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Mathematics 11 Online
OpenStudy (calculusxy):

MEDAL!!! Question will be posted below.

OpenStudy (calculusxy):

\[\huge v^2 - 20 = 0 \]

OpenStudy (calculusxy):

Solve with quadratic formula

Nnesha (nnesha):

ok have you tried anything yet ?

OpenStudy (calculusxy):

yes gimme a moment

Nnesha (nnesha):

okay

OpenStudy (anonymous):

\[\frac{ -b \pm \sqrt{b^2-4ac}}{ 2a } \]

OpenStudy (calculusxy):

\[\frac{ -0 \pm \sqrt{0 - 4(-20)} }{ 2 }\]

OpenStudy (anonymous):

yes

OpenStudy (calculusxy):

i thought the new equation would be like: \[v^2 + 0a - 20 = 0\]

OpenStudy (calculusxy):

\[\frac{ -0 \pm 8.9 }{ 2 }\]

OpenStudy (anonymous):

0v would be better, but yes. v^2 +0v -20 = 0

Nnesha (nnesha):

v^2 -20 is same as `v^2 -0v -20`

Nnesha (nnesha):

well would be great if get the final answer in radical

Nnesha (nnesha):

don't take square of 20 factor it out

Nnesha (nnesha):

square root*

OpenStudy (calculusxy):

the answer key for this in decimal form...

Nnesha (nnesha):

oh okay then.

OpenStudy (calculusxy):

\[\frac{ -0 \pm 8.9 }{ 2 } = 4.45 \]

OpenStudy (anonymous):

remember it's \[\pm \]

OpenStudy (calculusxy):

okay so -4.45 or 4.45

OpenStudy (anonymous):

yes

Nnesha (nnesha):

^ye right it's \[\rm a \pm b ~= a-b ~~ and ~a+b\]

OpenStudy (calculusxy):

is my answer correct?

OpenStudy (anonymous):

yes it is. if your key says 4.4721... it's only because you cut off some decimals from 8.9

Nnesha (nnesha):

i didn't get 4.4 `5`

OpenStudy (anonymous):

but you did solve it correctly.

OpenStudy (calculusxy):

ok thank you :)

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