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dolphi86 [110]
3 years ago
13

Find the product (7 lbs 12 oz) × 3

Mathematics
2 answers:
ohaa [14]3 years ago
7 0
I dont understand other than multiplying that. 23.25 lbs 
GenaCL600 [577]3 years ago
7 0

23 lbs 4oz is your answer

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Which of the following are square roots of —8 + 8i/3? Check all that apply.
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Answer:

Options (2) and (3)

Step-by-step explanation:

Let, \sqrt{-8+8i\sqrt{3}}=(a+bi)

(\sqrt{-8+8i\sqrt{3}})^2=(a+bi)^2

-8 + 8i√3 = a² + b²i² + 2abi

-8 + 8i√3 = a² - b² + 2abi

By comparing both the sides of the equation,

a² - b² = -8 -------(1)

2ab = 8√3

ab = 4√3 ----------(2)

a = \frac{4\sqrt{3}}{b}

By substituting the value of a in equation (1),

(\frac{4\sqrt{3}}{b})^2-b^2=-8

\frac{48}{b^2}-b^2=-8

48 - b⁴ = -8b²

b⁴ - 8b² - 48 = 0

b⁴ - 12b² + 4b² - 48 = 0

b²(b² - 12) + 4(b² - 12) = 0

(b² + 4)(b² - 12) = 0

b² + 4 = 0 ⇒ b = ±√-4

                     b = ± 2i

b² - 12 = 0 ⇒ b = ±2√3

Since, a = \frac{4\sqrt{3}}{b}

For b = ±2i,

a = \frac{4\sqrt{3}}{\pm2i}

  = \pm\frac{2i\sqrt{3}}{(-1)}

  = \mp 2i\sqrt{3}

But a is real therefore, a ≠ ±2i√3.

For b = ±2√3

a = \frac{4\sqrt{3}}{\pm 2\sqrt{3}}

a = ±2

Therefore, (a + bi) = (2 + 2i√3) and (-2 - 2i√3)

Options (2) and (3) are the correct options.

6 0
3 years ago
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3/4=6/8

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Step-by-step explanation: I just did a test and got 100%

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Find the surface area of regular prism with height 6 cm, if the base of the prism is a Regular quadrilateral with side 4 cm.
marshall27 [118]
A "regular quadrilateral" is a square, so the length and width are both 4 cm. The surface area of a rectangular prism is given by
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The surface area of the prism is 128 cm².


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