<span>difference of squares formula:
</span>a^2<span> – b^</span>2<span> = (a + b)(a – b)
</span>so answers are
<span>(3 + xz)(–3 + xz)
</span><span>(y2 – xy)(y2 + xy)
</span><span>(64y2 + x2)(–x2 + 64y2)
</span><span>
cause
</span><span>(3 + xz)(–3 + xz)
= </span><span>(xz + 3 )(xz - 3)
= x^2z^2 - 9
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</span>(y^2 – xy)(y^2 + xy)
= y^4 -x^2y^2
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<span>(64y^2 + x^2)(–x2 + 64y^2)
=</span>(64y^2 + x2)(64 - x^2)<span>
= 64^2 y^4 - x^4</span>
Sin(pi/6) would be appropriate as that is 1/2
Remember you can' do anything to boht sides, but when times negative, flip sign
distributive property
a(b+c)=ab+ac
distribute the -4
-8+4x
now we have
2x<8-8+4x
2x<4x
minus 2x both sides
0<2x
therefor
0<x
answer is x>0
There is a decrease in power. It can be concluded that there has also been a change in:
A) current or voltage.