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Nana76 [90]
3 years ago
9

Please help me. I've been stuck all day! I need to factor 10x^2-19x-51

Mathematics
2 answers:
frutty [35]3 years ago
3 0
10X^2 + 15x-34x -51
5x (2x+3)-17(2x+3)
(5x-17) (2x+3)
castortr0y [4]3 years ago
3 0
Well, you first need to look at what goes into 10 and 51.

10:
1*10 = 10
2*5 = 10
not a lot of numbers.

51:
1*51
3*17
not a lot of numbers either.

now, very rarely is it 1 * (whatever the number, so in this case let's use 51.)

We need both numbers from x to multiply to be 10.
The only logical one here is 5*2

So far:
(5x ? ?) (2x ? ?)

Your best safe bet to do for your next one is to not use the 1*57, but the 17*3.

So, now:
(5x ? 17) (2x ? 3)
You need both to be a negative ( negative times a negative is a positive), so...

Your Answer:
(5x - 17) (2x + 3)
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Answer:

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53/100= .53 --> Convert to percentage--> .53×100 = 53%.

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22/60≈0.367--> Convert to percentage --> 0.367×100≈37%

D) This percentage will be found out of who did NOT take the medicine:

25/40= 0.625 --> Convert to percentage --> 0.625×100≈63%

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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

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* 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.
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Answer:

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Will give brainliest answer
Aneli [31]

Answer:

a

Step-by-step explanation:

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