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Lorico [155]
3 years ago
11

Question Find the sum horizontally. (3x – x°- 2x?)+(-x° - 3x) =

Mathematics
1 answer:
Trava [24]3 years ago
6 0

Answer:

2x

Step-by-step explanation:

you take 3x-x you get 2x. then you take 2x_2x you get 0.then -x_3x you get 2x.you take 0+2x you get 2x that is the answer

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Please help so I can make my family proud
Levart [38]

Answer:

A = 83.5m^2

Step-by-step explanation:

9 * 19 = 171m for the complete rectangle

9 - 4 = 5

5 * 19 = 95

95 / 2 = <u>47.5m</u> for the triangles

9 * 4 = <u>36m</u>

36 + 47.5 = <u>83.5m</u>

<em><u>Plz mark as brainliest if correct! Have a nice day!!! </u></em>

<em><u>-Lil G</u></em>

8 0
2 years ago
The graph of the equation y=x^2-7 is symmetric with respect to which of the following?a.the line y=-x+7b.the y-axis c.the x axis
lys-0071 [83]

Answer:

b

Step-by-step explanation:

since coefficient of x = 0

5 0
3 years ago
Which of the following are square roots of —8 + 8i/3? Check all that apply.
8090 [49]

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.

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