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The triangle enclosed by the line x+y=m and the coordinate axes is a right-angled triangle with legs parallel to the coordinate axes. Let's call the length of the horizontal leg of the triangle x, and the length of the vertical leg y. Then, the area of the triangle is:

A = (1/2)xy

We can use the equation of the line to express x and y in terms of m:

x + y = m

x = m - y

y = m - x

Substituting these expressions into the equation for the area, we get:

A = (1/2)(m - y)y

A = (1/2)(m - x)x

Combining these two expressions and simplifying, we get:

A = (1/2)(m - x)x = (1/2)(m - y)y

A = (1/2)(m^2 - xy)

2A = m^2 - xy

We know that the area of the triangle is 200 square units, so we can substitute that into the equation and solve for m:

2A = m^2 - xy

2(200) = m^2 - xy

400 = m^2 - xy

We need to find values of x and y that satisfy the equation x + y = m and the equation 400 = m^2 - xy. We can substitute x = m - y into the second equation to get:

400 = m^2 - x(m - x)

400 = m^2 - mx + x^2

Rearranging and factoring, we get:

x^2 - mx + (m^2 - 400) = 0

This is a quadratic equation in x. We can use the quadratic formula to solve for x:

x = [m ± sqrt(m^2 - 4(m^2 - 400))] / 2

x = [m ± sqrt(1600 - 3m^2)] / 2

For the triangle to be a right-angled triangle, one of the legs must have length 10 times larger than the other leg. Let's assume that the horizontal leg is 10 times larger than the vertical leg, so x = 10y. Substituting this into the equation above, we get:

10y = [m ± sqrt(1600 - 3m^2)] / 2

20y = m ± sqrt(1600 - 3m^2)

We know that y is a positive number, so we can take the positive root:

20y = m + sqrt(1600 - 3m^2)

Squaring both sides, we get:

400y^2 = m^2 + 1600 - 2m sqrt(1600 - 3m^2)

Rearranging, we get:

2m sqrt(1600 - 3m^2) = m^2 - 400y^2 + 1600

Squaring both sides again, we get:

4m^2 (1600 - 3m^2) = (m^2 - 400y^2 + 1600)^2

Expanding and simplifying, we get:

-3m^4 + 400m^2 - 1600y^2 = 0

This is a quadratic equation in m^2. We can use the quadratic formula to solve for m^2:

m^2 = [-(400) ± sqrt(400^2 - 4(-3)(-1600y^2))] / (-6)

m^2 = [200 ± 10sqrt(200 + 3y^2)] / 3

=[200 ± 388]/3

We need the positive root, m^2 = (200 + 388)/3 = 196.  Therefore, m = 14.

x=What you started with
y=What your brother started with

[x - 5]=[y + 5]....................................(1)  
[x + 24] =2[y - 24]............................(2), solve for x, y

We can solve this system of equations by substitution.

From the first equation, we have:

x - 5 = y + 5

Simplifying, we get:

x - y = 10

Solving for x in terms of y, we get:

x = y + 10

Now we can substitute this expression for x into the second equation:

x + 24 = 2(y - 24)

Substituting, we get:

(y + 10) + 24 = 2(y - 24)

Simplifying, we get:

y + 34 = 2y - 48

Subtracting y from both sides, we get:

34 = y - 48

Adding 48 to both sides, we get:

y = 82 - what your brother started with

Now that we know y = 82, we can substitute this value back into the first equation to solve for x:

x - y = 10

x - 82 = 10

Adding 82 to both sides, we get:

x = 92 - what you started with

So the solution to the system of equations is x = 92 and y = 82. Therefore, your brother currently has 82 dollars and if you give him 5 dollars, you both will have 87 dollars each. If he gives you 24 dollars, then you will have 116 dollars(or twice what he has), and he will have 58 dollars.

To find the probability that the baseball player will get at least 5 hits in the game, we can use the binomial distribution, since the player either gets a hit or does not get a hit on each at bat, and the at bats are independent.

Let X be the number of hits the player gets in the game. Then X follows a binomial distribution with parameters n = 8 (the number of at bats) and p = 0.181 (the probability of getting a hit on each at bat).

The probability of getting at least 5 hits is equal to the sum of the probabilities of getting 5, 6, 7, or 8 hits:

P(X ≥ 5) = P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8)

Using the formula for the probability mass function of the binomial distribution, we can calculate each of these probabilities:

P(X = k) = (n choose k) p^k (1-p)^(n-k)

where (n choose k) is the binomial coefficient.

Therefore, we have:

P(X = 5) = (8 choose 5) (0.181)^5 (1-0.181)^(8-5) ≈ 0.222

P(X = 6) = (8 choose 6) (0.181)^6 (1-0.181)^(8-6) ≈ 0.079

P(X = 7) = (8 choose 7) (0.181)^7 (1-0.181)^(8-7) ≈ 0.013

P(X = 8) = (8 choose 8) (0.181)^8 (1-0.181)^(8-8) ≈ 0.001

Therefore, the probability of getting at least 5 hits in the game is:

P(X ≥ 5) = 0.222 + 0.079 + 0.013 + 0.001 = 0.315

Therefore, the probability that the baseball player will get at least 5 hits in the game is approximately 0.315 or 31.5% (rounded to one decimal place).

To get a product of 1 with two dice rolls, we must roll a 1 on the first die and a 1 on the second die.

To get a product of 2, we can either roll a 1 on the first die and a 2 on the second die, or roll a 2 on the first die and a 1 on the second die.

To get a product of 3, we can either roll a 1 on the first die and a 3 on the second die, or roll a 3 on the first die and a 1 on the second die.

So, the possible outcomes that give us a product of 1, 2, or 3 are:

Rolling a 1 on the first die and a 1 on the second die

Rolling a 1 on the first die and a 2 on the second die

Rolling a 2 on the first die and a 1 on the second die

Rolling a 1 on the first die and a 3 on the second die

Rolling a 3 on the first die and a 1 on the second die

This gives us a total of 5 possible outcomes out of 36:

(1, 1)

(1, 2)

(2, 1)

(1, 3)

(3, 1)

Therefore, the probability of getting a product of 1, 2, or 3 is:

P(product = 1, 2, or 3) = 5/36

So the probability of getting a product of 1, 2, or 3 is 5/36.

Since angle BAC is 60 degrees and angle ABC is 60 degrees, we know that angle ACB is also 60 degrees, making triangle ABC an equilateral triangle.

Let E be the foot of the perpendicular from point B to side AC, and let x be the length of BE. Then, the length of EC is also x, and the length of AB is 2x.

Since AD is the angle bisector of angle BAC, we know that the ratio of the length of AC to the length of AB is equal to the ratio of the length of DC to the length of BD:

AC / AB = DC / BD

Substituting the values we know, we get:

1 / 2 = DC / (x + 24)

Solving for DC, we get:

DC = (x + 24) / 2

Using the Pythagorean theorem, we can find x in terms of DC:

x^2 = AB^2 - AE^2 = (2x)^2 - DC^2 = 4x^2 - (x + 24)^2 / 4

Simplifying this equation, we get:

15x^2 - 12x - 576 = 0

Solving for x using the quadratic formula, we get:

x = (-b ± sqrt(b^2 - 4ac)) / 2a

where a = 15, b = -12, and c = -576. Taking the positive root, we get:

x = 12

Therefore, the length of AB is 2x = 24, and the area of triangle ABC is:

Area = (sqrt(3) / 4) AB^2 = (sqrt(3) / 4) (24)^2 = 144sqrt(3)

Therefore, the area of triangle ABC is 144sqrt(3).

Need help

(a) The expected value of the number shown when we draw a single slip of paper is given by the weighted average of the possible outcomes, where the weights are the probabilities of each outcome. Let X be the random variable representing the number shown on the slip of paper. Then we have:

X = { 3, with probability 8/10 = 0.8 5, with probability 2/10 = 0.2 }

So, the expected value of X is:

E[X] = 3(0.8) + 5(0.2) = 3.4

Therefore, the expected value of the number shown when we draw a single slip of paper is 3.4.

(b) If we add one additional 3 to the bag, then the number of slips with a 3 on them increases to 9, while the number of slips with a 5 on them remains at 2. The probabilities of each outcome are now:

X = { 3, with probability 9/11 5, with probability 2/11 }

So, the expected value of X is:

E[X] = 3(9/11) + 5(2/11) = 3.182

Therefore, the expected value of the number shown when we draw a single slip of paper after adding one additional 3 to the bag is 3.182.

(c) If we add two additional 3's to the bag, then the number of slips with a 3 on them increases to 10, while the number of slips with a 5 on them remains at 2. The probabilities of each outcome are now:

X = { 3, with probability 10/12 = 5/6 5, with probability 2/12 = 1/6 }

So, the expected value of X is:

E[X] = 3(5/6) + 5(1/6) = 2.833

Therefore, the expected value of the number shown when we draw a single slip of paper after adding two additional 3's to the bag is 2.833.

Let's denote the two points on the circle as A and B, and let's assume that A is fixed at the top of the circle (i.e., at the point (0,1)). We can then use polar coordinates to describe the position of point B on the circle. Let θ be the angle that the line segment AB makes with the positive x-axis, measured in radians.

We can then write the position of point B as (cos θ, sin θ), since it lies on the circle of radius 1 centered at the origin. Since we are choosing point B at random, the angle θ is uniformly distributed on the interval [0, 2π).

To find the probability that the distance between A and B is at most 1.5, we need to find the set of values of θ that satisfy this condition. The distance between A and B is given by the formula:

distance AB = sqrt((cos θ - 0)^2 + (sin θ - 1)^2) = sqrt(cos^2 θ + (sin θ - 1)^2)

So, the condition that the distance between A and B is at most 1.5 is equivalent to the inequality:

sqrt(cos^2 θ + (sin θ - 1)^2) ≤ 1.5

Squaring both sides and simplifying, we get:

cos^2 θ + (sin θ - 1)^2 ≤ 2.25

Expanding the second term and simplifying, we get:

sin^2 θ - 2sin θ + 2 ≤ 0

Using the quadratic formula to solve for sin θ, we get:

sin θ ≤ 1 ± sqrt(3) / 2

Since θ is uniformly distributed on the interval [0, 2π), the probability that sin θ satisfies this inequality is equal to the ratio of the length of the interval [0, 2π) for which sin θ satisfies this inequality to the length of the entire interval [0, 2π). The length of the interval [0, 2π) for which sin θ satisfies this inequality can be found by considering the two cases:

sin θ ≤ 1 + sqrt(3) / 2: In this case, we have:

0 ≤ θ ≤ pi/3 or 5pi/3 ≤ θ ≤ 2pi

The length of this interval is pi/3 + (2pi - 5pi/3) = 4pi/3.

sin θ ≤ 1 - sqrt(3) / 2: In this case, we have:

pi/3 ≤ θ ≤ 2pi/3 or 4pi/3 ≤ θ ≤ 5pi/3

The length of this interval is (2pi/3 - pi/3) + (5pi/3 - 4pi/3) = 2pi/3.

Therefore, the probability that the distance between A and B is at most 1.5 is:

(4pi/3 + 2pi/3) / 2pi = 1

So the probability that the distance between two randomly chosen points on the circle of radius 1 is at most 1.5 is 1 or 100%

Each individual circle is pi, so each semicircle is pi/2.  Therre are 9 semicircles, so the area is 9*pi/2.

The volume of the cube-shaped block of wood is given by:

V1 = (side length)^3
V1 = (11 in)^3
V1 = 1331 cubic inches

The volume of the square hole cut out of the block of wood is given by:

V2 = (side length)^2 * depth
V2 = (4 in)^2 * 11 in
V2 = 176 cubic inches

To find the volume of the remaining wood, we can subtract the volume of the hole from the volume of the original block:

V3 = V1 - V2
V3 = 1331 cubic inches - 176 cubic inches

V3 = 1155 cubic inches

Therefore, there are 1155 cubic inches of wood left after the hole is cut out.

Use the trig identities

\(\displaystyle \cos(A)+\cos(B)=2\cos\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right) \\ \text{and} \\ \displaystyle \sin(A)+\sin(B)=2\sin\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right).\)

The common 

\(\displaystyle \cos\left(\frac{A-B}{2}\right)\)

term will be equal to \(\cos(\pi/3)=1/2,\)

cancelling out the 2's.

Since the 10 paintings are DISTINCT and the 3 heirs are obviously DISTINCT, and since there ar NO restrictions, then the number of ways of disributing them is:

3^10 = 59,049 ways of distribting them among the 3 heirs.

e^(17*pi*i/60) + e^(57*pi*i/60)

= 0.629 + 0.777i - 0.987 + 156i

= -0.358 + 0.933i

A1 has 6 options to choose from. A2 (his/her sibling) has 5 - 1=4 options to choose from. B1 has the remaining 4 options to choose from. B2 has 3 - 1 = 2 options to choose from. The remaining 2 chairs are for the last pair C1 and C2.

So, in total, we have: 6 x 4 x 4 x 2 x 2 ==384 ways of seating the 3 pairs of children

To simplify this expression, we can factor out the common denominator of 60:

[e^(17pi*i)/60]  + [e^(57pi*i)/60] = (e^(17pi*i) + e^(57pi*i)) / 60

Now, we can use Euler's formula to write each exponential term in terms of sines and cosines:

e^(17pi*i) = cos(17pi) + i*sin(17pi) = -1 + 0i = -1
e^(57pi*i) = cos(57pi) + i*sin(57pi) = -1 + 0i = -1

Substituting these values into the expression, we get:

(e^(17pi*i) + e^(57pi*i)) / 60 = (-1 - 1) / 60 = -1/30

Therefore, the simplified expression is -1/30.

Here's a hint: what if you tried to glue two equilateral triangles together?

The expressions inside the radicals remind us of the Law of Cosines. For example, $\sqrt{a^2 - ab +b^2}$ can be interpreted as the length of the third side of a triangle in which two sides have lengths $a$ and $b$, and the angle between these two sides is $60^\circ$.

From these two glues triangles, what can we say about when equality occurs? When does A+B=C in a "triangle"?

a) You can make 8 chocolate fudge cookies.

b) You can make 26 cookies in total.

c) The question has no solutions.

We use the Binomial Theorem to expand the given expression. We have:

(2u - 3v + u^2 - v^2)^9 = ∑(k=0 to 9) C(9,k)(2u)^(9-k)(-3v)^k(u^2 - v^2)^(9-k)

To find the coefficient of u^2 v^9, we need to choose k such that (9-k) powers of (u^2 - v^2) multiply out to u^2 and k powers of (-3v) multiply out to v^9. From the first term (2u)^(9-k), we need to choose two powers of u to multiply out to u^2. Therefore, we choose (9-k-2) = (7-k) of the remaining terms to be (u^2 - v^2), and the remaining k terms to be (-3v). This gives us the equation:

2^(9-k) * (-3)^k * C(9,k) * (u^2 - v^2)^(7-k) * (-3v)^k = u^2 v^9 * coefficient

We need to find the value of the coefficient that satisfies this equation. Plugging in u^2 = v^9 = 1, we get:

2^(9-k) * (-3)^k * C(9,k) * (-3)^(k) = coefficient

Simplifying, we get:

coefficient = 2^(9-k) * 3^(2k) * C(9,k)

To find the coefficient of u^2 v^9, we need to find k such that (9-k-2) = (7-k) powers of (u^2 - v^2) multiply out to u^2, and k powers of (-3v) multiply out to v^9. This means that:

7 - k = 2
k = 5

Therefore, the coefficient of u^2 v^9 is:

2^(9-5) * 3^(2*5) * C(9,5) = 2^4 * 3^10 * 126 = 2,176,782,080

Therefore, the coefficient of u^2 v^9 in the expansion of (2u - 3v + u^2 - v^2)^9 is 2,176,782,080.

We can solve this problem using geometric probability. Let A be the area of triangle XYZ, which is (1/2)(12)(6) = 36. We can think of the point D as being chosen uniformly at random in the triangle XYZ. This means that any point in the triangle is equally likely to be chosen.

Let's consider the set of all possible points D that could be chosen. Since D can be any point in the triangle, this set is just the triangle XYZ itself. Let's call the set of all points within the triangle that make triangle XYD have area at most 20 as set S.

We want to find the probability that a randomly chosen point D is in set S. This probability is given by:

P(D in S) = (Area of S) / (Area of XYZ)

Let's find the area of S. We can think of S as being the region above line XY and below line DZ, where D is any point on line XY such that triangle XYD has area 20. Let's call the intersection of line XY and line DZ point E. Then triangle XYD has base 12 and height 20/12 = 5/3. This means that point D lies on a line parallel to YZ and 5/3 units above line XY. Let's call this line l.

The point E divides line YZ into two segments of length 2 and 4, since XY is perpendicular to YZ and XY = 12. Therefore, the distance from point E to line YZ is 2/3 of the length of YZ, or 4. We can use this information to find the equation of line l in slope-intercept form:

The slope of line l is -3/5 (since it is parallel to YZ and perpendicular to XY), and it passes through point (0, 5/3) (since it is 5/3 units above XY). Therefore, the equation of line l is:

y = (-3/5)x + 5/3

We want to find the x-coordinates of the points where line l intersects lines XY and DZ. Let's call these points F and G, respectively. To find the x-coordinate of point F, we plug in y = 0 into the equation of line l:

0 = (-3/5)x + 5/3

Solving for x, we get:

x = 5

Therefore, point F has coordinates (5, 0). To find the x-coordinate of point G, we plug in y = 6 into the equation of line l:

6 = (-3/5)x + 5/3

Solving for x, we get:

x = 19/3

Therefore, point G has coordinates (19/3, 6). The area of triangle XYD is (1/2)(12)(DF), where DF is the distance from point D to line XY. We want this area to be at most 20, so we have:

(1/2)(12)(DF) ≤ 20

Solving for DF, we get:

DF ≤ 10/3

This means that any point D on line l such that DF ≤ 10/3 will be in set S. We can find the x-coordinates of these points by solving the inequality:

(-3/5)x + 5/3 ≤ 10/3

Solving for x, we get:

x ≤ 25/3

Therefore, the set S is the region bounded by lines XY, DZ, and the line:

y = (-3/5)x + 5/3

in the triangle XYZ, where the x-coordinate of any point in this region is at most 25/3.

We can find the area of S by finding the area of the region bounded by lines XY, DZ, and the line y = (-3/5)x + 5/3 whose x-coordinate is at most 25/3. We can break this region into two triangles and a trapezoid. The area of the two triangles is:

(1/2)(12)(5/3) + (1/2)(25/3)(6-5/3) = 25/3

The area of the trapezoid is:

(1/2)(25/3 + 5)(5/3) = 100/9

Therefore, the area of S is:

25/3 + 100/9 = 275/27

Finally, we can find the probability that a randomly chosen point D is in set S by dividing the area of S by the area of XYZ:

P(D in S) = (275/27) / 36 = 275/972

Therefore, the probability that the area of triangle XYD is at most 20 is 275/972.

We can solve this problem using casework.

Case 1: The two same-colored beads are adjacent
There are 2 ways to choose the color of the same-colored beads. Once we choose the color, there are 2 ways to arrange the same-colored beads, and then 4! / 2! ways to arrange the other beads. Therefore, there are 2 * 2 * (4!/2!) = 48 ways to assemble the bracelet in this case.

Case 2: The two same-colored beads are not adjacent
There are 2 ways to choose the color of the same-colored beads. Once we choose the color, there are 2 ways to arrange the same-colored beads, and then 4! ways to arrange the other beads. However, we have overcounted the arrangements where the bracelet can be rotated and/or reflected to obtain another arrangement. There are 6 ways to rotate the bracelet, and then 2 ways to reflect it, for a total of 12 equivalent arrangements. Therefore, there are (2 * 2 * 4!) / 12 = 16 ways to assemble the bracelet in this case.

Adding up the two cases, we get a total of 48 + 16 = 64 ways to assemble the bracelet.

Therefore, there are 64 different ways for Joanna to assemble her bracelet, if two of the beads have the same color and the other four have the same color.

We can solve this problem by using the Principle of Inclusion-Exclusion (PIE). We first count the total number of ways to seat the six children without any restrictions.

There are 6 choices for the first seat, 5 choices for the second seat (since one child has already been seated), and so on, for a total of 6! = 720 possible arrangements.

Next, we count the number of arrangements where at least one pair of siblings sit next to each other in the same row. We can choose one of the three pairs of siblings to sit together, and then there are 2 ways to arrange these siblings within the pair. There are 4 choices for the first seat in the row containing the pair, and then 3 choices for the second seat (since one child is already seated), for a total of 3 * 2 * 4 * 3 * 2 = 144 arrangements. However, we have overcounted the arrangements where two pairs of siblings sit together, so we need to subtract them off.

There are 3 ways to choose which two pairs of siblings sit together, and then there are 2 ways to arrange each pair within their respective row. There are 2 choices for the first seat in the row containing the first pair, and then 1 choice for the second seat (since the other child in that pair is already seated), and 2 choices for the first seat in the row containing the second pair, and then 1 choice for the second seat (since the other child in that pair is already seated), for a total of 3 * 2 * 2 * 1 * 2 * 1 = 24 arrangements.

However, we have subtracted too much, since there is one arrangement where all three pairs of siblings sit together in the same row that we have subtracted twice. There are 3! = 6 ways to arrange the three pairs within the row, and then 3! = 6 ways to arrange the rows, for a total of 6 * 6 = 36 arrangements.

Therefore, the number of arrangements where at least one pair of siblings sit together in the same row is 144 - 24 + 36 = 156.

Finally, we subtract this number from the total number of arrangements to get the number of arrangements where no pair of siblings sit together in the same row:

720 - 156 = 564.

Therefore, there are 564 ways to seat the three pairs of siblings from different families in two rows of three chairs, if siblings may sit next to each other in the same row, but no child may sit directly in front of their sibling.

After 1 step, Franklin can end up at any of the four points that are one unit away from the starting point: (1,0), (-1,0), (0,1), and (0,-1). Therefore, there are 4 different points that Franklin could end up at after 1 step.