sandwich

We know that f(f(x)) = x for all x.  If x < 2, then f(x) = ax + b.  Substituting into the first equation, we get
f(ax + b) = x

Since f(ax + b) = 8 - 3(ax + b) if ax + b >= 2, then we must have ax + b >= 2.  This means that x >= -2/a.

If -2/a <= x < 2, then f(x) = ax + b.  Substituting into the first equation, we get
(ax + b) + b = x

This simplifies to ax + 2b = x.  Since ax + b >= 2, then x must equal 2.  However, we know that f(2) = 8 - 3(2) = 2, so this case is not possible.

Therefore, the only possible value of x is x = -2/a.  In this case, f(x) = f(-2/a) = 8 - 3(-2/a) = 10/a.  Substituting into the first equation, we get
10/a + b = -2/a

This simplifies to a + b = -10.  Therefore, a + b =  −10

First, we can divide both sides by 11:

sqrt(2y) = 3y - 5/11

We can then square both sides of the equation to get rid of the radical:'

2y = 9y^2 - 35y + 25/121

This simplifies to 9y^2 - 44y + 25/121 = 0

Then we factor: (3y - 5/11)(3y - 5/22) = 0

So the solutions are y = 5/33 or y = 5/66.

But if we plug these in, we find that only y = 5/33 works.

Let O be the center of the square, and let E, F, G, and H be the points where the sides of the square intersect the sides of the triangles. Since the triangles are equilateral, OE=OF=OG=OH=5. Let s be the side length of the largest square that can be inscribed in the space inside the square and outside of the triangles. Since O is the center of the square, s must be less than 5.

Let x=OE−s. Since s is the side length of the inscribed square, OE+s=OE−x+2s=OE+x=5. Therefore, x=2.

Since OE=5, the side length of the inscribed square is s=OE−2=5−2=3​.

The volume of a cone is given by the formula:

V = (1/3)πr^2h

Where:

V is the volume of the cone

r is the radius of the base of the cone

h is the height of the cone

We know that the volume of the cone is 12288*pi cubic inches and that the vertex angle of the vertical cross section is 60 degrees. This means that the cross section of the cone is an equilateral triangle.

The radius of the base of the cone is equal to the height of the equilateral triangle divided by the sine of 60 degrees.

r = h / sin(60)

r = h / (√3/2)

r = 2h / √3We can now plug this value of r into the formula for the volume of the cone:V = (1/3)π(2h/√3)^2hV = (4/3)πh^3We know that the volume of the cone is 12288*pi cubic inches, so we can set this equal to the right-hand side of the equation above:

12288*pi = (4/3)πh^3

h^3 = 3072

h = 17.7

To the nearest tenth, the height of the cone is 17.8 inches.

From |x - 2| = p,

x - 2 = p.

Then x - p = 2.

To find how many different sequences of rolls could there have been given that Catherine rolls a standard 6-sided die 5 times and the product of her rolls is 900 (the order of the rolls matters), we can use the following approach:

One way to find the sequences is to factorize the number 900 into its prime factors 12. Since 900 = 2^2 * 3^2 * 5^2, we need to find all the possible combinations of five numbers chosen from the set {2, 2, 3, 3, 5, 5}. We can use the formula for combinations with repetition and consider each number as a distinct item. The number of sequences is then given by:

(5 + 6 - 1)! / (5! * (6 - 1)!) = 126

(1) So simple!  Minor TU is 180 - 32 = 148 degrees.

(2) I will have to think about this one...

(3) The external tangent gives us similar triangles OTC and PUC.  Let x = PC.  Then x/(x + 5) = 3/8, so x = 3.  The answer is PC = 3 + 3 + 5 = 11.