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3 years ago

42a2-72a-24 Factoring

Mathematics

2 answers:

Colt1911 [192]3 years ago

8 0

Answer:

6(7a+2) (a-2)

Step-by-step explanation:

Add and subtract the second term to the expression and Factor by grouping.

I'm not really sure but I hope this is the answer

Svetlanka [38]3 years ago

6 0

9514 1404 393

Answer:

6(a -2)(7a +2)

Step-by-step explanation:

Removing a common factor of 6 can make the quadratic factors easier to find.

= 6(7a^2 -12a -4)

We want to find factors of (7)(-4) that have a total of -12. It doesn't take much search to see they are (-14)(2), so the expression can be written as ...

= 6(7a^2 -14a +2a -4) = 6((7a^2 -14a) +(2a -4)) . . . . rewrite 'a' term; group pairs

= 6((7a(a -2) +2(a -2)) . . . . factor pair groups

= 6(a -2)(7a +2) . . . . . . . . finish factoring

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Como simplificar 120/45

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Answer:

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3 years ago

A high school coach needs to buy new athletic shorts for the 15 members of the basketball team. The coach must spend less than $

AfilCa [17]

If the budget is $200 and he have 15 members then we have divide the two. 200 / 15 = $13.33 per shorts. 15x =< $200. x represents 13.33. So the solution represents the coach may spend up to $13.33 per pair of shorts. If it was even 1 cent more than $13.33 than he wouldn't have enough.So he can spend up to $13.33 or less per pair of shorts.

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3 years ago

Read 2 more answers

2. Given a quadrilateral with vertices (−1, 3), (1, 5), (5, 1), and (3,−1):

zlopas [31]

<h2>Explanation:</h2>

In every rectangle, the two diagonals have the same length. If a quadrilateral's diagonals have the same length, that doesn't mean it has to be a rectangle, but if a parallelogram's diagonals have the same length, then it's definitely a rectangle.

So first of all, let's prove this is a parallelogram. The basic definition of a parallelogram is that it is a quadrilateral where both pairs of opposite sides are parallel.

So let's name the vertices as:

$A(-1,3) \\ \\ B(1,5) \\ \\ C(5,1) \\ \\ D(3,-1)$

First pair of opposite sides:

<u>Slope:</u>

$\text{For AB}: \\ \\ m=\frac{5-3}{1-(-1)}=1 \\ \\ \\ \text{For CD}: \\ \\ m=\frac{1-(-1)}{5-3}=1 \\ \\ \\ \text{So AB and CD are parallel}$

Second pair of opposite sides:

<u>Slope:</u>

$\text{For BC}: \\ \\ m=\frac{1-5}{5-1}=-1 \\ \\ \\ \text{For AD}: \\ \\ m=\frac{-1-3}{3-(-1)}=-1 \\ \\ \\ \text{So BC and AD are parallel}$

So in fact this is a parallelogram. The other thing we need to prove is that the diagonals measure the same. Using distance formula:

$d=\sqrt{(y_{2}-y_{1})^2+(x_{2}-x_{1})^2} \\ \\ \\ Diagonal \ BD: \\ \\ d=\sqrt{(5-(-1))^2+(1-3)^2}=2\sqrt{10} \\ \\ \\ Diagonal \ AC: \\ \\ d=\sqrt{(3-1)^2+(-5-1)^2}=2\sqrt{10} \\ \\ \\$