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Alex_Xolod [135]
3 years ago
5

Simplify 3m-2(m+1) and -3(3x+1)-(6x-3)

Mathematics
1 answer:
Maslowich3 years ago
3 0
M-2 and -15xbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb
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Describe the translation below from ABC to A’B’C. In terms of X and Y directions
4vir4ik [10]

Answer: 6 units to the right and 6 units down

Step-by-step explanation:

Every point of the original figure is translated 6 units right and 6 units down to form triangle A’B’C’. So, the translation that maps the original figure to the translated figure is 6 units right and 6 units down.

5 0
4 years ago
Determine the next step for solving the quadratic equation by completing the square.
nexus9112 [7]

\qquad \textit{perfect square trinomial} \\\\ (a\pm b)^2\implies a^2\pm \stackrel{\stackrel{\text{\small 2}\cdot \sqrt{\textit{\small a}^2}\cdot \sqrt{\textit{\small b}^2}}{\downarrow }}{2ab} + b^2

the idea behind the completion of the square is simply using a perfect square trinomial,  hmmm usually we do that by using our very good friend Mr Zero, 0.

if we look at the 2nd step, we have a group as x² - x, hmmm so we need a third element, which will be squared.

keeping in mind that the middle term of the perfect square trinomial is simply the product of the roots of "a" and "b",  so in this case the middle term is "-x", and the 1st term is x², so we can say that

\stackrel{middle~term}{2(\sqrt{x^2})(\sqrt{b^2})~}~ = ~~\stackrel{middle~term}{-x}\implies 2xb~~ = ~~~~ = ~~-x \\\\\\ b=\cfrac{-x}{2x} \implies b=-\cfrac{1}{2}

so that means that our missing third term for a perfect square trinomial is simply 1/2, now we'll go to our good friend Mr Zero, if we add (1/2)², we have to also subtract (1/2)², because all we're really doing is borrowing from Zero, so we'll be including then +(1/2)² and -(1/2)², keeping in mind that 1/4 - 1/4 = 0, so let's do that.

-3~~ = ~~-2\left[ x^2-x+\left( \cfrac{1}{2} \right)^2 ~~ - ~~\left( \cfrac{1}{2} \right)^2\right]\implies -3=-2\left(x^2-x+\cfrac{1}{4}-\cfrac{1}{4} \right) \\\\\\ -3=-2\left(x^2-x+\cfrac{1}{4} \right)+(-2)-\cfrac{1}{4}\implies -3=-2\left(x^2-x+\cfrac{1}{4} \right)+\cfrac{1}{2} \\\\\\ -3-\cfrac{1}{2}=-2\left(x^2-x+\cfrac{1}{4} \right)\implies -\cfrac{7}{2}=-2\left(x-\cfrac{1}{2} \right)^2\implies \cfrac{7}{4}=\left(x-\cfrac{1}{2} \right)^2

~\dotfill\\\\ \pm\sqrt{\cfrac{7}{4}}=x-\cfrac{1}{2}\implies \cfrac{\pm\sqrt{7}}{2}=x-\cfrac{1}{2}\implies \cfrac{\pm\sqrt{7}}{2}+\cfrac{1}{2}=x \implies \cfrac{\pm\sqrt{7}+1}{2}=x

8 0
2 years ago
Pls helpppppppppl!!!!
Troyanec [42]

Answer:

I dont really understand it but I think it's the first one cuz that's the only one that makes sense to me hope it helped tho:))))))))

8 0
3 years ago
Read 2 more answers
63, 70, 68, 73, 58, 67 <br> I need the mean absolute deviation for this one too.
sweet-ann [11.9K]

Answer:

66.5

Step-by-step explanation:

I took a test and got it right

3 0
3 years ago
Read 2 more answers
PLEASE HELP FOR GEOMETRY
djyliett [7]
( pie over 2 - x)

Because we know that cos (pie over 2 ) =0 and sin (pie over 2) = 1
So we get
Sin (pie over 2 -x) =cos(x)
4 0
3 years ago
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