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vesna_86 [32]
3 years ago
8

Which products result in a perfect square trinomial? Select three. (-x+9)(-x-9), (xy+x)(xy+x), (2x-3)(-3+2x), (16-x^2)(x^2-16),

(4y^2+25)(25+4y^2)
Mathematics
1 answer:
GrogVix [38]3 years ago
3 0

Answer:

(2x-3)(-3+2x),(16-x^2)(x²-16),4y^2+25)(25+4y^2)

Step-by-step explanation:

I know that the rest of the products didn't make a perfect square cause they don´t aren´t squared,I also know that they are the odd ones out among the rest of the product so yeah,I am not too sure if I got it right...sorry if I didn´t.

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What are the vertices of AA'B'C'if AABC is dilated by a scale factor of 4?
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Answer:

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Step-by-step explanation:

Assuming the dilatation is centred at the origin, then multiply each of the coordinates by 4 , that is

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Josh and Corey met up for lunch. When they were done, Josh traveled 8 blocks east and Corey traveled 4 blocks west. How many blo
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Answer:

12

Step-by-step explanation:

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3 years ago
Please help? I’m super lost...
babunello [35]

Answer:

Step-by-step explanation:

In all of these problems, the key is to remember that you can undo a trig function by taking the inverse of that function.  Watch and see.

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Take the inverse sin of both sides.  When you do that, you are left with just 2theta on the left.  That's why you do this.

sin^{-1}(sin2\theta)=sin^{-1}(-\frac{\sqrt{3} }{2} )

This simplifies to

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We look to the unit circle to see which values of theta give us a sin of -square root of 3 over 2.  Those are:

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7a=tan^{-1}(1)

We look to the unit circle to find which values of <em>a</em> give us a tangent of 1.  They are:

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3\beta=cos^{-1}(\frac{1}{2})

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3\alpha =\frac{7\pi}{6},3\alpha  =\frac{11\pi}{6}

Dividing both of those equations by 3 gives us

\alpha =\frac{7\pi}{18},\frac{11\pi}{18}

And we're done!!!

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