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3 years ago

Hair color. |. Clothing

Mathematics

1 answer:

Andru [333]3 years ago

7 0

Answer:

16 possible hair/clothing combinations

Step-by-step explanation:

1) Brown-dress; 2)Brown-jumpsuit; 3)Brown-pajamas ; 4)Brown-Jeans; 5)Black-Dress; 6)black jumpsuit; 7)black pajamas; 8)black jeans; 9)blonde dress; 10)blond jumpsuit; 11)blonde pajamas; 12)blond jeans; 13)red dress; 14)red jumpsuit; 15)red pajamas; 16)red jeans

Or just multiply 4x4 because there were four of each.

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3 years ago

A cell phone provider offers a plan that costs $30 per month plus $0.10 per text message sent or received. A compatible plan cos

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3 years ago

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igor_vitrenko [27]

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3 years ago

PLEASE HELP ASAP!!!!! IM CONFUSED AND GOT IT WRONG

wariber [46]

Answer:

Area of the gym floor = 100.8 square meters

Step-by-step explanation:

Area of the gym floor = Area of rectangle A + Area of rectangle B - Area of the overlap

Area of rectangle A = 105 square meters

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4 0

3 years ago

Read 2 more answers

(1-i)^2 find 4 th root

sergiy2304 [10]

By de Moivre's theorem,

$1 - i = \sqrt2\,e^{-i\pi/4} \implies (1-i)^2 = 2\,e^{-i\pi/2}$

$\implies \sqrt[4]{(1 - i)^2} = \sqrt[4]{2}\,e^{i(2\pi k-\pi/2)/4} = \sqrt[4]{2}\,e^{i(4k-1)\pi/8}$

where $k\in\{0,1,2,3\}$ . The fourth roots of $(1-i)^2$ are then

$k = 0 \implies \sqrt[4]{2}\,e^{-i\pi/8}$

$k = 1 \implies \sqrt[4]{2}\,e^{i3\pi/8}$

$k = 2 \implies \sqrt[4]{2}\,e^{i7\pi/8}$

$k = 3 \implies \sqrt[4]{2}\,e^{i11\pi/8}$

or more simply

$\boxed{\pm\sqrt[4]{2}\,e^{-i\pi/8} \text{ and } \pm\sqrt[4]{2}\,e^{i3\pi/8}}$

We can go on to put these in rectangular form. Recall

$\cos^2(x) = \dfrac{1 + \cos(2x)}2$

$\sin^2(x) = \dfrac{1 - \cos(2x)}2$

Then

$\cos\left(-\dfrac\pi8\right) = \cos\left(\dfrac\pi8\right) = \sqrt{\dfrac{1 + \cos\left(\frac\pi4\right)}2} = \sqrt{\dfrac12 + \dfrac1{2\sqrt2}}$

$\sin\left(-\dfrac\pi8\right) = -\sin\left(\dfrac\pi8\right) = -\sqrt{\dfrac{1 - \cos\left(\frac\pi4\right)}2} = -\sqrt{\dfrac12 - \dfrac1{2\sqrt2}}$

$\cos\left(\dfrac{3\pi}8\right) = \sin\left(\dfrac\pi8\right) = \sqrt{\dfrac12 - \dfrac1{2\sqrt2}}$

$\sin\left(\dfrac{3\pi}8\right) = \cos\left(\dfrac\pi8\right) = \sqrt{\dfrac12 + \dfrac1{2\sqrt2}}$

and the roots are equivalently

$\boxed{\pm\sqrt[4]{2}\left(\sqrt{\dfrac12 + \dfrac1{2\sqrt2}} - i\sqrt{\dfrac12 - \dfrac1{2\sqrt2}}\right) \text{ and } \pm\sqrt[4]{2}\left(\sqrt{\dfrac12 + \dfrac1{2\sqrt2}} + i \sqrt{\dfrac12 - \dfrac1{2\sqrt2}}\right)}$

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2 years ago