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3 years ago

Answer correct for brainliest

Mathematics

1 answer:

AVprozaik [17]3 years ago

3 0

Answer:

D. Both a relation & a function.

Step-by-step explanation:

A relation is a set of one or more ordered pairs. A function is a relation in which each input has exactly one output. A set of ordered pairs represents a function if, for every value of x, there is only one value of y. In the given situation, the x-coordinate is the number of weeks since she starts recording and the y-coordinate is the number of books read per week. It can be seen from the given set that each x-value corresponds with exactly one y-value. Therefore, the set of ordered pairs is best described as both a relation and a function.

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(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

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

7 0

2 years ago

I made 4 cakes using 4 cups of flour. How many cups of flour is using for<br> each cake?

maria [59]

Since it’s 4 cupcakes and 4 flours it’s 1 cup of flour per cupcake

7 0

3 years ago

PLEASE HELP!! S is the center of the circle. Suppose that JK=12, LK=12, NS=8, and PS= 2x - 6. Fine the following

Kazeer [188]

Answer:

x = 7

LP = 6

Step-by-step explanation:

JK = 12

LK = 12

NS = 8

PS = 2x - 6

radius bisects the chords JK and LK because it is perpendicular to them.

JK = JN + NK = 6 + 6 = 12

LK = LP + PK = 6 + 6 = 12

LP = 6

radius= root(6^2 + 8^2) = 10 then PS = 8 = 2x -6,

8 = 2x - 6

14 = 2x

x = 7

8 0

4 years ago

The circumference of a circle is 36 cm.

dusya [7]

18cm (A). Remember to get the radius when you have the circumference is to ALWAYS divide it by 2. 36/2 = 18

6 0

3 years ago

X^2+14x+-55=0 help me ☺️

vodomira [7]

\left[X _{1}\right] = \left[ \left( -1\,i \right) \,\sqrt{\left( -55+14\,x\right) }\right][X1]=[(−1i)√(−55+14x)] I hope helping with u

3 0

3 years ago