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2 years ago

Holly Cuts 3 pans of brownies into eighths. How many 1/8 size brownie pieces does she have?

Mathematics

2 answers:

Anuta_ua [19.1K]2 years ago

5 0

She has 24 pieces, 8x3=24.
Plz rate 5 stars

s344n2d4d5 [400]2 years ago

3 0

She has 24 because 8x3 =24

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5. Simplify the expression, (-2 - 5i) - (-4+6i) 2 - 11i -2+i -6 + 11i -5i

monitta

$\qquad\qquad\huge\underline{{\sf Answer}}♨$

Let's simplify ~

$\qquad \sf \dashrightarrow \:( - 2 - 5i) - ( - 4 + 6i)$

$\qquad \sf \dashrightarrow \: - 2 - 5i + 4 - 6i)$

$\qquad \sf \dashrightarrow \: - 2 + 4 - 5i- 6i)$

$\qquad \sf \dashrightarrow \:2 - 11i$

Therefore, A is the Correct choice !

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2 years ago

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2 years ago

All factors of x cubed minus 6x squared plus 11x minus 6

miv72 [106K]

Answer:

Step-by-step explanation:

x³-6x²+11x-6

put x=1

1³-6×1²+11×1-6=1-6=11-6=0

by synthetic division

1| 1 -6 11 -6

| 1 -5 6

|----------------

| 1 -5 6 |0

x²-5x+6=0

x²-2x-3x+6=0

x(x-2)-3(x-2)=0

(x-2)(x-3)=0

x=1

x-1=0

so x³-6x²+11x-6=(x-1)(x-2)(x-3)

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3 years ago

Find the complex fourth roots of 81(cos(3pi/8) + i sin(3pi/8))

BartSMP [9]

By using De Moivre's theorem:

If we have the complex number ⇒ z = a ( cos θ + i sin θ)
∴ $\sqrt[n]{z} = \sqrt[n]{a} \ (cos \ \frac{\theta + 360K}{n} + i \ sin \ \frac{\theta +360k}{n} )$
k= 0, 1 , 2, ..... , (n-1)

For The given complex number ⇒ z = 81(cos(3π/8) + i sin(3π/8))


Part (A) find the modulus for all of the fourth roots

∴ The modulus of the given complex number = l z l = 81

∴ The modulus of the fourth root = $\sqrt[4]{z} = \sqrt[4]{81} = 3$

Part (b) find the angle for each of the four roots

The angle of the given complex number = $\frac{3 \pi}{8}$
There is four roots and the angle between each root = $\frac{2 \pi}{4} = \frac{\pi}{2}$
The angle of the first root = $\frac{ \frac{3 \pi}{8} }{4} = \frac{3 \pi}{32}$
The angle of the second root = $\frac{3\pi}{32} + \frac{\pi}{2} = \frac{19\pi}{32}$
The angle of the third root = $\frac{19\pi}{32} + \frac{\pi}{2} = \frac{35\pi}{32}$
The angle of the fourth root = $\frac{35\pi}{32} + \frac{\pi}{2} = \frac{51\pi}{32}$

Part (C): find all of the fourth roots of this

The first root = $z_{1} = 3 ( cos \ \frac{3\pi}{32} + i \ sin \ \frac{3\pi}{32})$
The second root = $z_{2} = 3 ( cos \ \frac{19\pi}{32} + i \ sin \ \frac{19\pi}{32})$

The third root = $z_{3} = 3 ( cos \ \frac{35\pi}{32} + i \ sin \ \frac{35\pi}{32})$
The fourth root = $z_{4} = 3 ( cos \ \frac{51\pi}{32} + i \ sin \ \frac{51\pi}{32})$

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3 years ago

Are the ratios 6:18 and 2:6 equivalent?

Ksju [112]

Yes because 6/18=1/3 and 2/6=1/3. Then 6/18=2/6

7 0

3 years ago

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