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german
3 years ago
11

Which Pairs are additive inverses? a. 6,-(-6) b. -7,7 c.-7,-7 d.7,7 e.6,-6

Mathematics
2 answers:
Goshia [24]3 years ago
8 0
B and E are the corect answers. For a pair of numbers to be additive inverses, that means that the sum of the two numbers must be 0. And only a pair of opposite nunbers, such as 2 and -2, can have a sum of 0.
xz_007 [3.2K]3 years ago
7 0

I was never sure of what the "additive inverse" is. 
So, just now, just for you, I went and looked it up.

The additive inverse of any number ' A ' is the number
that you need to ADD to A to get zero.  That's all !

So now, let's check out the choices:

a), 6, -(-6)

That second number, -(-6), is the same as +6 .
So the two numbers are the same.
Do you get zero when you add them up ?  No.

b). -7, 7
What do you get when you add -7 and 7 ?
You get zero.
So these ARE additive inverses.

c). -7, -7
What do you get when you add -7 to -7 ?
You get -14 .  That's not zero, so these
are NOT additive inverses.

d).  7, 7
What do you get when you add 7 to 7 ?
You get 14.  That's NOT zero, so these
are NOT additive inverses.

e).  6, -6
What do you get when you add 6 to -6 ?
You get zero.
So these ARE additive inverses.

What do we end up with from the list of choices:

a).,  c).,  and d). are NOT additive inverses.

b). and e). ARE additive inverses.

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Determine the zeros of the function <br> 5) y=-x² + 2x + 1
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<h3>Zeros of function are x = 1 + \sqrt{2} \text{ and } x = 1 - \sqrt{2}</h3>

<em><u>Solution:</u></em>

<em><u>We have to find the zeros of the function</u></em>

y = -x^2 + 2x+1

Find the zeros of function:

-x^2 + 2x+1 = 0\\\\\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}\\\\x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}

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Simplify\\\\x=\frac{-2 \pm \sqrt{4+4}}{-2}\\\\x =\frac{-2 \pm \sqrt{8}}{-2}\\\\Simplify\\\\x =\frac{-2 \pm 2 \sqrt{2}}{-2}\\\\x = 1 \pm \sqrt{2}

We have two zeros

x = 1 + \sqrt{2} \text{ and } x = 1 - \sqrt{2}

Thus zeros of function are x = 1 + \sqrt{2} \text{ and } x = 1 - \sqrt{2}

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