpoints. That is:

C(x,y,z) = \left(\frac{-9+11}{2},\frac{-12+8}{2},\frac{-6+14}{2}   \right)

C(x,y,z) = (1,-2,4)

The radius can be found by halving the length of diameter, which can be determined by knowning location of endpoints and using Pythagorean Theorem:

r = \frac{1}{2}\cdot \sqrt{(-9-11)^{2}+(-12-8)^{2}+(-6-14)^{2}}

r = 10\sqrt{3}

The general formula of a sphere centered at (h, k, s) and with a radius r is:

(x-h)^{2}+(y-k)^{2}+(z-s)^{2} = r^{2}

Hence, the equation of a sphere with one of its diameters with endpoints (-9, -12, -6) and (11, 8, 14) is (x-1)^{2}+(y+2)^{2}+(z-4)^{2} = 30.

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3 years ago
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y = m(x -x1) + y1 \\  \\ y = 2(x - ( - 2)) + 5 \\  \\ y = 2(x + 2) + 5 \\  \\ y = 2x + 4 + 5 \\  \\ y = 2x + 9
8 0
3 years ago
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