All Answers 353099 Answers

-4 + 3i + (-2 -i) = -6 + 2i. Plot that easily and connect the points.

Tthe number of ways of choosing three people in a circular table with 12 people, so that no two people are next to each other.

Solution:

There are 12 ways to choose 3 persons who are adjacent (next to each other).

There are 12 ways to choose 2 persons who are adjacent.

There are 8 ways to choose 1 person who is not adjacent to the other two (2).

There are nCr(12,3) to choose a set of three persons from set of 12.

Subtract the persons who are adjacent from the total sets.

\(\text{So then ... } \dbinom {12}{3} - 12 - (12*8) = 112\\ \small \text{ There are 112 ways to select three persons from a set of 12 seated in a circle, such that no two persons are adjacent. }\\\)

GA

--. .-

the answer to this question is 180

The area of a trapezoid is $80$. The length of one base is $16$ units greater than the other base, and one base of the trapezoid is $12$. Find the length of the median of the trapezoid.  

The length of the median is equal to the average of the two bases.  

So, with one base 12, and the other base 16 longer, we have  

                  12 + (12 + 16)        40  

median  =  ––––––––––––  =  ––––  =  20  

                             2                   2  

The short base has to be 12, because 12 can't be 16 greater than another positive number.  

The area of the trapezoid has no bearing on the solution to the problem.  

.

I can give you a hint with problem 2!

If we have 3 pieces, and 9 places to put them, then how many places can we put them in? Try to make a diagram to help you :)

\(27\)

In a triangle, the sum of the angles is always 180 degrees. So, we can set up the following equation:

A+B+C=180

Substituting the given expressions for A, B, and C, we get:

(p+2q)+(6p−5q)+(14p+8q)=180

Simplifying, we get:

21p+5q=180

Solving for p, we find:

p=(180−5q)/21​

So, p in terms of q is (180−5q)/21​ degrees.

I have $3$ different mathematics textbooks, $2$ different psychology textbooks, and $2$ different chemistry textbooks. In how many ways can I place the $7$ textbooks on a bookshelf, in a row?  

                 7!             5040  

₇P₇  =  –––––––  =  ––––––  =  5040       

             (7 – 7)!           1   

You might be alarmed by seeing zero in the denominator,   

which is the most verboten condition known in math,   

but zero factorial is defined as 1 so it's okay.  

_.

200 =  2^3 * 5^2 * 1   =  2*2*2*5*5*1

No. of distinct  sequences  =

6! / [ No. of times 2 appears ! * No. of times 5 appears ! ]

6! / (3! * 2!)  =   60

By  " brute force " .....

Laverne  says     2   7   12    17      22       27      32

Progressive sums     9      21    38        60       87      119

Another (more mathematical )  way

First term Laverne says = 2

Let the last term she says = 2 + 5n

Sum  > 100

[ 2 + (2 + 5n) ]  * n  / 2   > 100

[ 5n + 4 ] * n  >  200

5n^2 + 4n - 200 > 0      (1)            set =  0   and solve for positive n

5n^2 + 4n - 200 = 0

n = [ -4 + sqrt [ 16 + 4000] ] / 10    ≈  5.9

≈ 5.9 is a root of 5n^2 + 4n - 200....so....the next largest integer also  makes (1)  true

n = 6

And the number that Laverne says  is  2 + 5(6)  =    32

Not great at "counting" probs....here's my best attempt....corrections welcome !!!

1 3      5 6 7 8 9 10 11                7

2 4      6 7 8 9 10 11 12              7

3 5      7 8 9 10 11 12                 6

4 6      8 9 10 11 12 1                 6

5 7      9 10 11 12 1  2                6

6 8     10 11 12 1 2  3                 6

7 9     11 12 1  2  3  4                 6

8 10    12 1 2 3 4 5                     6

9 11     1 2 3 4 5 6                      6

10 12   2 3 4 5 6 7                      6

11  1    4 5 6 7 8                         5

12  2    5 6 7 8 9                         5

2(5) + 8(6) + 2(7)  =  72 ways

  See here

There are no students  only  in the Math club

3 Students are in  both  - 3 who are in the Science club are also in the Math club

There are 3 students  only in the  Science club

tan (7pi/4)

Note that this  really     tan ( 2pi - pi/4)  

So...it's just  a 45°  angle in the 4th quadrant =   tan (- pi / 4)

tan (pi/4)  =  1

But in the 4th quadrant the tangent  is  -

So

tan (7pi/4)  = tan (- pi /4)  =   -1

The bottom left corner is (-2,-5)! I just checked

thanks, cphill!

OPO  = (3/5)(2/4)(2/3)  = 1/5

POP  = (2/5)(3/4)(1/3)  =  1/10

1/5 + 1/10  =   3 / 10

1 step  =  4 points

2 steps  = 9 points

3 steps = 16 points

n steps  = (n + 1)^2  points

12 steps  = (12 + 1)^2  =  169 points

If \(S_1\) is 1/2

that means \(a_1\) is 1/2.

If \(a_1 \) is 1/2, then \(a_2\) is 0

So \(a_{15}\) is 1/2 - 14(1/2) = 13/2

So 1/2 + 13/2 = 7

7 times 15/2 = 105/2

1% x 200 = 2

so 2kg of berries did not change.

198kg were water.

10% of the water were gone.

10% x 200 = 20

That means now 178kg were now water

So 178+2 = 180

The 23rd term is equal to: \(a_1\times r^{22}\)

The 26th terms is equal to: \(a_1\times r^{25}\)

So then \(r^3= \)\(12\over{16}\)

r^3 = 3/4

r = \(\sqrt[3]{3\over4}\)

32nd term would be 23rd term times r^9

\((\sqrt[3]{3\over4})^9\)=3^3/4^3 = 27/64

16\(\times\)27/64 = 27/4

Max, your monographic summary analysis on binomial theory is amazing! It’s one of the best I’ve read anywhere. 

It’s wonderful you have returned! I hope you stay around for awhile.

I will let you figure out the rest.

I don’t know if the OP student will choose to figure out the rest, but I sure as hell will.

------------

You may find this formula interesting.

\(\sum \limits_{k=0}^{m}\binom{n}{k} (-1)^k \binom{p – s*k – 1}{p – s*k – n} \leftarrow \text {where p is the point (sum) target, n is the # of die, }\\ \hspace {47mm} \text{s is the number of sides, and } m=\lfloor \dfrac{p-n}{s}\rfloor \\\)

This formula is from J. V. Uspensky’s Introduction to Mathematical Probability (1937).

This calculates the number of combinations for any given sum of pips on (N) number of dice with (S) sides, where each die-side has a unique number of pips from 1 to (S)