Find the point P on the line yequals=66x that is closest to the point (74 comma 0 )(74,0). What is the least distance between P and (74 comma 0 )(74,0)?

The point P that is closest to the point (74, 0) is P(7.4, 22.2), and the least distance between P and (74, 0) is 70.2

Step-by-step explanation:

Given

y = 66x

And a point A(74, 0)

We are required to find the point on the line that is closest to A, and then find the distance between these points.

Let the point be P(s, t)

Because it's coordinate satisfies the given equation

y = 3x

Let t = 3s, then we have

p(s, t) = P(s, 3s)

Next, we find the distance between P and A

Let the distance be

D = √[(x1 - y1)² + (x2 - y2)²]

Here x1 = 74, x2 = 0, y1 = s, y2 = 3s.

D = √[(74 - s)² + (0 - 3s)²]

= √(s² - 148s + 5476 + 9s²)

= √(10s² - 148s + 5479)

To maximize the distance, we find the derivative of D with respect to s, and set it to 0.

That is dD/ds = 0

dD/ds =

(20s - 148)/2√(10s² - 148s 5479) = 0

=> 20s - 148 = 0

20s = 148

s = 148/20

s = 7.4

Since t = 3s, we have

t = 22.2

So, the point is P(7.4, 22.2)

Finally, we find the distance D.

D = √[(74 - s)² + (0 - 3s)²]

D = √[(74 - 7.4)² + (0 - 3×7.4)²]

D = √(4435.56 + 492.84)

= √4928.4

D = 70.2

Therefore, the distance is 70.2

f(x) = x^3 -2x^2 -3x

x( x^2 -2x-3)

x( x^2 +x -3x -3)

x( x(x+1) -3(x+1))

x( x-3)( x+1)

so zeroes are 0, 3,-1

its (2) option

geometry: triangles

solvers

lessons

answers archive

call the larger triangle abc

then the smaller triangle cde is similar to abc (all angles are equal, aaa).  

ab is the side that is 36, and we wish to find de, the side corresponding to ab

area_cde = area_abc - area_cde --> area_abc / area_cde = 2

let x=length of de

ratio of areas = (ratio of sides)^2

+2+=+%2836%2fx%29%5e2+

+2+=+1296%2fx%5e2+

++x%5e2+=+1296%2f2+=+648+

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