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Mathematics 7 Online
OpenStudy (anonymous):

solve sqrt 6x-1= sqrt 5

OpenStudy (anonymous):

$$\sqrt{6x-1}=\sqrt5$$Notice we can transform the equation by squaring both sides:$$6x-1=5$$

OpenStudy (anonymous):

6x-1=25

OpenStudy (anonymous):

Square both sides:\[\bf \sqrt{6x-1} = \sqrt{5} \rightarrow 6x-1=5\]Re-arrange and solve for x:\[\bf 6x = 6 \implies x = 1\]In this case we don't have any extraneous solutions since there is only one x-value so we don't need to worry about that.

OpenStudy (anonymous):

i have none othe above x = 5 x = 4 x = –1 None of the above

OpenStudy (anonymous):

That will be correct.

OpenStudy (anonymous):

how about simplify i^11

OpenStudy (anonymous):

-1?

OpenStudy (anonymous):

Do you remember the pattern? Observe the following:\[\bf i^1 = (\sqrt{-1})^1=i\]\[\bf i^2=(\sqrt{-1})^2=-1\]\[\bf i^3=(\sqrt{-1})^3=i^2*i=-1*\sqrt{-1}=-i\]\[\bf i^4 = (\sqrt{-1})^4=i^2*i^2=-1*-1=1\]\[\bf i^5=(\sqrt{-1})^5=i^4*i=1*i=i\]And we are back to where we started. Do you see the pattern? Can you predict the 11th power of i? @topstryker

OpenStudy (anonymous):

i ?

OpenStudy (anonymous):

@topstryker \(i^{11}=i^6i^4i=(i^2)^3(i^2)^2i=(-1)^3(-1)^2i=(-1)i=-i\)

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