Can someone solve this?

Step-by-step explanation:

The relevant relation is that the product of distances from the point of intersection of secants to the two points of intersection of each with the circle is a constant. The point of tangency counts as both points of intersection.

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By way of example, the product of distances to the circle for the tangent segment is 10·10 = 100

This is also the product of the distances on the secant that intersects the tangent:

  100 = 5(5+x+9)

  20 = x +14 . . . . . divide by 5

  x = 6 . . . . . . . . . . subtract 14

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The same is true when the secants intersect inside the circle:

  6·9 = y·3

  y = 18 . . . . . . divide by 3

answer: 204

step-by-step explanation:

we are told that the revenue is

x⋅(4080−10x)=−10x^2+4080x

this is a quadratic function, so to maximize it we should find the x-value of its vertex. recall that the x-value of the vertex of the parabola ax^2+bx+c is given by the formula −b/2a.

so in this case, we have a=−10 and b=4080, so the x-value of the vertex is

−b/2a= −(4080)/2(−10)=204

see explanation

step-by-step explanation:

a

(6r + 7) + (13 + 7r) ( remove brackets and collect like terms )

= 6r + 7 + 13 + 7r

= 13r + 20

b

remove brackets and collect like terms

13 - r - 1 + r

= 12 - r

c

remove brackets and collect like terms

= - 8 - r + 2r - 4

= - 12 + r

d

remove brackets and collect like terms

= 7r - - - 6r

[-   - = - - = - ]

= r -