Identify the excluded values of this product. Then rewrite the product in simplest form.

Mathematics

1 answer:

lara [203]2 years ago

8 0

$\frac{6 {y}^{2} + 18y - 60 }{3 {y}^{2} - 12y } \times \frac{ {y}^{2} - 16 }{ {y}^{2} + 2y - 8} \\ \\ \frac{ {6y + 30y - 12y - 60}^{2} }{3y(y - 4)} \times \frac{ {(y)}^{2} - ( {4)}^{2} }{ {y}^{2} + 4y - 2y - 8 } \\ \\ \frac{6y(y + 5) - 12(y + 5)}{3y(y - 4)} \times \frac{(y + 4)(y - 4)}{y(y + 4) - 2(y + 4)} \\ \\ \frac{(6y - 12)(y + 5)}{3y(y - 4)} \times \frac{(y + 4)(y - 4)}{(y + 4)(y - 2)} \\ \\ \frac{6(y - 2)(y + 5)}{3y(y - 2)} \\ \\ \frac{2(y + 5)}{y} .$

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I don't know to find the answer to 8. Can someone explain to me?

lapo4ka [179]

Perhaps the easiest way to find the midpoint between two given points is to average their coordinates: add them up and divide by 2.

A) The midpoint C' of AB is
.. (A +B)/2 = ((0, 0) +(m, n))/2 = ((0 +m)/2, (0 +n)/2) = (m/2, n/2) = C'
The midpoint B' is
.. (A +C)/2 = ((0, 0) +(p, 0))/2 = (p/2, 0) = B'
The midpoint A' is
.. (B +C)/2 = ((m, n) +(p, 0))/2 = ((m+p)/2, n/2) = A'

B) The slope of the line between (x1, y1) and (x2, y2) is given by
.. slope = (y2 -y1)/(x2 -x1)
Using the values for A and A', we have
.. slope = (n/2 -0)/((m+p)/2 -0) = n/(m+p)

C) We know the line goes through A = (0, 0), so we can write the point-slope form of the equation for AA' as
.. y -0 = (n/(m+p))*(x -0)
.. y = n*x/(m+p)

D) To show the point lies on the line, we can substitute its coordinates for x and y and see if we get something that looks true.
.. (x, y) = ((m+p)/3, n/3)
Putting these into our equation, we have
.. n/3 = n*((m+p)/3)/(m+p)
The expression on the right has factors of (m+p) that cancel*, so we end up with
.. n/3 = n/3 . . . . . . . true for any n

_____
* The only constraint is that (m+p) ≠ 0. Since m and p are both in the first quadrant, their sum must be non-zero and this constraint is satisfied.

The purpose of the exercise is to show that all three medians of a triangle intersect in a single point.

7 0

3 years ago

Solve |p + 2| = 10Solve |p + 2| = 10<br><br> {-12}<br> {-8, 8}<br> {-12, 8}

valentinak56 [21]

By definition, we have

$|p+2| = \begin{cases} p+2 &\text{ if } p+2 \geq 0 \\-p-2 &\text{ if } p+2 < 0 \end{cases}$

So, we have to solve two different equations, depending of the possible range for the variable. We have to remember about these ranges when we decide to accept or discard the solutions:

Suppose that $p+2\geq 0 \iff p \geq -2$