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OpenStudy (anonymous):
:) & no can you explain
OpenStudy (anonymous):
its B
jigglypuff314 (jigglypuff314):
like how √5 √3 can be combined to equal √15
√(14q) 2√(4q) can be combined to equal 2√(14 * 4 * q * q)
which by the same rule can be separated into 2 √(4) √(q*q) √(14)
and yeah it's B :)
OpenStudy (anonymous):
oh thank you, yay can you help me some more :/
OpenStudy (anonymous):
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jigglypuff314 (jigglypuff314):
so to solve for a for a = √(-2a) square both sides
to get a² = -2a then add 2a to both sides
so a² + 2a = 0 then factor out an a
so a(a+2) = 0 what would a equal?
OpenStudy (anonymous):
D
jigglypuff314 (jigglypuff314):
mmm not quite :)
once you find what a equals
plug it into the original equation
if you get something like 2 = -2 then the number you plugged in is extraneous
OpenStudy (anonymous):
hmmm i think A
jigglypuff314 (jigglypuff314):
mmm
if you plug in 0 into a = √(-2a)
you would get 0 = √(0)
simplifies to 0 = 0 which is true
so not A :)
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OpenStudy (anonymous):
hmm :/
OpenStudy (anonymous):
C :D
jigglypuff314 (jigglypuff314):
correct! :)
OpenStudy (anonymous):
yayy :D
OpenStudy (anonymous):
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jigglypuff314 (jigglypuff314):
for this, think of the square roots like variables
what would you get if you added 4x + 5x = ?
OpenStudy (anonymous):
9
OpenStudy (anonymous):
D
jigglypuff314 (jigglypuff314):
yep! :D
OpenStudy (anonymous):
yayyyy :D
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jigglypuff314 (jigglypuff314):
√(10/121) can be like (√10) / (√121)
so then what is √121 = ?
OpenStudy (anonymous):
11
jigglypuff314 (jigglypuff314):
yep :)
so you're left with √10 / 11
OpenStudy (anonymous):
C
jigglypuff314 (jigglypuff314):
yep :)
and I gtg now, talk to ya later! It's been nice working with you <3
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