if i= (sqrt)-1, then which expression is equal to 9i(13i)
A. -117
B. 117i
C. 117
D. -117i
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Nnesha (nnesha):
distribute parentheses by outside term
OpenStudy (anonymous):
so do you equal them like: 9i(13i)=(sqrt)-1
Nnesha (nnesha):
\[\huge\rm i = \sqrt{-1} \]
just i equal sqrt{-1}
multiply 9i(13i) it's not going to = sqrt-1}
OpenStudy (anonymous):
im confused...
Nnesha (nnesha):
okay first multiply 9i(13i) = ?
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OpenStudy (anonymous):
117i
Nnesha (nnesha):
or first you can replace i by sqrt -1
Nnesha (nnesha):
i times i = ??
OpenStudy (anonymous):
i^2
Nnesha (nnesha):
9 i times 13 i
9 times 13 and i times i
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Nnesha (nnesha):
yes right i^2 = -1
OpenStudy (anonymous):
117i^2 is the answer??
Nnesha (nnesha):
\[\huge\rm 9\sqrt{-1}(13\sqrt{-1})\]
okay if we replace i by sqrt{-1}
right now multiply
what would u get when you multiply \[\huge\rm \sqrt{-1} \times \sqrt{-1}= ?\]
Nnesha (nnesha):
\[\huge\rm i^2 = i \times i \]
\[\large\color{red}{\rm i = \sqrt{-1}}\]
because i equal sqrt of -1
you can substitute sqrt -1 for i
when you multiply sqrt{-1} by sqrt{-1} both square would cancel each other out
so \[\sqrt{-1} \times \sqrt{-1} = -1\]
so that's why i^2 = -1
OpenStudy (anonymous):
okay.
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Nnesha (nnesha):
so replace i^2 bu -1
Nnesha (nnesha):
y*
OpenStudy (anonymous):
so that answer is 117i
Nnesha (nnesha):
i^2 = ???
OpenStudy (anonymous):
-1
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