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a = 2 - sqrt(sqrt(159) - 12) and  b = 2 + sqrt(sqrt(159) - 12)    [Real
a = 2 + sqrt(sqrt(159) - 12) and  b = 2 - sqrt(sqrt(159) - 12)    [Real

a = 2 - i sqrt(12 + sqrt(159)) and  b = 2 + i sqrt(12 + sqrt(159))   [Imaginary
a = 2 + i sqrt(12 + sqrt(159)) and  b = 2 - i sqrt(12 + sqrt(159))   [Imaginary

There's a saying in poker: "If you look around the table and you don't see the sucker, you're the sucker".

its just 7.4 * 10^3

Wow. 4th incomplete question ive seen in 4 minutes. PUT SOME BRAIN TO YOUR QUESTIONS PLZZ

*Clears throat." WHERE IS THE GRAPH. Im dying. This is the third time i had to look at an incomplete question

Wow. You just copy pasted this from alcumus. Do you not realize that alcumus's latex is missing in this question. You are lazy, and i will not answer your question. That is the outcome of asking a nonsense question and expecting answers and not giving gratitude

To find the constant k, we can use the given information and the distance formula. The distance formula states that for two points (x1, y1) and (x2, y2), the distance between them is given by the equation d = sqrt((x2-x1)^2 + (y2-y1)^2).

Given that A = (4, -1), B = (-3, 1), and C = (5, -3), we can calculate the distances between A and Q, B and Q, and C and Q as follows:

dAQ = sqrt((xQ-4)^2 + (-1-yQ)^2) dBQ = sqrt((-3-xQ)^2 + (1-yQ)^2) dCQ = sqrt((5-xQ)^2 + (-3-yQ)^2)

Now, since PA^2 + PB^2 + PC^2 = 3PQ^2 + k, we can substitute in the distances and solve for k. Let F be the expression we want.  This gives us the following equation:

dAQ^2 + dBQ^2 + dCQ^2 = 3PQ^2 + F

Substituting in the distances we calculated earlier and simplifying, we get the following equation:

(xQ-4)^2 + (-1-yQ)^2 + (-3-xQ)^2 + (1-yQ)^2 + (5-xQ)^2 + (-3-yQ)^2 = 3PQ^2 + F

Rearranging and simplifying, we get the equation:

F = 6PQ^2 - 3(xQ-4)^2 - 3(1-yQ)^2 - 3(-3-xQ)^2 - 3(-3-yQ)^2

Since the points A, B, and C are given, we can now substitute in their coordinates to get the value of k. This gives us the equation:

F= 3PQ^2 - 3(-3)^2 - 3(4)^2 - 3(-1)^2 - 3(-3)^2

Simplifying, we get the final result:

F = 3PQ^2 + 27

Therefore, the constant k is 27.

This question was auto blocked because you used the word 'suck.'

You need to be careful of your language online but  I have unblocked it for you. 

Is there a specific theorem to prove a^2 + b^2 > ab?

I will assume that a and be are in the set of real numbers.

If and a and b are both  0 then it is not true

If one is zero but not the other than LHS is positive and RHS is 0 so it will be true

If a or b (but not both) is negative then the RHS is positive and the LHS is negative so it will be true

So I am just left to prove it is true if a and b both have the same sign. (and not 0)

\((a-b)^2=a^2+b^2-2ab\\so\\ LHS=a^2+b^2\\ LHS=(a-b)^2+2ab\\ LHS=ab+[(a-b)^2+ab]\\ LHS=ab + \text{a positive number}\\ LHS >ab\\ LHS>RHS\)

Let the price of 1 shirt ==S

Let the price of 1 pair of pants ==P

2S  +  5P ==154.00..................(1)

3S  +  4P ==168.00..................(2), solve for S, P

Use elimination or substitution to get:

S ==$32.00 -  price of 1 shirt

P ==$18.00 -  price of 1 pair of pants.

There are 6!/6 = 5! = 120 ways.

First calculate all the possible cases which is \(\frac{6!}{2!2!2!}\)= 90 

Then calculate all the possible cases which this condition occurs. The first time you pull there is three possible ways to pull a matching pair, 3*1, The second time you pull there is 2 possible ways to get a matching pair, 2*!, after you do this, you have to pull a matching pair. So (3*2)/90 = 1/15 

You could also do this with choose, (3/(6 choose 2))*(2/4 choose 2) * (1/2 choose 2) which also gives 1/15 

https://web2.0calc.com/questions/counting_95164

This has already been asked and answered

Day 1: It had 45

Day 2: After selling 1/3 of them it has 2/3 * 45 = 30 magazines.

Day 3: After selling 1/5 of them it has 4/5 * 30 = 24 magazines.

It has 24 magazines

This question is incomplete, please enter a value for the number of yellow tiles there are. 

To win the jackpot, you must match either:

- Two of the three white balls

- The red SuperBall

Let's calculate the probability for each of these cases and then add them up.

Case 1: Matching two of the three white balls

There are \(C(12,3) = 220\) ways to choose three white balls out of 12. And there are \(C(3,2) = 3\) ways to choose two of these three balls. So, the number of favorable outcomes for this case is \(3 * C(12,3) = 660\).

Since there are C(12,3) ways to choose three white balls from your ticket, the probability of matching two of the three white balls is\( 660 / C(12,3) = 660 / 220 = 3 / 10\).

Case 2: Matching the red SuperBall

There are 20 - 13 + 1 = 8 ways to choose the red SuperBall. And there is only 1 way to choose the red SuperBall from your ticket. So, the number of favorable outcomes for this case is 8.

Since there are 20 - 13 + 1 = 8 ways to choose the red SuperBall from your ticket, the probability of matching the red SuperBall is 8 / 8 = 1.

Adding the probabilities for these two cases, we get the total probability of winning the jackpot:

P = 3/10 + 1 = 7/10

So, the probability of winning the jackpot is 7/10.

Since QN is perpendicular to PR, we know that triangle QNO is a right triangle. Using the Pythagorean theorem, we can find the length of QO:

\(QO = sqrt(QN^2 + PR^2) = sqrt(12^2 + 14^2) = sqrt(288) = 12√2\)

Now, we can find the length of RO by adding the length of QO and NO:

\(RO = QO + NO = 12√2 + 7 = 19√2\)

Since triangle PQR is isosceles with midpoints M and N, we know that PM = PR/2 = 7 and PQ = PR = 14.

Let's call the height of triangle PQR "h". Then, using the area formula for triangle PQR, we get:

\(Area = (1/2)bh = (1/2)(14)(h)\)

Using the Pythagorean theorem, we can find h in terms of RO:

\(h^2 = PQ^2 - RO^2 = 14^2 - (19√2)^2\)

So,

\(h = sqrt(14^2 - (19√2)^2) = sqrt(196 - 368) = sqrt(-172)\)

Since the square of a real number is always non-negative, we know that this value of h is not possible. Therefore, triangle PQR does not exist.

There are a total of 17 people attending the meeting, so if the scientists are seated together in five consecutive seats, the remaining 12 people can be seated in the other 12 seats.

The number of ways to seat the scientists is just 1 (since they must all sit together), and the number of ways to seat the remaining 12 people is simply 12! (the number of permutations of 12 items).

However, we must divide by 2 to account for rotations of the table, since two seatings are considered equivalent if one can be obtained from rotating the other.

Therefore, the total number of different arrangements is:

1 * 12! / 2 = 72576

So there are 72,576 different arrangements possible.

here are a total of 17 people attending the meeting, so if the scientists are seated together in five consecutive seats, the remaining 12 people can be seated in the other 12 seats.

The number of ways to seat the scientists is just 1 (since they must all sit together), and the number of ways to seat the remaining 12 people is simply 12! (the number of permutations of 12 items).

However, we must divide by 2 to account for rotations of the table, since two seatings are considered equivalent if one can be obtained from rotating the other.

Therefore, the total number of different arrangements is:

1 * 12! / 2 = 72576

So there are 72,576 different arrangements possible.

Let's call the area of triangle ABC "A".

Since angle BAC is equal to 60 degrees and angle ABC is equal to 60 degrees, triangle ABC is an equilateral triangle.

Since angle bisector BD divides the side AC into segments of length AD and CD, we know that CD = AC - AD = AC - 24.

Let's call the length of AC "c". Then, we can use the Pythagorean Theorem to find the length of BD:

BD = sqrt(c^2 - 24^2) = sqrt(c^2 - 576)

Now, we can use the formula for the area of a triangle:

A = (1/2)bh

where b is the length of the base (AC in this case) and h is the height (BD in this case).

So,

A = (1/2)c * sqrt(c^2 - 576)

We can simplify this expression to get the final answer.

This system of equations can be written as an augmented matrix and transformed into an echelon form or reduced row-echelon form using row operations. If the reduced row-echelon form has a row of zeros except for the last entry (which is non-zero), then the system has no solutions.

Here's the augmented matrix of the system:

[1 1 3 10]

[-4 8 5 7]

[k 0 1 3]

Using row operations, we can transform the matrix into row-echelon form:

[1 1 3 10]

[0 9 2 11]

[0 0 1-k 3-3k]

From the last row, we see that the system has no solutions if and only if 1-k is non-zero and 3-3k is non-zero. That is, if 3k ≠ 3 and k ≠ 1.

So the value of k that makes the system have no solutions is k = 1.

x = k/r 

Given  5 = k/25 then k = 5*25

Hence x = 5*25/40

remainder = 45*(1-1/3)*(1-1/5)

You can crunch the numbers.

Where is the truth table?