mathrokus

I'm not sure if there is an easier way to do this, but one way would involve trigonometric functions.  Since AB = 6, we have an isoceles triangle 5-5-6 at Triangle PAB.  Extend the perpendicular bisector of AB and you have two 3-4-5 right triangles.  Now you can use sin^-1(4/5) to find Angle PAB.  Then you can construct a second right triangle with the hypotenuse being the radius of the circle and one leg being 3, which was the line segment of A to the midpoint of AB.  Then you can use tan^-1 to find the radius.

I am not sure if this is the easiest way, but this was all I could think of.

For polygons with a, b, and c sides to surround a point, 1/a + 1/b + 1/c should equal 1/2.

For example, two regular pentagons and a regular decagon surround a point because 1/5 + 1/5 + 1/10 = 2/10 + 2/10 + 1/10 = 5/10 = 1/2

For an equilateral triangle and a dodecagon,

1/3 + 1/12 + 1/n = 1/2

4/12 + 1/12 + 1/n = 1/2

5/12 + 1/n = 6/12

1/n = 1/12

n = 12

This means that an equilateral triangle and two regular dodecagons completely surround a point.

Since Angle TPO is 33 degrees, then Angle OPU also is 33 degrees, since PO is the angle bisector of Angle TPU.

Therefore, Angle TPU is 66 degrees.

Since tangent lines intercect the circle at 90 degrees, we can form quadrilateral TOUP with Angles T and U being 90 degrees and angle P 66 degrees.

The 4 angles of a quadrilateral add up to 360 degrees, so 360 - 2(90) - 66 = 180 - 66 = 114 degrees.

Central angle TOU is 114 degrees, so the intercepted minor arc TU is 114 degrees.

5x^2 - kx + 8 -2x^2 + 25 = 0

Simplifying,

3x^2 - kx + 33 = 0

a = 3, b = k, c = 33

The formula for the discriminant is b^2 - 4ac.

For the quadratic to have no real solutions, the discriminant must be negative.

So, b^2 - 4ac < 0

Plugging in the values,

k^2 - 4(3)(33) < 0

k^2 -396 < 0

k^2 < 396

For this question, we have to find the largest value of k that is a solution to k^2 < 396.

To maximize k, k should also be positive, (-k)^2 would always have the same solution as k^2

We know that 396 ~ 400, so we can test squares.

20^2 = 400

19 ^ 2 = 361

Therefore, our solution is k = 19.

10 members: 1000 ft^2 wall: 10 min

5 members: 1000 ft^2 wall: 20 min

Half the workforce means double the time.

10 members: 1000 ft^2 wall: 10 min

5 members: 1000 ft^2 wall: 20 min

Half the workforce means double the time.

5x^2 - 2x + 8 - 2x^2 + 6x + 2

Combining Like Terms,

3x^2 + 4x + 10, so

a = 3, b = 4, and c = 10

The formula for the discrimant is b^2 - 4ac

Plugging in the values,

4^2 - 4(3)(10) = 16 - 120 = -104

This means that the quadratic has no real roots.