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aniked [119]
3 years ago
12

Find the point, M, that divides segment AB into a ratio of 5:2 if A is at (1, 2) and B is at (8, 16).

Mathematics
1 answer:
d1i1m1o1n [39]3 years ago
4 0
Since M divides segment AB into a ratio of 5:2, we can say that M is 5/(5+2) of the length of AB. Therefore 5/7 × AB.
distance of AB = d
5/7×(x2 - x1) for the x and 5/7×(y2 - y1) for the y
5/7×(8 - 1) = 5/7 (7) = 5 for the x
and 5/7×(16 - 2) = 5/7 (14) = 10 for the y
But remember the line AB starts at A (1, 2),
so add 1 to the x: 5+1 = 6
and add 2 to the y: 10+2 = 12
Therefore the point M lies exactly at...
A) (6, 12)


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Complete parts a through c for the given function.
victus00 [196]

Answer:

a. The critcal points are at

x=0,-5,3

b. Then, x = -5   is a maximum and x=3 is a minimum

c. The absolute minimum is at   x = 3  and the absolute maximum is at  x = -5.

Step-by-step explanation:

(a)

Remember that you need to find the points where

f'(x)=0

Therefore you have to solve this equation.

20x^4  + 40x^3 - 300x^2 = 0

From that equation you can factor out    20x^2  and you would get

20x^2 (  x^2  +2x - 15)  = 0

And from that you would have   20x^2 = 0  , so x = 0.

And you would also have  x^2 +2x-15 = 0.

You can factor that equation as    x^2 +2x -15 = (x+5)(x-3) = 0

Therefore   x=-5 ,   x=3.

So the critcal points are at

x=0,-5,3

b.  

Remember that a function has a maximum at a critical point if the second derivative at that point is negative. Since

f''(x) = 80x^3 + 120x^2 -600x\\f''(-5) = 80(-5)^3 + 120(-5)^2 -600(-5) = -4000 < 0\\\\f''(3) = 80(3)^3 + 120(3)^2 -600(3) = 1440 > 0 \\

Then, x = -5   is a maximum and x=3 is a minimum

c.

The absolute minimum is at   x = 3  and the absolute maximum is at  x = -5.

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Answer:

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Step-by-step explanation:

$1.20 + 2.00 = 3.20

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TEA [102]
With these, always write out the multiples first.

Start like this:
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