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aleksandrvk [35]
4 years ago
15

Help who ever helps me i will mark them as brain liest i promise!!!​

Mathematics
1 answer:
castortr0y [4]4 years ago
8 0

Mini inequality crash course:

<em />

<em>Is it greater than or less then?</em>

This can be a bit tricky to remember. How I do this is looking at the shape of the sign?

So the number that is BIGGER is on the side where the mouth opens or the WIDDER side. Biggest number goes on widest side

Bigger number goes here--->   __ >

< __ <--- Bigger number goes here

The number that is the SMALLEST goes on the side where the "pointy" part is

Smallest number goes here---> ___<

> ___ <---Smallest number goes here

***There's an image bellow that also explains it. Hopefully it makes sense

(This all applies the same way to ≥ and ≤ )

You read and say inequalities from left to right. For example I will use example 6:

s < -6 ...............….......S is less then -6

-6 > s..........................-6 is greater then s

^^^Notice that these show the exact same thing yet they are said differently

Here is another example, but with some easy numbers I made up:

3 < 6.........Three is less then six

6 > 3.........Six is greater then three

There are two kind of inequalities:

≥ / ≤

Or

> / <

≥ / ≤

^^^ These two are the greater then or equal to / less then or equal to signs

This means that it includes the number.

***How I remember is the little bar underneath the symbol is like half an equal sign, so the numbers can be equal or not***

For example in number 5:

x ≤ 3 ---> x is less than or equal to 3

^^^This means that x can be three OR any number SMALLER then three

Another example:

x ≥ -34 ---> x is any number greater then OR equal to -34

Graphing inequalities on a number line:

The circles will always be placed on the number that the variable is greater then, less the, greater then or equal to, or less then or equal to

<em>How to tell if the circle is colored or not:</em>

If the inequality has either > or < then the circle is hollow/empty/not colored in

If the inequality has either ≥ or ≤ then the circle is shaded/colored in

***How I remember this is that the symbol has the extra bar at the bottom so the circle will have a little extra something in it

Which direction will the arrow go?

If it is x < # or x ≤ # or # > x or # ≥ x then the arrow will go to the left of the circle.  It signifies that x is any number less then/ less then or equal to the #

****How I remember it is if the variable is<em> L</em>ess then it goes <em>L</em>eft

If it is x ># or x ≥ #  or # < x or # ≤ xthen the arrow will go to the right of the circle.  It signifies that x is any number more then/ more then or equal to the #

16. She has 3,187 signatures but she still needs more, so the expression would be 3,187 + x (x being how many more signitures she needs to reach her goal)

She needs at least 5,000. It has to be more or equal to 5,000

So we have the equation:

5,000 ≤ 3,187 + x    or 3,187 + x ≥ 5,000

Now solve for x by subtracting 3,187 to both sides

x ≥ 1, 813

To reach her goal she has to still get 1,813 or more signitures

I really hope this helped and that you understand now!

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Z^4-5(1+2i)z^2+24-10i=0
mixer [17]

Using the quadratic formula, we solve for z^2.

z^4 - 5(1+2i) z^2 + 24 - 10i = 0 \implies z^2 = \dfrac{5+10i \pm \sqrt{-171+140i}}2

Taking square roots on both sides, we end up with

z = \pm \sqrt{\dfrac{5+10i \pm \sqrt{-171+140i}}2}

Compute the square roots of -171 + 140i.

|-171+140i| = \sqrt{(-171)^2 + 140^2} = 221

\arg(-171+140i) = \pi - \tan^{-1}\left(\dfrac{140}{171}\right)

By de Moivre's theorem,

\sqrt{-171 + 140i} = \sqrt{221} \exp\left(i \left(\dfrac\pi2 - \dfrac12 \tan^{-1}\left(\dfrac{140}{171}\right)\right)\right) \\\\ ~~~~~~~~~~~~~~~~~~~~= \sqrt{221} i \left(\dfrac{14}{\sqrt{221}} + \dfrac5{\sqrt{221}}i\right) \\\\ ~~~~~~~~~~~~~~~~~~~~= 5+14i

and the other root is its negative, -5 - 14i. We use the fact that (140, 171, 221) is a Pythagorean triple to quickly find

t = \tan^{-1}\left(\dfrac{140}{171}\right) \implies \cos(t) = \dfrac{171}{221}

as well as the fact that

0

\sin\left(\dfrac t2\right) = \sqrt{\dfrac{1-\cos(t)}2} = \dfrac5{\sqrt{221}}

(whose signs are positive because of the domain of \frac t2).

This leaves us with

z = \pm \sqrt{\dfrac{5+10i \pm (5 + 14i)}2} \implies z = \pm \sqrt{5 + 12i} \text{ or } z = \pm \sqrt{-2i}

Compute the square roots of 5 + 12i.

|5 + 12i| = \sqrt{5^2 + 12^2} = 13

\arg(5+12i) = \tan^{-1}\left(\dfrac{12}5\right)

By de Moivre,

\sqrt{5 + 12i} = \sqrt{13} \exp\left(i \dfrac12 \tan^{-1}\left(\dfrac{12}5\right)\right) \\\\ ~~~~~~~~~~~~~= \sqrt{13} \left(\dfrac3{\sqrt{13}} + \dfrac2{\sqrt{13}}i\right) \\\\ ~~~~~~~~~~~~~= 3+2i

and its negative, -3 - 2i. We use similar reasoning as before:

t = \tan^{-1}\left(\dfrac{12}5\right) \implies \cos(t) = \dfrac5{13}

1 < \tan(t) < \infty \implies \dfrac\pi4 < t < \dfrac\pi2 \implies \dfrac\pi8 < \dfrac t2 < \dfrac\pi4

\cos\left(\dfrac t2\right) = \dfrac3{\sqrt{13}}

\sin\left(\dfrac t2\right) = \dfrac2{\sqrt{13}}

Lastly, compute the roots of -2i.

|-2i| = 2

\arg(-2i) = -\dfrac\pi2

\implies \sqrt{-2i} = \sqrt2 \, \exp\left(-i\dfrac\pi4\right) = \sqrt2 \left(\dfrac1{\sqrt2} - \dfrac1{\sqrt2}i\right) = 1 - i

as well as -1 + i.

So our simplified solutions to the quartic are

\boxed{z = 3+2i} \text{ or } \boxed{z = -3-2i} \text{ or } \boxed{z = 1-i} \text{ or } \boxed{z = -1+i}

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Answer:

A

Step-by-step explanation:

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