Question

@ikram002p

Answer 1

Now what?

Answer 2

do u know where this formula come from ?
its from phythatgorus

24^2 = x^2 + x^2
right ?

Answer 3

so
2x^2 - 24 ^2 =0

Answer 4

\(\Huge (\sqrt2 x - 24) (\sqrt2 x + 24)=0 \)
right ?

Answer 5

Then distribute?

Answer 6

Huh

Answer 7

no just take
\(\Huge \sqrt2 x - 24) =0 \) and
\(\Huge \sqrt2 x + 24) =0 \) ( but this one ignore it since it gives u negative length)

Answer 8

I don't know what you mean by just take that?

Answer 9

ohh
u have this quadratic equation
\(\Huge 2x^2-24^2=0\)
how would u solve it ?

Answer 10

square -24

Answer 11

2x^2-576

Answer 12

2x^2=576.. divide by 2? the square root?

Answer 13

16.97

Answer 14

yeah u right !
u got it !

Answer 15

So would I write sqrt288 or 16.97 as my answer?

Answer 16

@ikram002p

Answer 17

it says radical
so let it sqrt288
but simplify it can u do that ?

Answer 18

\[\Large a^2 + b^2 = c^2\]


\[\Large x^2 + x^2 = 24^2\]


\[\Large 2x^2 = 24^2\]


\[\Large 2x^2 = 576\]


\[\Large x^2 = \frac{576}{2}\]


\[\Large x^2 = 288\]


\[\Large x = \sqrt{288}\]


\[\Large x = \sqrt{144*2}\]


\[\Large x = \sqrt{144}*\sqrt{2}\]


\[\Large x = 12*\sqrt{2} \ ... \ \text{Exact Answer}\]


\[\Large x \approx 16.97 \ ... \ \text{Approximate Answer}\]

Answer 19

12*sqrt2 is what I would put?

Answer 20

@jim_thompson5910
she need it radical

Answer 21

yeah ur correct @vshiroky !

Answer 22

ok so ignore the last line of my steps

Answer 23

hopefully you see how I simplified that radical

Answer 24

do I have to put the * sign? of just 12sqrt2

Answer 25

\[\Large 12\sqrt{2}\] works too

Answer 26

since the 12 near the root implies multiplication

Answer 27

Can you help with this jim? I'm running out of time on my homework, i spent too long on that last one

Answer 28

@jim_thompson5910

Answer 29

again we'll use the pythagorean theorem

\[\Large a^2 + b^2 = c^2\]

\[\Large 3^2 + (\sqrt{55})^2 = c^2\]

\[\Large 9 + 55 = c^2\]

\[\Large 64 = c^2\]

\[\Large c^2 = 64\]

\[\Large c = \sqrt{64}\]

\[\Large c = 8\]

Answer 30

So the first is sqrt64 and the second is 8?

Answer 31

Can I post more here @jim_thompson5910

Answer 32

no they are both 8 since 8 is an exact answer and because it's a whole number, there's no real need to have an approximate answer

Answer 33

but that box is there if you happened to get something like 8.23 or something

Answer 34

how many more do you have?

Answer 35

Is that right? I have a few still

Answer 36

good, the punctation in there may need to be removed though (depending on how picky the computer is)

Answer 37

punctuation*

Answer 38

were you able to get anywhere?

Answer 39

No cause it has 4a that confuses me

Answer 40

\[\Large a^2 + b^2 = c^2\]

\[\Large a^2 + (4a)^2 = 9^2\]

\[\Large a^2 + 16a^2 = 81\]

how about now?

Answer 41

Still no

Answer 42

ok I'll keep going

Answer 43

\[\Large a^2 + 16a^2 = 81\]

\[\Large 17a^2 = 81\]

\[\Large a^2 = \frac{81}{17}\]

\[\Large a = \sqrt{\frac{81}{17}}\]

\[\Large a = \frac{\sqrt{81}}{\sqrt{17}}\]

\[\Large a = \frac{9}{\sqrt{17}}\]

\[\Large a = \frac{9*\sqrt{17}}{\sqrt{17}*\sqrt{17}}\]

\[\Large a = \frac{9*\sqrt{17}}{\sqrt{17*17}}\]

\[\Large a = \frac{9*\sqrt{17}}{\sqrt{17^2}}\]

\[\Large a = \frac{9\sqrt{17}}{17}\]

Answer 44

if \[\Large a = \frac{9\sqrt{17}}{17}\] then

\[\Large b = 4a\]

\[\Large b = 4*\frac{9\sqrt{17}}{17}\]

\[\Large b = \frac{4*9\sqrt{17}}{17}\]

\[\Large b = \frac{36\sqrt{17}}{17}\]

Answer 45

get anywhere?

Answer 46

No I think I'm just stressed cause I am running out of time

Answer 47

well you have to check to see if a^2 + b^2 = c^2 is true

Answer 48

c is always the longest side

Answer 49

if a^2 + b^2 = c^2 is true, then the triangle is a right triangle

Answer 50

I got 576+49=625

Answer 51

so is 24^2 + 7^2 = 25^2?

Answer 52

the left side turns into 625, so we have 625 = 625

Answer 53

So not a right angle and 625?

Answer 54

making 24^2 + 7^2 = 25^2 true

Answer 55

so we definitely have a right triangle

Answer 56

Ok and 625 is the next answer?

Answer 57

c is always the longest side

c is the hypotenuse

Answer 58

not c^2