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natali 33 [55]
3 years ago
12

Plz, help me if u cant answer both could you just answer one of them for me.

Mathematics
2 answers:
Serggg [28]3 years ago
8 0

Answer: the answer is 3.14 squared

Then divide by 6

Step-by-step explanation:

labwork [276]3 years ago
3 0

Answer:

for 2. I don't think it can be a triangle

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120<br> -5<br><br><br> Integer hellp me
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Answer: 115 is the answer

Step-by-step explanation:

7 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
Please help! due tomorrow
Soloha48 [4]
6v +12 -5v=v +12.
4b +32+2= 4b+34
like that?
5 0
3 years ago
Read 2 more answers
Help get all this points pls
balandron [24]
It would be b.) y=5x+5
8 0
3 years ago
Read 2 more answers
Find the circumference pls help.
Triss [41]
Equation for circumference is c = 2(pi)r

r = 1/2(d)

d = 20

So…. r = 1/2(20) = 20/2 = 10

Going back to circumference equation

c = 2(pi)(10) = 2(3.14)(10) = 62.8
6 0
2 years ago
Read 2 more answers
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