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Since BAC=60 degrees and ABC=45 degrees, then ACB=60 + 45 = 105 degrees.

Since the angle bisector of BAC meets BC at D, we know that BD/CD = AB/AC. We also know that AB = 24 + 15 = 39 and AC = 24. Substituting these values, we get:

BD/CD = AB/AC BD/CD = 39/24 BD = (39/24) * CD

We can use the Law of Sines to solve for CD:

sin(ABC) / AC = sin(ACB) / CD sin(45) / 24 = sin(105) / CD CD = (24 * sin(105)) / sin(45) CD = 30

Now that we know BD = (39/24) * CD = (39/24) * 30 = 52.5 and CD = 30, we can use the formula for the area of a triangle to find the area of ABC:

Area of ABC = (1/2) * BD * CD Area of ABC = (1/2) * 52.5 * 30

Area of ABC = 787.5

Therefore, the area of triangle ABC is 787.5 square units.

The amount that the springs are stretched depends on the force that is pulling them apart. In each scenario, the force pulling the springs apart is the weight of Grogg and the sled.

In scenario A, the sled is stationary, so there is no force pulling the springs apart. Therefore, the springs are not stretched at all.

In scenario B, the sled is moving forward at constant speed, so there is no net force acting on it. The force of friction is counteracted by the force of the sled pushing back on the ground. Therefore, the springs are not stretched in this scenario either.

In scenario C, the sled is slowing down, so there is a net force acting on it. The force of friction is greater than the force of the sled pushing back on the ground. Therefore, the springs are stretched in this scenario.

In scenario D, Grogg is going down a ramp, so there is a force pulling him down the ramp. This force is greater than the force of friction, so the springs are stretched in this scenario.

In scenario E, Grogg is going up a ramp, so there is a force pulling him up the ramp. This force is less than the force of friction, so the springs are not stretched in this scenario.

Therefore, the ranking of the scenarios in terms of how much the springs are stretched is:

A. The sled is stationary and the ground is flat.

B. The sled slides forward across the ground at constant speed; there is negligible friction.

E. Grogg is going [i]up[/i] a ramp of θ=π/8 and friction is negligible. He is slowing down.

C. The sled slides forward, but is slowing down with acceleration a=−2g​x^ due to friction.

D. Grogg is going down a ramp of θ=π/8 and friction is negligible, so he is accelerating down the incline.

(A) > (B) > (E) = (C) > (D)

The volume of revolution of a shape can be found using the disc method:

V = \int_a^b \pi [r(x)]^2 dx

where r(x) is the distance from the curve to the axis of rotation at x.

If we break the shape down into squares of side length s, we can approximate the volume of each square using the disc method:

V_i \approx \pi [r_i]^2 s^2

where ri​ is the distance from the center of the square to the axis of rotation.

The volume of revolution of the entire shape is then the sum of the volumes of all the squares:

V \approx \sum_{i = 1}^n V_i = \sum_{i = 1}^n \pi [r_i]^2 s^2

In the case of the circle, the axis of rotation is pointing in the y^​ direction. So, the distance from the center of each square to the axis of rotation is just the y-coordinate of the center of the square. We can write this as ri​=yi​.

Therefore, the summation for the volume of revolution of the circle is

V \approx \sum_{i = 1}^n \pi r_{xi} r_{yi} s

where n is the number of squares, s is the side length of each square, and (yi​,0) is the center of the $i$th square.

There are two cases to consider:

The two cards have the same suit.

The two cards have different suits.

Case 1: The two cards have the same suit.

There are 13 ways to choose the first card, and then 12 ways to choose the second card (since the first card cannot be chosen again), for a total of 13×12=156 ways.

Case 2: The two cards have different suits.

There are 4 ways to choose the first card's suit, and then 13 ways to choose the first card within that suit, and then 12 ways to choose the second card (from the remaining 51 cards), for a total of 4×13×12=624 ways.

Summing the cases, we get a total of 156+624=780​ ways to pick two different cards from a standard deck of cards.

Rotating the icosagon by 360 degrees is the same as doing nothing, so we want the number of positive angles less than 360 degrees that are a divisor of 360. The divisors of 360 are 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 18, 20, 30, 36, 60, 90, 180, and 360. Since there are 14 divisors, there are 14​ different positive angles less than 360 degrees that can be used to rotate the icosagon to coincide with its original position.

Since the square can never cover the upper left and lower left corners of the rectangle, the square can never cover the blue region shown below.

The blue region consists of 4 smaller rectangles: two rectangles of size 2×8 and two rectangles of size 8×2. Thus, the area of the blue region is 4⋅2⋅8+4⋅8⋅2=128+64=192​.

You can see a solution here: https://web2.0calc.com/questions/please-help_91446

Let O be the center of the semicircle. Since △AOB is a right triangle, we have AO=OB=r, where r is the radius of the semicircle.

Also, since CD is a diameter of the semicircle, we have OC=r.

Triangle OCD is a right triangle, so we have CD^2=OC^2+OD^2. Substituting r for OC and 3 for OD, we get:

7^2 = r^2 + 3^2

r^2 = 49 - 9 = 40

r = \sqrt{40} = 2\sqrt{10}

Therefore, the radius of the semicircle is 2*sqrt(10).

Hope this helps!

The denominator of the function is equal to:

(x^2 - 6x + 8) - (x^2 + x - 6) = (x^2 - 6x + 8 - x^2 - x + 6) = -2x

The function is undefined when the denominator is 0, so the real numbers x that are not in the domain of the function are the solutions to the equation −2x=0.

The solutions to this equation are x=0.

Therefore, the real numbers that are not in the domain of the function are 0​.

The system of equations can be rewritten as:

xy = x + y xz = 2x + 2z yz = 4y + 4z

Subtracting the first equation from the second equation, we get:

xz - xy = x + 2z - (x + y) = z

Subtracting the second equation from the third equation, we get:

yz - xz = 4y + 4z - (2x + 2z) = 2y + 2z

Adding these two equations, we get:

3y = 3z

Therefore, y=z. Substituting this into the equation yz=4y+4z, we get:

y^2 = 5y

Solving for y, we get y=0 or y=5.

If y=0, then the system of equations has no solution.

If y=5, then the system of equations becomes:

5x = 6 5z = 10

Solving for x and z, we get x=3/2 and z=2.

Therefore, the unique solution to the system of equations is (3/2,2,2)​.

The total number of ways to distribute 8 indistinguishable balls among 5 distinguishable boxes is 10C8=45.

However, we need to subtract the number of ways that only one of the boxes has 0 balls, which is 5C1=5.

So the answer is 45−5=40​.