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4 years ago

PLEASEEEE HELPPPPP!!!

Mathematics

1 answer:

Taya2010 [7]4 years ago

4 0

(1, 3.5 ) and (3, 3.5 )

the endpoints of the midsegment are at the midpoints of DE and DF

using the midpoint formula with D(0, 7), E(2, 0), F(6, 0)

midpoint of DE = [ $\frac{1}{2}$ (0 + 2), $\frac{1}{2}$ (7 + 0)] = (1, 3.5)

midpoint of DF = [ $\frac{1}{2}$ (0 + 6), $\frac{1}{2}$ (7 + 0)] = (3, 3.5)

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4 years ago

What is the product?

DENIUS [597]

Answer:

14 x^5 - 21 x^4 - 266 x^3 - 315 x^2

Step-by-step explanation:

Expand the following:

7 x^2 (2 x + 5) (x^2 - 4 x - 9)

| | | | 2 x | + | 5

| | x^2 | - | 4 x | - | 9

| | | | -18 x | - | 45

| | -8 x^2 | - | 20 x | + | 0

2 x^3 | + | 5 x^2 | + | 0 | + | 0

2 x^3 | - | 3 x^2 | - | 38 x | - | 45:

7 x^2 2 x^3 - 3 x^2 - 38 x - 45

7 x^2 (2 x^3 - 3 x^2 - 38 x - 45) = 7 x^2 (2 x^3) + 7 x^2 (-3 x^2) + 7 x^2 (-38 x) + 7 x^2 (-45):

7 2 x^2 x^3 - 3 7 x^2 x^2 - 38 7 x^2 x - 45 7 x^2

7 (-45) = -315:

7 2 x^2 x^3 - 3 7 x^2 x^2 - 38 7 x^2 x + -315 x^2

7 x^2 (-38) x = 7 x^(2 + 1) (-38):

7 2 x^2 x^3 - 3 7 x^2 x^2 + -38×7 x^(2 + 1) - 315 x^2

2 + 1 = 3:

7 2 x^2 x^3 - 3 7 x^2 x^2 - 38 7 x^3 - 315 x^2

7 (-38) = -266:

7 2 x^2 x^3 - 3 7 x^2 x^2 + -266 x^3 - 315 x^2

7 x^2 (-3) x^2 = 7 x^4 (-3):

7 2 x^2 x^3 + -3×7 x^4 - 266 x^3 - 315 x^2

7 (-3) = -21:

7 2 x^2 x^3 + -21 x^4 - 266 x^3 - 315 x^2

7 x^2×2 x^3 = 7 x^(2 + 3)×2:

7×2 x^(2 + 3) - 21 x^4 - 266 x^3 - 315 x^2

2 + 3 = 5:

7 2 x^5 - 21 x^4 - 266 x^3 - 315 x^2

7×2 = 14:

Answer: 14 x^5 - 21 x^4 - 266 x^3 - 315 x^2

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4 years ago