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not enough information  

In secondary school, the adjugate strategy (otherwise called the adjoint) and the Gauss-Jordan end technique are usually taught. Here, I will depict a somewhat intriguing technique that utilizes the Cayley-Hamilton hypothesis. 

For an n×n 

network, this technique requires finding the determinant of one n×n lattice, containing a vague x, and the result of (n−1) 

indistinguishable networks. 

Here is a bit by bit record of this technique. We expect that the n×n 

I guess you are referring to combinations, and want 6 choose 4

₆C₄ = 6!/(4!*[6-4]!  = 6*5/2 = 15

This question was asked and answered here: https://web2.0calc.com/questions/i-m-stuck-on-this-problem#r1 

Let the matrix \(\mathbf{A} = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}\).
Calculate \(\mathbf{A}^{10} \begin{pmatrix}1 \\ 0 \end{pmatrix},\ \mathbf{A}^{10} \begin{pmatrix}0 \\ 1 \end{pmatrix}\).

\(\begin{array}{|lrcll|} \hline \mathbf{A}^{\color{red}1} & &=& \begin{pmatrix} 1 & 2\cdot \color{red}1 \\ 0 & 1 \end{pmatrix} \\\\ \mathbf{A}^{\color{red}2} & = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}\begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix} &=& \begin{pmatrix} 1 & 2\cdot \color{red}2 \\ 0 & 1 \end{pmatrix} \\\\ \mathbf{A}^{\color{red}3} & = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}\begin{pmatrix} 1 & 2\cdot 2 \\ 0 & 1 \end{pmatrix} &=& \begin{pmatrix} 1 & 2\cdot \color{red}3 \\ 0 & 1 \end{pmatrix} \\\\ \mathbf{A}^{\color{red}4} & = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}\begin{pmatrix} 1 & 2\cdot 3 \\ 0 & 1 \end{pmatrix} &=& \begin{pmatrix} 1 & 2\cdot \color{red}4 \\ 0 & 1 \end{pmatrix} \\\\ \cdots \\ \mathbf{A}^{\color{red}10} & &=& \begin{pmatrix} 1 & 2\cdot \color{red}10 \\ 0 & 1 \end{pmatrix} \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline && \mathbf{A}^{\color{red}10}\begin{pmatrix}1 \\ 0 \end{pmatrix} \\ &=& \begin{pmatrix} 1 & 20 \\ 0 & 1 \end{pmatrix}\begin{pmatrix}1 \\ 0 \end{pmatrix} \\ &=& \begin{pmatrix}1 \\ 0 \end{pmatrix} \\ \hline \end{array} \begin{array}{|rcll|} \hline && \mathbf{A}^{\color{red}10} \begin{pmatrix}0 \\ 1 \end{pmatrix} \\ &=& \begin{pmatrix} 1 & 20 \\ 0 & 1 \end{pmatrix}\begin{pmatrix}0 \\ 1 \end{pmatrix} \\ &=& \begin{pmatrix}20 \\ 1 \end{pmatrix} \\ \hline \end{array}\)

This question has already been done to death.

Consider the matrices \(\begin{align*} \mathbf{A} &= \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}, \\ \mathbf{B} &= \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}, \\ \mathbf{C} &= \begin{pmatrix} 1 & 2 \\ 2 & -1 \end{pmatrix}. \end{align*}\)
Let \(v\) be a unit vector.
Find the magnitudes \(\|\mathbf{A}\mathbf{v}\|,\ \|\mathbf{B}\mathbf{v}\|,\ \|\mathbf{C}\mathbf{v}\|\),

or state if any of these cannot be uniquely determined from the information given.

\(\text{Let unit vector $v =\dbinom10$ }\)

\(\begin{array}{|rcll|} \hline \mathbf{A}\mathbf{v} &=& \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}\cdot\dbinom10 \\ \mathbf{A}\mathbf{v} &=& \dbinom11 \\ \|\mathbf{A}\mathbf{v}\| &=& \sqrt{1^2+1^2} \\ \mathbf{\|\mathbf{A}\mathbf{v}\|} &=& \mathbf{\sqrt{2}} \\ \hline \end{array} \begin{array}{|rcll|} \hline \mathbf{B}\mathbf{v} &=& \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}\cdot\dbinom10 \\ \mathbf{B}\mathbf{v} &=& \dbinom11 \\ \|\mathbf{B}\mathbf{v}\| &=& \sqrt{1^2+1^2} \\ \mathbf{\|\mathbf{B}\mathbf{v}\|} &=& \mathbf{\sqrt{2}} \\ \hline \end{array} \begin{array}{|rcll|} \hline \mathbf{C}\mathbf{v} &=& \begin{pmatrix} 1 & 2 \\ 2 & -1 \end{pmatrix}\cdot\dbinom10 \\ \mathbf{C}\mathbf{v} &=& \dbinom12 \\ \|\mathbf{C}\mathbf{v}\| &=& \sqrt{1^2+2^2} \\ \mathbf{\|\mathbf{C}\mathbf{v}\|} &=& \mathbf{\sqrt{5}} \\ \hline \end{array}\)

Let \(A\) be a matrix, and let \(x\) and \(y\) be vectors such that neither is a scalar multiple of the other and such that
\(\mathbf{A} \mathbf{x} = \mathbf{y},\ \mathbf{A} \mathbf{y} = \mathbf{x} + 2\mathbf{y}\).
Then we have that \(\mathbf{A}^{-1} \mathbf{x} = a \mathbf{x} + b\mathbf{y}\) for some scalars \(a\) and \(b\).
Find \(a\) and \(b\).

\(\begin{array}{|rcll|} \hline \mathbf{A} \mathbf{x} &=& \mathbf{y} \quad &| \quad \times \mathbf{A}^{-1} \\ \mathbf{A}^{-1}\mathbf{A}\mathbf{x} &=& \mathbf{A}^{-1} \mathbf{y} \quad &| \quad \mathbf{A}^{-1}\mathbf{A} = \text{Identity matrix} \\ \mathbf{x} &=& \mathbf{A}^{-1}\mathbf{y} \\\\ \hline \mathbf{A} \mathbf{y} &=& \mathbf{x} + 2\mathbf{y} \quad &| \quad \times \mathbf{A}^{-1} \\ \mathbf{A}^{-1}\mathbf{A}\mathbf{y} &=& \mathbf{A}^{-1}\mathbf{x} + 2\mathbf{A}^{-1} \mathbf{y} \quad &| \quad \mathbf{A}^{-1}\mathbf{A} = \text{Identity matrix} \\ \mathbf{y} &=& \mathbf{A}^{-1}\mathbf{x} + 2\mathbf{A}^{-1} \mathbf{y} \quad &| \quad \mathbf{A}^{-1}\mathbf{y} = \mathbf{x} \\ \mathbf{y} &=& \mathbf{A}^{-1}\mathbf{x} + 2\mathbf{x} \\ \mathbf{A}^{-1}\mathbf{x} &=& -2\mathbf{x}+\mathbf{y} \\ \hline \end{array}\)

\(\mathbf{a=-2,\ b=1} \)

done

If \(n\) is a positive integer and \((x+1)^n\) is expanded in decreasing powers of \(x\),
three consecutive numerical coefficients are in the ratio \(2:15:70\). Compute \(n\).

\(\begin{array}{|rcll|} \hline \mathbf{\dbinom{n}{k + 1}} &=& \dfrac{n!}{(k+1)!(n-k-1)!} \quad &| \quad k!(k+1)=(k+1)! \\\\ &=& \dfrac{1}{k+1} *\dfrac{n!}{k!(n-k-1)!} \quad &| \quad (n-k-1)!(n-k)=(n-k)! \\\\ &=& \dfrac{n-k}{k+1} * \dfrac{n!}{k!(n-k)!} \quad &| \quad \dfrac{n!}{k!(n-k)!} = \dbinom{n}{k} \\\\ &=& \mathbf{ \dfrac{n-k}{k+1} \dbinom{n}{k} } \\ \hline \end{array} \)

similarly \( \mathbf{\dbinom{n}{k - 1} = \dfrac{k}{n-k+1} \dbinom{n}{k} }\)

\(\begin{array}{|rcll|} \hline \dbinom{n}{k - 1} : \dbinom{n}{k} : \dbinom{n}{k + 1} &=& 2:15:70 \\\\ &&\boxed{ \mathbf{\dbinom{n}{k - 1} = \dfrac{k}{n-k+1} \dbinom{n}{k} }\\ ~\\ \mathbf{\dbinom{n}{k + 1} = \dfrac{n-k}{k+1} \dbinom{n}{k} } } \\\\ \dfrac{k}{n-k+1} \dbinom{n}{k} : \dbinom{n}{k} : \dfrac{n-k}{k+1} \dbinom{n}{k} &=& 2:15:70 \\ \hline \dfrac{\dbinom{n}{k}}{\dfrac{k}{n-k+1} \dbinom{n}{k}} &=& \dfrac{15}{2} \\\\ \dfrac{n-k+1}{k} &=& \dfrac{15}{2} \\\\ n-k+1 &=& \dfrac{15}{2}k \\ n &=& \dfrac{15}{2}k +k-1\\ \mathbf{n} &=& \mathbf{\dfrac{17}{2}k -1} \qquad (1) \\ \hline \dfrac{\dfrac{n-k}{k+1} \dbinom{n}{k} } {\dbinom{n}{k}} &=& \dfrac{70}{15} \\\\ \dfrac{n-k}{k+1} &=& \dfrac{70}{15} \\\\ 70(k+1) &=& 15(n-k) \\ 70k+70 &=& 15n-15k \\ 85k &=& 15n-70 \\ \mathbf{k} &=& \mathbf{ \dfrac{15n-70}{85} } \qquad (2) \\ \hline n &=& \dfrac{17}{2}k -1 \quad | \quad k=\dfrac{15n-70}{85} \\\\ n &=& \dfrac{17(15n-70)}{2*85}-1 \\\\ n &=& \dfrac{17(15n-70)}{170}-1 \\\\ n &=& \dfrac{ 15n-70 }{10}-1 \quad | \quad *10 \\\\ 10n &=& 15n-70-10 \\ 10n &=& 15n-80 \\ 5n &=& 80 \\ n &=& \dfrac{80}{5}\\ \mathbf{n} &=& \mathbf{16} \\\\ k &=& \dfrac{15n-70}{85} \quad | \quad n=16 \\\\ k &=& \dfrac{15*16-70}{85} \\ k &=& \dfrac{170}{85} \\ \mathbf{k} &=& \mathbf{2} \\ \hline \dbinom{n}{k - 1} : \dbinom{n}{k} : \dbinom{n}{k + 1} &=& 2:15:70 \\\\ \dbinom{16}{1} : \dbinom{16}{2} : \dbinom{16}{3} &=& 2:15:70 \\\\ 16 : 120 : 560 &=& 2:15:70 \\ \hline \end{array}\)

In △ABC, D lies on BC⎯⎯⎯⎯⎯⎯⎯⎯ such that BD=32CD. If AB=2AC and AD=32CD, what is the value of cos∠ADC (enter answer in fraction form)?

AB = 62.225

AD = BD = 32

CD = 1

AC = 31.1125

∠DAB = ∠DBA = 13.523277°

∠ADC = 27.046554°

cos∠ADC = 0.89063735   ≈  89 / 100   

Let's say all of them are $10 then the total is $470

$710 - $470 = $240

Because changing one $10 coin to a $20 coin adds $10,

There are 24 $20 coins

and 23 $10 coins

why is math  

So you can make sure the bank doesn't make a mistake when you cash your pay check.  Except, it you aren't founded in maths you won't be able to earn a pay check large enough to bother with accuracy anyway. 

Gosh, I'm so sorry! I totally forgot about that.

The rest of the question: Find the set of all possible values of xy.

Numbers under 100:

Since you are not allowed to divide by primes LESS THAN 2, 3, and 5, that means you must consider prime factors of 7 and higher. As a consequence, you must divide 100 / 7 =floor(14). Then you must count ALL prime number between 7 and 14. And you have:7, 11 and 13 =3. Next:floor(100/11)=9 and 9 < 11, therefore there are no prime to consider.

So, the total number of "quiteprimes" is the number of prime numbers as we calculated them above =3

because

Let's start by looking at the "restrictions" imposed within the problem. First is that there must be a chemistry textbook in the middle. At this stage of the problem, let's envision the 9 textbooks as 9 blanks, with the middle one having to be chemistry:

_ _ _ _ C _ _ _ _ 

Since we know it must be a chemistry book, that gives us 4 options for how many chemistry books could go into that particular "blank".

Next, let's look at the next restriction, which is that there must be a physics textbook at each end. The implication of this is that the blanks would look something like this:

P _ _ _ 4 _ _ _ P

With P being the physics textbooks. Since we know there are 5 physics textbooks total, the possibilites for the first physics "blank" are 5 books to choose from, with the next one being 4 to choose from. At this stage of the problem, we have :

5 _ _ _ 4 _ _ _ 4

, with the numbers representing how many ways there are to satisfy each respective restriction. After these 2 restrictions, we are left with 6 books(9-3) which we are free to place however. As such, the rest of the problem is just simple multiplication, with the final result looking like this:

6 6 5 4 4 3 2 1 4. Multiple these, and you get 6*6*5*4*4*3*2*1*4 = 69120

Ok I'm gonna assume some things about this question because you didn't fully explain. I'm answering based on the assumption that you have to find a*b*c in the equation ax^2+bx+c = 0. Let's take a look at this parabola first. Realize that the equation is 

1/2 * x^2. Look at how "fast" the graph increases. At an x value of 2, the y value is 2. At an x value of 4, the value is 8. Compare that to a normal graph (y = x^2) in which an x value of 2 produces a y value of 4, and an x value of 4 produces 16, and it's clear that the parabola has a coefficient of 1/2 in front of the x^2 term. With that being said, the problem becomes quite simple. We know the vertex of the parabola which is -3, -2. With that in mind, we can rewrite the parabola into vertex form, which is

a(x -h)^2 + k   with the vertex being (h,k), and a being the coefficient

This gives us 1/2(x+3)^2 - 2, which simplifies to 1/2x^2 + 3x + 5/2. Since the question(presumably) asks us for a * b * c, we get 1/2 * 3 * 5/2, which gives us an answer of 15/4.(Please correct me if I'm wrong)

Ok let's use some counting knowledge:

(N c 0) + (N c 1) + (N c 2) . . . .. . . . . . .  (N c N) is equal to 2^n    (If you don't believe me, you can try it out yourself). By that logic, (13c0) + (13c1) ......(13c13) = 2^13. Now, how does this relate? That's because realize that (13c0) + (13c1)......(13c6) is equal to (13c7) + (13c8)......(13c13) by symmetry (Think about the definition of N choose k). Therefore, the solution is simply 1/2 * 2^13, which gives us 2^12, and thus an answer of 4096

forgive me if i am wrong but i dont see the !

I don't think so, because 0! is one

isnt 13/0 error so it doesn't work

workin' on it.....

https://web2.0calc.com/questions/help-please-thx_4#r4

YZ ≈ 9.428