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Misha Larkins [42]
3 years ago
7

Part A: Determine if the expression is one of our special products. If so, identify it by its general form:

Mathematics
1 answer:
kondaur [170]3 years ago
6 0

Answer:

1.

Part A: Yes, it is (a - b)².

Part B: a² - 2ab + b² => (x - 6)² = x² - 12x + 36.

Part C: x² - 12x + 36.

2.

Part A: Not a special product.

Part B: Binomial distribution => (x + 8)(x + 1) = x² + 9x + 8.

Part C: x² + 9x + 8.

3.

Part A: Yes, it is (a + b)²

Part B: a² + 2ab + b² => (3x + 2)² = 9x² + 12x + 4.

Part C: 9x² + 12x + 4.

4.

Part A: Yes, it is (a + b)(a - b), a difference of squares.

Part B: a² - b² => 4x² - 49

Part C: 4x² - 49

5.

Part A: Not a special product.

Part B: Binomial distribution => (x - 5)(2x - 5) = 2x² - 15x + 25.

Part C: 2x² - 15x + 25

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Answer: y=x+2

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ok, so the points from the graph in the picture that i used were (0, 2) and (-2, 0).

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2 years ago
A sheet of paper 90 cm-by-66 cm is made into an open box (i.e. there's no top), by cutting x-cm squares out of each corner and f
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Answer:

26 - \sqrt{181} cm

Step-by-step explanation:

The volume of the box is:

V = height * length * width

V = x*(66 - 2*x)*(90 - 2*x)

V = (66*x - 2*x^2)*(90 - 2*x)

V = 5940*x - 132*x^2 - 180*x^2 + 4*x^3

V = 4*x^3 - 312*x^2 + 5940*x

where x is the length of the sides of the squares,  in cm.

The mathematical problem is :

Maximize: V = 4*x^3 - 312*x^2 + 5940*x

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x > 0

2*x < 66 <=> x < 33

In the maximum, the first derivative of V, dV/dx, is equal to zero

dV/dx = 12*x^2 - 624*x + 5940

From quadratic formula

x = \frac{-b \pm \sqrt{b^2 - 4(a)(c)}}{2(a)}

x = \frac{624 \pm \sqrt{(-624)^2 - 4(12)(5940)}}{2(12)}

x = \frac{624 \pm \sqrt{104256}}{24}

x = \frac{624 \pm \sqrt{2^6*3^2*181}}{24}

x = \frac{624 \pm 8*3*\sqrt{181}}{24}

x_1 = \frac{624 + 24*\sqrt{181}}{24}

x_1 = 26 + \sqrt{181}

x_2 = \frac{624 - 24*\sqrt{181}}{24}

x_2 = 26 - \sqrt{181}

But x_1 > 33, then is not the correct answer.

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