(1)

258  =  129 * r^2

r^2  =  268 / 129

r^2  = 2

r  = sqrt 2

N = 129 * (sqrt 2)^t        where t is the number of years after 2011

(2)

C =  160 * (1/2)^t        where t  =  day t

https://web2.0calc.com/questions/question8

y  = -x + 3k  ......the slope of this line   = -1

x^2 + 4x + y^2 + 6y =  k      complete the square on x,y

x^2 + 4x + 4 + y^2 + 6y + 9  =  k + 4 + 9

(x + 2)^2  + (y + 3)^2  =  13 + k

The center of this  circle is  (-2, -3)

A line with a slope = 1  through this point will intersect the tangent line

y = (x +2) - 3

y = x -1

So

x -1 = -x + 3k

2x = 3k+1

x = (3k + 1) / 2

y = (3k+ 1)  / 2  - 1

y = (3k + 1 - 2) / 2

y = (3k - 1) / 2

Putting these values into the equation of the circle to find k

[ (3k + 1) / 2 + 2]^2   +[ (3k -1)/2 + 3[^2  =  13 + k

Simplifying this we  get

9k^2  + 28k - 1  =  0

Use the Quad Formula

We have two values for k

k ≈ .035313

and

k ≈ -3.1464

some one PLS HALP

Five CDs will cost 55 dollars.

16r = 1/4

r = (1/4) / 16  = 1/64

30th term  is    (1/4) r^6  = (1/4)(1/64)^6   = (1/4) ( 1/4^3) ^6  =  (1/ 4)^19 = 0.000000000003638

Rearrange as

8a + b =  -3       (1)

9a + (2 + m)b = k     (2)

This system has infinite solutions when the equations are the  same

Multiply (1) by 9  and (2) by 8

72a  + 9b = -27

72a + 8(2 + m)b  = 8k

So

9 =  8(2+m)                  -27  =   8k

9 = 16 + 8m                   k = -27/8

8m  = -7

m = -7/8 

(s/2) + 5t = 3 - 7t + 8s    → s + 10t = 6 - 14t + 16s  → -15s + 24t  = 6    (1)

3t - 6s  = 9 -2t  →  6s + 5t  =  9   (2)

Multiply (1)  by  6  and  (2)  by 15

-90s +  144t  = 36

90s + 75t  =  135     add these

219 t   =  171

t = 171 / 219  =  57 / 73

6s + 5(57/73)  = 9

6s + 285/73  = 9

6s = 9 - 285/73

6s = 372/73

s = 62 / 73

1/4   +  10d  =  -1/5

10d =  -1/5 - 1/4  =   -9/20

d =  -9/200

43rd  term =   -1/5  + 10d  =  -1/5 + 10 (-9/200)  =  -1/5 - 9/20  =  -65/100 = -13 / 20

Let m=37. We can show that n is divisible by 37 if and only if the result of this process is divisible by 37 as follows:

If n is divisible by 37, then the alternating sum of the two-digit chunks is divisible by 37.

This is because the alternating sum of the two-digit chunks is just the sum of the digits of n, modulo 37. Since n is divisible by 37, the sum of its digits is also divisible by 37, so the alternating sum of the two-digit chunks is also divisible by 37.

If the alternating sum of the two-digit chunks is divisible by 37, then n is divisible by 37.

This is because the sum of the digits of n is congruent to the alternating sum of the two-digit chunks, modulo 37.

Therefore, if the alternating sum of the two-digit chunks is divisible by 37, then the sum of the digits of n is also divisible by 37, so n is divisible by 37.

Therefore, the alternating sum of the two-digit chunks is a divisibility test for 37.

To see this in action, let's consider the number 354764. The alternating sum of the two-digit chunks of 354764 is 64−47+35=52. Since 52 is divisible by 37, we know that 354764 is also divisible by 37.

The wording of this problem is weird, but I think it's basically asking to find the total amount of students. 

Because there are 3 students in math club, science club has double that with 6!

Ok, so now we know there are 3 kids in math club, 3 kids in both, and 6 kids in science. Are we done? 

The answer is ofcourse not. We have to subtract the duplicates, which is the kids in both. We subtract 3 from both clubs so that we now have kids ONLY in math and ONLY in science. The kids ONLY in math are 0, and the kids ONLY in science is 3. 

\(3+3+0=6\)

So there are 6 students in total!

I hoped I answered your question! 

Thanks! :) :) :)

There are three cases to consider, since all three marbles can be red, all white, or all blue.

Case 1: All Red Marbles

Favorable cases: We need to choose 3 red marbles from 4 available. This can be done in ⁴C₃ ways (combinations, not permutations).

Total cases: We can choose any 3 marbles from a total of 4 red + 5 white + 6 blue = 15 marbles in ¹⁵C₃ ways.

Case 2: All White Marbles

Favorable cases: We need to choose 3 white marbles from 5 available. This can be done in ⁵C₃ ways.

Total cases: Same as Case 1 (¹⁵C₃).

Case 3: All Blue Marbles

Favorable cases: We need to choose 3 blue marbles from 6 available. This can be done in ⁶C₃ ways.

Total cases: Same as Case 1 (¹⁵C₃).

Total Probability

The probability that all three marbles are the same color (considering all three cases) is the sum of the probabilities of each case:

P(all same) = P(all red) + P(all white) + P(all blue)

Calculate Each Probability:

P(all red) = ⁴C₃ / ¹⁵C₃ P(all white) = ⁵C₃ / ¹⁵C₃ P(all blue) = ⁶C₃ / ¹⁵C₃

We can simplify each term using the fact that nCr = n! / (r! * (n-r)!):

P(all red) = (4!) / ((3!) * (1!)) / ((15!) / ((3!) * (12!))) P(all white) = (5!) / ((3!) * (2!)) / ((15!) / ((3!) * (12!))) P(all blue) = (6!) / ((3!) * (3!)) / ((15!) / ((3!) * (12!)))

Notice that (3!) and (12!) terms appear in all the denominators and numerators (except the first factorials in the numerators). These terms will cancel out, leaving:

P(all red) = 4 / (15 * 14 * 13) P(all white) = 10 / (15 * 14 * 13) P(all blue) = 20 / (15 * 14 * 13)

Final Probability:

Add the probabilities of each case:

P(all same) = 4/210 + 10/210 + 20/210 = 34/210

Simplify the fraction:

P(all same) = 17/105

Therefore, the probability that all three marbles drawn are the same color is 17/105.

To find the probability of winning a super prize in the SuperLottery, we need to consider two mutually exclusive events:

1. **Winning by matching at least two of the white balls.**

2. **Winning by matching the red SuperBall.**

We'll calculate the probabilities for each event and then add them together.

1. **Winning by matching at least two of the white balls:**

   To calculate this probability, we can find the probability of not matching any of the white balls and subtract it from 1.

   The probability of not matching any of the white balls on a single draw is:

   \[\frac{{9 \choose 3}}{{12 \choose 3}}\]

   So, the probability of matching at least two of the white balls is:

   \[1 - \frac{{9 \choose 3}}{{12 \choose 3}}\]

2. **Winning by matching the red SuperBall:**

   The probability of matching the red SuperBall is simply \( \frac{1}{8} \) since there's only one SuperBall drawn from 8 possibilities.

Now, let's calculate these probabilities:

1. Probability of winning by matching at least two of the white balls:

   \[1 - \frac{{9 \choose 3}}{{12 \choose 3}} = 1 - \frac{84}{220} = 1 - \frac{21}{55} = \frac{34}{55}\]

2. Probability of winning by matching the red SuperBall:

   \[P(\text{Red SuperBall}) = \frac{1}{8}\]

Finally, to find the total probability of winning a super prize, we add the probabilities of the two mutually exclusive events:

\[P(\text{Winning super prize}) = P(\text{White balls}) + P(\text{Red SuperBall}) = \frac{34}{55} + \frac{1}{8}\]

\[= \frac{272}{440} + \frac{55}{440}\]

\[= \frac{272 + 55}{440}\]

\[= \frac{327}{440}\]

So, the probability of winning a super prize in the SuperLottery is \( \frac{327}{440} \).

To solve this problem, let's first consider the structure of the cube.

When the large cube is cut into \(6^3\) smaller cubes, each small cube will have one face painted black and the remaining faces unpainted (since each face of the large cube was painted black initially).

Now, if we roll one of these small cubes, regardless of which face is on top initially, it has a \(1/6\) chance of landing with each face facing upward, since all faces are identical.

However, the probability that the face on top after rolling is black depends on whether the cube was rolled along an axis perpendicular to a black face or not. If it was, the black face will still be on top; if not, it won't be.

Considering that the black faces are on the outside of the original cube, any roll along the x, y, or z-axis will result in the black face being on top. There are 3 such axes.

So, the probability of the black face being on top after rolling is:

\[P(\text{Black face on top}) = \frac{3}{6} = \frac{1}{2}\]

Hence, the probability that when the small cube is rolled, the face on the top is black is \( \frac{1}{2} \).

We can find the values of b and c using the relationship between the quadratic formula and the roots of the equation.

Roots and Quadratic Formula:

The quadratic formula relates the coefficients (a, b, and c) of the quadratic equation to its roots (r1 and r2):

x = (-b ± sqrt(b^2 - 4ac)) / (2a)

In this case, we know the roots (5 + 3i and 5 - 3i) and want to find b + c.

Properties of Complex Roots:

Complex numbers as roots of quadratic equations always come in conjugate pairs. This means that if one root is a + bi (where i is the imaginary unit), the other root will be a - bi.

In this case, our roots are 5 + 3i and 5 - 3i, which confirms they are complex conjugates.

Sum of Roots and Coefficients:

There's a useful relationship between the roots (r1 and r2) of a quadratic equation and its coefficients (a, b, and c):

Sum of roots (r1 + r2) = -b / a

Product of roots (r1 * r2) = c / a

Since we're looking for b + c, and the quadratic has a leading coefficient of 1 (a = 1), we can use these relationships directly.

Finding b + c:

Sum of Roots:

The sum of the roots (5 + 3i) and (5 - 3i) is:

(5 + 5) + (3i - 3i) = 10

Since the sum of roots is also -b / a (and a = 1), we have:

-b = 10

Therefore, b = -10

Product of Roots:

The product of the roots (5 + 3i) and (5 - 3i) is:

(5 + 3i) * (5 - 3i) = 25 + 25 = 50

Since the product of roots is also c / a (and a = 1), we have:

c = 50

b + c:

Finally, add b and c to find the desired value:

b + c = -10 + 50 = 40

Therefore, b + c = 40.

We can solve for x and y in this system of equations by using substitution or the method of completing the square. Here, we'll use substitution:

Solve one equation for one variable:

From the first equation, we can express one variable in terms of the other:

x = 10 - y (Equation 1)

Substitute into the second equation:

Substitute this expression for x in the second equation:

(10 - y)^2 + y^2 = 64

Expand and solve for y:

Expand the squared term:

100 - 20y + y^2 + y^2 = 64 Combine like terms: 2y^2 - 20y + 36 = 0 Factor the equation: 2(y^2 - 10y + 18) = 0

This factors further as: 2(y - 6)(y - 3) = 0

Therefore, the possible values for y are:

y = 6 or y = 3

Substitute y back to find x:

Substitute each value of y back into Equation (1) to find the corresponding value of x:

If y = 6:

x = 10 - 6 = 4

If y = 3:

x = 10 - 3 = 7

Solutions:

Therefore, the ordered pairs (x, y) that satisfy the system of equations are:

(x, y) = (4, 6)

(x, y) = (7, 3)

To calculate the probability of winning the jackpot, we need to find the probability of two independent events happening: 

1. Matching the three white balls.

2. Matching the red SuperBall.

Let's calculate each probability separately:

1. **Matching the three white balls:**

   Since the white balls are drawn without replacement, the probability of matching the first number on your ticket is 1 out of 12, the second number is 1 out of 11, and the third number is 1 out of 10. This is because each time a ball is drawn, there is one fewer ball left in the pool.
   
   So, the probability of matching the three white balls is:

   \[\frac{1}{12} \times \frac{1}{11} \times \frac{1}{10}\]

2. **Matching the red SuperBall:**

   There is only one red SuperBall drawn, and you have one chance out of 8 (from 13 to 20) to match it.
   
   So, the probability of matching the red SuperBall is:
   
   \[\frac{1}{8}\]

Now, to find the probability of winning the jackpot, we multiply the probabilities of both events happening:

\[P(\text{Winning jackpot}) = P(\text{Matching three white balls}) \times P(\text{Matching red SuperBall})\]

\[= \left(\frac{1}{12} \times \frac{1}{11} \times \frac{1}{10}\right) \times \frac{1}{8}\]

\[= \frac{1}{12 \times 11 \times 10 \times 8}\]

\[= \frac{1}{10560}\]

So, the probability of winning the jackpot is \( \frac{1}{10560} \).

Problem 1: The answer is 16/3.

Problem 2: The answer is 72.

Problem 3: The answer is 120 degrees.

Here are the answers.

Problem 1: 122 degrees

Problem 2: 60

Problem 3: 12/7

Here are the answers.

Problem 1: 378*sqrt(3) - 144

Problem 2: 64

Problem 3: 3/8

Case 1: \(c \leq -5\)

Then 

\(-c-5-3c = 10 - 2c+8+6c\\ -8c - 23 = 0\\ c = -\dfrac{23}8\)

But then c is not less than or equal to -5. This root is rejected.

Case 2: \(-5 < c \leq 0\)

Then

\(c + 5 - 3c = 10 - 2c + 8 +6c\\ -6c - 13 = 0\\ c = -\dfrac{13}6\)

This root is in the range -5 < c <= 0. So we keep it.

Case 3: \(0 < c \leq 4\)

Then

\(c + 5 - 3c = 10 - 2c + 8 -6c \\ 6c = 13\\ c = \dfrac{13}6\)

This root is in range 0 < c <= 4. So we keep it.

Case 4: \(c > 4\)

Then

\(c + 5 - 3c = 10 + 2c - 8 - 6c\\ 2c + 3 = 0\\ c = -\dfrac32\)

But then c is not greater than 4. This root is rejected.

Hence, the solutions are: \(c = \dfrac{13}6\), \(c = -\dfrac{13}6\).

\(\quad \log_{1/2} \left(16\cdot 5 \cdot \dfrac1{10}\right)\\ = \log_{1/2} 8\\ = \dfrac{\log_2 8}{\log_2 \dfrac12}\\ =\dfrac3{-1}\\ = -3\)

\(\displaystyle xy + x = \frac{3x + 2y + z + y + 2z}{3}\\ xy + x = x + \dfrac{2y}3 + \dfrac z3 + \dfrac y3 + \dfrac{2z}3\\ xy + x = x + y + z\\ xy = y + z\\ x = \dfrac{y + z}y \)

\(\displaystyle \frac{a}{3} + 1 = \frac{a + 3}{a} + 1\\ \dfrac a3 + 1 = 2 + \dfrac 3a\\ \dfrac a3 - \dfrac3a - 1 = 0\\ \left(\dfrac a3\right)^2 - \dfrac a3 - 1 = 0\\ \dfrac a3 = \dfrac{1 \pm \sqrt 5}2\\ a = \dfrac{3(1 \pm \sqrt 5)}2\)

So the values of a are 3 times the golden ratio, or 3 times the negative reciprocal of the golden ratio. Nice!