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Anettt [7]
3 years ago
8

This is correct right?↓

Mathematics
1 answer:
Doss [256]3 years ago
7 0

Answer:

udud8ducififidid

dudcucucuc

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2+2 how do i add this???????????????
emmasim [6.3K]

Answer:

4

Step-by-step explanation:

You must simply combine the two

6 0
3 years ago
  For the dilation with center​ (0,0) shown on the​ graph, your friend says the scale factor is 8/3
olchik [2.2K]

Answer:

It actually depends on where the point of letter is

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 this fraction?
e-lub [12.9K]

\sqrt{361}=19\ \text{because}\ 19^2=361

\dfrac{361}{\sqrt{361}}=\dfrac{361}{19}=19

19 is

A: real

B: whole

C: integer

D: rational

5 0
3 years ago
For f(x) = -4x+2 find f(x) when x = -1 A. -2. B. 6 C. -4 D. 2
djverab [1.8K]

Answer:

The answer is letter B which is 6.

f(x) = -4x+2

f(x)=-4(-1)+2

f(x)=4+2

f(x)=6

5 0
3 years ago
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