Question

Need Help!

Answer 1

D

Answer 2

it is D @zab505

Answer 3

Why?

Answer 4

@tkhunny

Answer 5

This is an eyeball problem. Just look at them. (1,1) is WAY too far out.

Answer 6

the Unit Circle, recall is a "Unit" circle, thus it has a radius of 1 thus any x, y on it will always give 1 as hypotenuse thus keep in mind that \(\bf c^2=a^2+b^2\implies r^2=x^2+y^2\implies {\color{blue}{ r}}=\sqrt{x^2+y^2}\implies {\color{blue}{ 1}}=\sqrt{x^2+y^2}\)

Answer 7

hmm anyhow \(\bf c^2=a^2+b^2\implies r^2=x^2+y^2\implies {\color{blue}{ r}}=\sqrt{x^2+y^2}\\\quad\\ \implies {\color{blue}{ 1}}=\sqrt{x^2+y^2}\)

so you're given 2 coordinates... both of them squared and summed up, should give a 1 once square rooted

Answer 8

I tried that, but it only worked with one and there is always 2 answers

Answer 9

well... + and - really

Answer 10

so you're expected to have 2 non-matching ones?

Answer 11

well... I assume that just means \(\large \bf {\color{blue}{ 1}}={\color{red}{ \pm}}\sqrt{x^2+y^2}\)

Answer 12

I'm expected to have 2 correct answers

Answer 13

so... which one did you get?

Answer 14

did you try all 4 choices?

Answer 15

https://www.google.com/search?client=opera&q=sqrt((0.8)%5E2%2B(-0.6)%5E2)&sourceid=opera&ie=utf-8&oe=utf-8&channel=suggest&gws_rd=ssl <----

Answer 16

Yeah I got the first one, but only that one

Answer 17

ok let us try the 2nd one then one sec

Answer 18

\(\bf \large {
\left(-\frac{2}{3},\frac{\sqrt{5}}{3}\right)
\\ \quad \\
{\color{blue}{ r}}=\pm \sqrt{\left(-\frac{2}{3}\right)^2+\left(\frac{\sqrt{5}}{3}\right)^2}\implies {\color{blue}{ r}}=\pm \sqrt{\frac{2^2}{3^2}+\frac{\sqrt{5^2}}{3^2}}
\\ \quad \\
{\color{blue}{ r}}=\pm \sqrt{\frac{4}{9}+\frac{\sqrt{5}}{9}}\implies {\color{blue}{ r}}=\pm \sqrt{?}
}\)

Answer 19

hmmm lemme fix that a bit

Answer 20

\(\bf \large {
\left(-\frac{2}{3},\frac{\sqrt{5}}{3}\right)
\\ \quad \\
{\color{blue}{ r}}=\pm \sqrt{\left(-\frac{2}{3}\right)^2+\left(\frac{\sqrt{5}}{3}\right)^2}\implies {\color{blue}{ r}}=\pm \sqrt{\frac{2^2}{3^2}+\frac{\sqrt{5^2}}{3^2}}
\\ \quad \\
{\color{blue}{ r}}=\pm \sqrt{\frac{4}{9}+\frac{5}{9}}\implies {\color{blue}{ r}}=\pm \sqrt{?}
}\)

Answer 21

1

Answer 22

yeap

Answer 23

Wait so what it is asking for is that they =1?

Answer 24

and if so then they are the correct answers in this question?

Answer 25

well... pretty much yes recall that \(\bf {\color{blue}{ r}}=\pm \sqrt{\frac{9}{9}}\to \pm\sqrt{1}\to \pm 1\)

Answer 26

\(\bf {\color{brown}{ 1^2\to 1\cdot 1\to 1}}
\\ \quad \\

\sqrt{1}\to \sqrt{1^2}\to 1\)

Answer 27

So A and B

Answer 28

yeap the other ones are either below or above 1

Answer 29

Thank you so much!

Answer 30

yw

Answer 31

Is it asking for which one is or isn't?

Answer 32

well.. is asking on "which of the following points could \(\bf {\color{brown}{ not}}\) be points in the circle?"

Answer 33

so... yes...A and B ARE IN the circle...the other ones are \(\bf {\color{brown}{ not}}\)

Answer 34

So the answers would be c and d

Answer 35

?

Answer 36

yeap