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Since R1, R2, R3, R4 are the four regions into which the unit circle is divided by the lines x=6 and y=4, we see that regions R1 and R3 are congruent, and regions R2 and R4 are congruent. Therefore, R1+R2+R3+R4=2(R1+R2).

To compute the area of R1, we can construct sector OAC, where O is the center of the circle and A is the point at which the circle intersects the line y=4. Sector OAC has central angle 90 degrees, and its radius is 10, so its area is \frac{1}{4}(10^2)\pi = \frac{25}{2}\pi.

Since regions R1 and R3 are congruent, and region R1 is the upper right quadrant of sector OAC, the area of R1 is \boxed{\frac{25\pi}{4}}.

(B)

We can solve the system of equations for x, y, and z by using the following steps:

Multiply the first equation by −6 and the second equation by 5.

Add the two equations together to get −6xy+30yz+30xz=22.

Divide both sides of the equation by −6.

Simplify the equation to get xy−5yz−5xz=−11.

Add the equation xyz=6 to the equation xy−5yz−5xz=−11 to get 6xy−5yz−5xz=−5.

Factor the equation to get (x−1)(6y−5z)=5.

Since x, y, and z are all non-zero, we must have x−1=5 and 6y−5z=−1.

Solve for x and y to get x=6 and y=65z−1​.

Substitute these values into the equation xyz=6 to get 6z=636z−6​.

Solve for z to get z=21​.

Substitute z=21​ into the equation y=65z−1​ to get y=31​.

Therefore, the only possible value of x+y is 6+1/3​ = 19/3.

(A) For any number a, we can substitute a into the cubic equation f(x)=x3+px2+qx+r, and we get: \begin{align*} f(a)&= a^3 + pa^2 + qa + r \ &= a(a^2 + pa + q) + r \end{align*}The term (a2+pa+q) is the quadratic expression that has the roots a and −p. We can also think of it as the sum of the roots of the quadratic expression. So if we want a cubic expression with the roots a, b, and c, then the quadratic expression with the roots a and −p is (a+b+c). We can use the same logic for the other two quadratic expressions that we need to create. We get: \begin{align*} a(a+b+c) + r = a^3 + ab + ac + r\ b(a+b+c) + r = ab^2 + bb + bc + r\ c(a+b+c) + r = ac^2 + bc + cc + r \end{align*}Subtracting the first equation from the second equation, we get: \begin{align*} b(a+b+c) + r - (a(a+b+c) + r) &= 0\ b(a+b+c) - a(a+b+c) &= 0\ (b-a)(a+b+c) &= 0\ (b-a)\left(t+5\right)&=0 \end{align*}We can do the same for the other pair of equations: \begin{align*} (c-a)\left(t+6\right)&=0\ (c-b)\left(t+5\right)&=0 \end{align*}Multiplying these three equations together, we get the monic cubic expression that we want. \begin{align*} &\left(t+5\right)\left(t+6\right)\left(t+5\right) \ &= (t^2+11t+30)(t+5)\ &= t^3+16t^2+65t+150 \end{align*}Therefore, the monic cubic polynomial in terms of a variable t whose roots are t=a, t=b, and t=c is: \boxed{t^3+16t^2+65t+150}

a b^4 = 48 a^2 b 

b^3 = 48 a

b^3 / 48 = a 

Sub into second equation 

  b^6 / 2304 * b^5  =  6 b^9 /110592 * b 

b^11/ 2304 = 6b^10 / 110592 

b = 1/8     then      a = 1/24576

\(\boxed{x =\dfrac{ (34.14-5.76)}{6}}\)

Explanation: The total cost minus the eggs cost = the total apple cost. (28.38). Since the bags of apples are the same size and value, and you have 6 bags, you divide the total apple cost by 6 to get \(\boxed{4.73}\) dollars per bag of apples. \(\square\)

To find the sine of angle PAQ, we need to find the length of PQ and the length of AP.

To find the length of PQ, we can use the Pythagorean Theorem on triangle BPQ.

PQ2=BP2+PC2=42+12=17

PQ=17​

To find the length of AP, we can use the Pythagorean Theorem on triangle APQ.

AP2=AQ2+PQ2=52+17=42

AP=42​=67​

Now that we know the length of PQ and AP, we can find the sine of angle PAQ using the definition of sine:

sin(PAQ) = opposite/hypotenuse​ = PQ/AP = ​sqrt(17)/(6*sqrt(7)) = sqrt(119)/42.

thanks for trying, but the correct answer I found out on my own. I guess ill post an explanation here

We take a cross-section along the great circle. Let O be the center of the sphere, and let M be the midpoint of line AB

Since OA = OB = 1, angle OMA = 90, Am is sqrt3/2 so triangle AOM is a 30 60 90 triangle

So AOM is 60 degrees, BOM is also 60 degrees, therefore AOB is 120 degrees, therefore arc AB has length

120/360 *2pi=2/3pi

First, we try to "clear fractions" and see what we get:

x2−7x+10Bx−11​=x−2A​+x−53​ Bx−11=A(x−5)+3(x−2) Expanding, we get: Bx−11=Ax−5A+3x−6 Bx−Ax=3x−5A−6 (B−A)x=−5A−6 If B−A=0, then we would need −5A−6=0, which means A=−56​. But in this case, the equation (B−A)x=−5A−6 simplifies to 0=0, which means that the equation is true for any value of x. Since we know that A and B are unique values, we must have B−A=0. Therefore, we can divide both sides of the equation by B−A to get: x=B−A−5A−6​ Setting x=2, we get: 2=B−A−5A−6​ Solving for A, we get: A=52B−2​ Substituting this value of A into the equation Bx−11=A(x−5)+3(x−2), we get: Bx−11=52B−2​(x−5)+3(x−2) 5Bx−55=2Bx−10+3x−6 3Bx−55=3x−16 3Bx=3x−39 2Bx=−36 B=−x18​ Since B is a unique value, we must have x=0. Therefore, B is a non-zero real number.

Setting x=5, we get: 5B−11=A(5−5)+3(5−2) 5B−11=3(5−2) 5B−11=9 5B=20 B=4 Substituting this value of B into the equation A=52B−2​, we get: A=52(4)−2​ A=56​

Therefore, A+B=6/5​+4=26/5.

Since ∠AOB=90∘, we have [\angle BOM = \frac{1}{2}\angle ABC] and [\angle AON = \frac{1}{2}\angle BAC.] Since N is the midpoint of AC, we have [\angle ACN = 90^\circ - \frac{1}{2}\angle BAC.] Similarly, since M is the midpoint of BC, we have [\angle BCM = 90^\circ - \frac{1}{2}\angle ABC.] Therefore, [\angle MON = \angle BCM + \angle ACN = 180^\circ - \frac{1}{2}\angle ABC - \frac{1}{2}\angle BAC = \boxed{90^\circ + \frac{1}{2}\angle A}.]

There are 52 cards in a deck, so there are 51 pairs of adjacent cards. Each pair has an equal probability of being Aces or Kings, which is 524​⋅513​=22112​. So, the expected number of pairs of adjacent cards that are both Aces or Kings is:

Expected number = 51 * (\frac{12}{221}) = \boxed{\frac{40}{221}}

The distance between two points on a sphere along the minor arc of a great circle is equal to the central angle between the two points in radians. The central angle between two points on a sphere is equal to the arccosine of the dot product of the unit vectors pointing to the two points.

Therefore, to find the distance from A to B along the minor arc of a great circle, we first need to find the unit vectors pointing to A and B. We can do this by dividing the coordinates of A and B by the radius of the sphere, which is 1.

Let a and b be the unit vectors pointing to A and B, respectively. Then the distance from A to B along the minor arc of a great circle is equal to the central angle between a and b, which is given by:

theta = arccos(\mathbf{a} \cdot \mathbf{b})

We know that the space distance between A and B is 3​. We can use this information to find the dot product of a and b.

\mathbf{a} \cdot \mathbf{b} = \cos(theta) = \frac{\sqrt{3}}{2}

Therefore, the distance from A to B along the minor arc of a great circle is given by:

theta = arccos(\mathbf{a} \cdot \mathbf{b}) = arccos(\frac{\sqrt{3}}{2})

theta = \frac{\pi}{3}

Therefore, the distance from A to B along the minor arc of a great circle is pi/3.

1. If the balls and boxes are both distinguishable, then we can place the balls in the boxes in 84 ways.

2. If neither the balls nor the boxes are distinguishable, then we are simply putting 4 indistinguishable objects into 8 distinguishable boxes. This can be done in (48+4−1​)=(411​) ways.

3. If the balls are indistinguishable, and the boxes are distinguishable, then we can put the balls in the boxes in 8 ways. We can then permute the balls within each box in 4! ways. This gives us a total of 8⋅4!=192 ways.

4. If the balls are distinguishable, and the boxes are indistinguishable, then we can simply put the balls in the boxes in 4! ways.

Therefore, the answers are:

1.84=4096​2.(411​)=330​3.8⋅4!=192​4.4!=24​

Letxbe the fraction of the swim team that are sophomores. Then, the fraction of the swim team that are freshmen is1−x.

We are given that the probability that a sophomore is elected is4/7times the probability that a freshman is elected. This can be expressed mathematically as follows:

P(\text{sophomore}) = 4/7 \cdot P(\text{freshman})

We are also given that each member is equally likely to be elected, which means that the probability of a sophomore being elected is equal to the probability of a freshman being elected plus the probability of a sophomore being elected. This can be expressed mathematically as follows:

P(\text{sophomore}) = P(\text{freshman}) + P(\text{sophomore})

Substituting the first equation into the second equation, we get:

4/7 \cdot P(\text{freshman}) = P(\text{freshman}) + P(\text{sophomore})

Combining like terms, we get:

3/7 \cdot P(\text{sophomore}) = P(\text{freshman})

Dividing both sides byP(freshman), we get:

3/7 \cdot x = 1-x

Solving forx, we get:

x = \frac{7}{10}

Therefore, 7/10 ​​of the swim team are sophomores.

Having an even number of steps actually makes this problem simpler.

The maximum distance traveled by going in any direction is 12. Also note that any pair of coordinates (x, y) in which x + y is odd is not valid because an even number of steps should either cancel out the direction or produce an even sum.

Counting, we see that there are 6(4) + 1 = 25 such points on the x and y axes.

On any line y = n, we should have n + x as an even number so for odd n's we can do ceil(n/2). For quadrant 1 this is:

ceil(11/2) + ceil(9/2) + ceil(7/2) + ceil(5/2) + ceil(3/2) + ceil(1/2) = 5 + 4 + 3 + 2 + 1 = 15.

For even n's, we can just do n/2 which is also, in quadrant 1, equal to 15.

So we should have 4(15+15) + 25 = 145.

Please let me know if I did anything wrong.