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How do you multiply #(-13u+v^2)^2#?

Answer 1

Genesis Allen

#v^4-26uv^2+169u^2#

#(v^2-13u)^2# #(v^2)^2-2(v^2)(13u)+(13u)^2# #v^4-26uv^2+169u^2#

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Answer 2

Hazel Anglin

The formula for #(a+b)^2# is #a^2+2ab+b^2#. Let's apply this to our problem.

#(-13u+v^2)^2# = #(-13u)^2+2(-13u)(v^2)+(v^2)^2#

#-13u*-13u = 169u^2# #2(-13u)(v^2) = -26uv^2# #v^2*v^2 = v^4#

Therefore, our answer is #169u^2-26uv^2+v^4#.

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Answer 3

Nathan Axel

#169u^2-26uv^2+v^4#

In general #color(white)("XXX")(color(red)a+color(blue)b)^2=color(red)a^2+2color(red)acolor(blue)b+color(blue)b^2#

So #color(white)("XXX")(color(red)(-13u)+color(blue)v^2))^2#

#color(white)("XXXXXXXX")=(color(red)(-13u))^2+2 * (color(red)(-13u)) * color(blue)(v^2) + (color(blue)(v^2))^2#

#color(white)("XXXXXXXX")=169u^2-26uv^2+v^4#

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Answer 4

Ethan Adkins

To multiply ((-13u+v^2)^2), you expand the expression using the formula for squaring a binomial: ((a+b)^2 = a^2 + 2ab + b^2). So, the result is (169u^2 - 26uv^2 + v^4).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.