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Taya2010 [7]
3 years ago
12

Please help me I need the help (with options)

Mathematics
2 answers:
Anna [14]3 years ago
7 0

Answer:

d

Step-by-step explanation:

diamong [38]3 years ago
6 0

Answer:

option=d.

hkm=jkl( by vertical opposite angle)

155°=(5x+15)°

155-15=5x

140=5x

140/5=x

28=x

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please mark this answer as brainlist

3 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.

6 0
3 years ago
What is the value of 4.5 x 10^5 written in standard form?
soldier1979 [14.2K]

Answer:

4500000

Step-by-step explanation: Let me know if this helped

3 0
3 years ago
I PROMISE BRAINIEST IF YOU HELP ME
Elan Coil [88]

Answer:

The associative property

Step-by-step explanation:

:/ wait its u again lol

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What is the cumulative area from the left under the curve for a z-score of -0.875? What is the area on the right of that z-score
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Step-by-step explanation:

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3 years ago
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