Which expression is equivalent to 21 + square root(-36)
help
Which? Do we have some options?
yes
You can simplify the radical. \[\Large\rm \sqrt{-36}=\sqrt{-1\cdot 36}=\sqrt{-1}\cdot \sqrt{36}=\mathcal i \sqrt{36}\]Is 36 a prefect square? Can we take it's root?
27i 21 + 6i 21 - 6i 21i
no and yes
lol that should have been yes and yes you silly billy ^^
6*6 is 36 yes? So sqrt36 = 6
lol
thats it
Hmm so which option does it look like? :U
c
21-6i
Hmm why the subtraction? :o
u change signs
\[\Large\rm =21 + \sqrt{-36}\]\[\Large\rm =21 + \sqrt{-1}\cdot \sqrt{36}\]The negative is what's turning into the i,\[\Large\rm =21 + \mathcal i\cdot \sqrt{36}\]So we shouldn't have a negative anymore. Not sure what you meant by `change signs` :d
okay thanks thats much better
Simplify the following expression: i2 X i5
Are those exponents or not?\[\Large\rm \mathcal i^2\cdot \mathcal i^5\]
yes
sorry
Applying rules of exponents:\[\Large\rm \color{royalblue}{x^a\cdot x^b=x^{a+b}}\]See how we can use this rule?
yes
how can we use this rule
You said yes -_- I thought you had it... \[\Large\rm \mathcal i^2\cdot \mathcal i^5=\mathcal i ^{2+5}=\mathcal i^{7}\] Now this next part is a little tricky... Since we have a power on i greater than 4, it means this can be simplified.
okay so it will be i
Let's back up to the original problem... \[\Large\rm \mathcal i^2\cdot \mathcal i^5\]Powers of i will repeat over and over. Every 4th power. So i^5 is the same as i^1 That tells us that i^5=i Our problem simplifies to: \[\Large\rm \mathcal i^2\cdot \mathcal i\]Do you understand how to calculate i^2?
yes ik that part
i^2 is -1 right? So it looks like we end up with:\[\Large\rm -1\cdot \mathcal i=-\mathcal i\]Powers of i can be pretty tricky! :(
yes ik i had trouble with these problems
Which expression is equivalent to 21 + square root(-36)
A sphere has a diameter of 4 meters. What is the approximate volume of the sphere.
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