Need help, URGENT maths question: (1 Viewer)

InteGrand · May 17, 2016

1008 said:
Thanks InteGrand. I've another question:

$$\noindent 1. Urn 1 contains 2 red balls and 3 black balls. Urn 2 contains 4 red balls and 5 black balls.$\\$a) Suppose a ball is drawn at random from urn 1 and placed into Urn 2. If a ball is then drawn at random from the 10 balls in urn 2, what is the probability that it is red? (I get this part) $\\$b) In the previous part, given that the ball drawn from urn 2 is red, what is the proba- bility that the ball transferred from urn 1 was black?$\\$2. Establish, using mathematical induction, Bonferoni's inequality: $\\P(A_{1}\cap A_{2}\cap ...\cap A_{n}) \geq 1-[P(A_{1}^{c})+P(A_{2}^{c}+...+P(A_{n}^{c}))$

I think these were both answered on this page: http://community.boredofstudies.org...320/introductory-probability.html#post7154145 .

1008 · May 17, 2016

InteGrand said:
I think these were both answered on this page: http://community.boredofstudies.org...320/introductory-probability.html#post7154145 .

Aaah...found them, thanks again!

$$\noindent 16. Let $B$ be a point in $\mathbb{R}^{n}$ with coordinate vector b. Let $x = a + \lambda d, \lambda \in \mathbb{R}$ be the equation of a line. $\\$(a) Show that the square of the distance from B to an arbitrary point x on the line is given by $\\q(\lambda) = |b-a|^{2}-2\lambda(b-a).d+\lambda^{2}|d|^{2}\\$(b) Find the shortest distance between the point B and the line by minimising $q(\lambda)\\$(c) If P is the point on the line closest to B, show that: $\\ \overrightarrow{PB} = b-a-proj_{d}(b-a),$ and show that $ \overrightarrow{PB}$ is orthogonal to the direction d of the line. $\\\\$

I get how to do the first two parts, How would you do c?

InteGrand · May 17, 2016

1008 said:
Aaah...found them, thanks again!

$$\noindent 16. Let $B$ be a point in $\mathbb{R}^{n}$ with coordinate vector b. Let $x = a + \lambda d, \lambda \in \mathbb{R}$ be the equation of a line. $\\$(a) Show that the square of the distance from B to an arbitrary point x on the line is given by $\\q(\lambda) = |b-a|^{2}-2\lambda(b-a).d+\lambda^{2}|d|^{2}\\$(b) Find the shortest distance between the point B and the line by minimising $q(\lambda)\\$(c) If P is the point on the line closest to B, show that: $\\ \overrightarrow{PB} = b-a-proj_{d}(b-a),$ and show that $ \overrightarrow{PB}$ is orthogonal to the direction d of the line. $\\\\$

I get how to do the first two parts, How would you do c?

This was answered here: http://community.boredofstudies.org/238/extracurricular-topics/350045/dot-product.html#post7148468 .

1008 · May 19, 2016

InteGrand said:
This was answered here: http://community.boredofstudies.org/238/extracurricular-topics/350045/dot-product.html#post7148468 .

Thanks

$$\noindent Convert $r=6sin \theta (0\leq \theta \leq \pi)$ into Cartesian form. Do this for $ r=2cos \theta (-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}).$

InteGrand · May 19, 2016

1008 said:
Thanks

$$\noindent Convert $r=6sin \theta (0\leq \theta \leq \pi)$ into Cartesian form. Do this for $ r=2cos \theta (-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}).$

$\noindent The first one is a circle with diameter $6$ centred on the $y$-axis, tangential to the $x$-axis; its Cartesian equation is $x^2 + \left(y-3\right)^2 = 9$. The second is a circle centred on the $x$-axis, with diameter 2, tangential to the $y$-axis; its Cartesian equation is $\left(x-1\right)^2 + y^2 = 1$.$

1008 · May 21, 2016

InteGrand said:
$\noindent The first one is a circle with diameter $6$ centred on the $y$-axis, tangential to the $x$-axis; its Cartesian equation is $x^2 + \left(y-3\right)^2 = 9$. The second is a circle centred on the $x$-axis, with diameter 2, tangential to the $y$-axis; its Cartesian equation is $\left(x-1\right)^2 + y^2 = 1$.$

Thanks.
$\noindent Let D, E, F be the points with coordinate vectors$\\d = \begin{pmatrix}5\\6\\7\end{pmatrix}, e = \begin{pmatrix}6\\7\\8\end{pmatrix}, f = \begin{pmatrix}7\\8\\10\end{pmatrix} \\$(a) Caculate $cos(\angle DEF)$ as a surd.$\\$(b) Calculate the area of $\Delta DEF$ as a surd.$

The problem I'm having is that I get cos(DEF) as a number greater than 1, so that the angle is undefined when I take its inverse cos. And for the second part, I plan to use the formula Area = 1/2 ab sin(c) to calculate the area, but can't do this...Maybe I'm doing something wrong???

InteGrand · May 21, 2016

1008 said:
Thanks.
$\noindent Let D, E, F be the points with coordinate vectors$\\d = \begin{pmatrix}5\\6\\7\end{pmatrix}, e = \begin{pmatrix}6\\7\\8\end{pmatrix}, f = \begin{pmatrix}7\\8\\10\end{pmatrix} \\$(a) Caculate $cos(\angle DEF)$ as a surd.$\\$(b) Calculate the area of $\Delta DEF$ as a surd.$

The problem I'm having is that I get cos(DEF) as a number greater than 1, so that the angle is undefined when I take its inverse cos. And for the second part, I plan to use the formula Area = 1/2 ab sin(c) to calculate the area, but can't do this...Maybe I'm doing something wrong???

$\noindent There's an easier way to find the area of a triangle in $\mathbb{R}^{3}$: take the length of the cross product of two adjacent vectors making up the triangle, and then halve it (without halving, it's the area of the \textsl{parallelogram} those vectors span rather than the triangle).$

$$\noindent For the first part, you must have made a mistake somewhere, the cosine of the angle between the vectors has to be between $-1$ and $1$.$

leehuan · May 21, 2016

1008 said:
Thanks.
$\noindent Let D, E, F be the points with coordinate vectors$\\d = \begin{pmatrix}5\\6\\7\end{pmatrix}, e = \begin{pmatrix}6\\7\\8\end{pmatrix}, f = \begin{pmatrix}7\\8\\10\end{pmatrix} \\$(a) Caculate $cos(\angle DEF)$ as a surd.$\\$(b) Calculate the area of $\Delta DEF$ as a surd.$

The problem I'm having is that I get cos(DEF) as a number greater than 1, so that the angle is undefined when I take its inverse cos. And for the second part, I plan to use the formula Area = 1/2 ab sin(c) to calculate the area, but can't do this...Maybe I'm doing something wrong???

MATH1151 ALG Ch4 Q22

a) Vec(DE) . Vec(FE) = (e-d).(e-f) = [1,1,1]^T . [-1,-1,2]^T = -4

|DE|²=sqrt(1+1+1)=sqrt(3)

|FE|²=sqrt(1+1+4)=sqrt(6)

Hence cos<DEF = (-4)/sqrt(3*6) = -4/(3sqrt2)

But yeah use the method InteGrand gave for part b)

1008 · May 21, 2016

InteGrand said:
$\noindent There's an easier way to find the area of a triangle in $\mathbb{R}^{3}$: take the length of the cross product of two adjacent vectors making up the triangle, and then halve it (without halving, it's the area of the \textsl{parallelogram} those vectors span rather than the triangle).$

Thanks, it is indeed easier!

$$\noindent For the first part, you must have made a mistake somewhere, the cosine of the angle between the vectors has to be between $-1$ and $1$.$

I found my mistake, so when I calculate DE = (1,1,1) and EF = (1,1,2), I find cos(DEF) = 4/sqrt(18). Previously, I found 4/(sqrt(12). I feel so dumb....

leehuan said:
MATH1151 ALG Ch4 Q22

a) Vec(DE) . Vec(FE) = (e-d).(e-f) = [1,1,1]^T . [-1,-1,2]^T = -4

|DE|²=sqrt(1+1+1)=sqrt(3)

|FE|²=sqrt(1+1+4)=sqrt(6)

Hence cos<DEF = (-4)/sqrt(3*6) = -4/(3sqrt2)

But yeah use the method InteGrand gave for part b)

Thanks!

Also, there isn't an answer provided at the back of the problems book for this, could you please explain if I am right or not? Here is the question:

The answer I get is:
Equation of altitude through A (l1) = (0,1,2) + t(2,-6,-14)
Equation of altitude through B (l2) = (-1,4,1) + u(0,3,3)
I found u and t by the cross product of BC and CD as well as AD and AC respectively.
Then, to determine if the lines intersect or not:

$\begin{pmatrix}0\\1\\2\end{pmatrix} + t \begin{pmatrix}2\\-6\\-14\end{pmatrix} = \begin{pmatrix}-1\\4\\1\end{pmatrix} + u \begin{pmatrix}0\\3\\3\end{pmatrix} \\ \Longrightarrow t\begin{pmatrix}-2\\6\\14\end{pmatrix} +u \begin{pmatrix}0\\3\\3\end{pmatrix}= \begin{pmatrix}1\\-3\\1\end{pmatrix} \\\text{Writing this as an augmented matrix Ax=b: }\\\begin{bmatrix}\left.\begin{matrix} -2& 0& \\ 6& 3& \\ 14& 3& \end{matrix}\right|\begin{matrix}1& \\ -3& \\ 1& \end{matrix}\end{bmatrix}$

As evident, b has to be a leading column, and the leading element in b doesn't equal 0 as lines aren't parallel. So altitudes don't intersect as there is no solution to this matrix.

InteGrand · May 21, 2016

1008 said:
Thanks, it is indeed easier!

I found my mistake, so when I calculate DE = (1,1,1) and EF = (1,1,2), I find cos(DEF) = 4/sqrt(18). Previously, I found 4/(sqrt(12). I feel so dumb....

Thanks!

Also, there isn't an answer provided at the back of the problems book for this, could you please explain if I am right or not? Here is the question:

The answer I get is:
Equation of altitude through A (l1) = (0,1,2) + t(2,-6,-14)
Equation of altitude through B (l2) = (-1,4,1) + u(0,3,3)
I found u and t by the cross product of BC and CD as well as AD and AC respectively.
Then, to determine if the lines intersect or not:

$\begin{pmatrix}0\\1\\2\end{pmatrix} + t \begin{pmatrix}2\\-6\\-14\end{pmatrix} = \begin{pmatrix}-1\\4\\1\end{pmatrix} + u \begin{pmatrix}0\\3\\3\end{pmatrix} \\ \Longrightarrow t\begin{pmatrix}-2\\6\\14\end{pmatrix} +u \begin{pmatrix}0\\3\\3\end{pmatrix}= \begin{pmatrix}1\\-3\\1\end{pmatrix} \\\text{Writing this as an augmented matrix Ax=b: }\\\begin{bmatrix}\left.\begin{matrix} -2& 0& \\ 6& 3& \\ 14& 3& \end{matrix}\right|\begin{matrix}1& \\ -3& \\ 1& \end{matrix}\end{bmatrix}$

As evident, b has to be a leading column, and the leading element in b doesn't equal 0 as lines aren't parallel. So altitudes don't intersect as there is no solution to this matrix.

A justification like that for why there's no solution would require us to put the augmented matrix into row-echelon form first, and then inspect it.

Another way to see there's no solution is to note that if there is a solution, the coefficient t needs to be -1/2, in order for the first entries to match. Then looking at second entries, we see that since t = -1/2, u needs to be 0. But with t = -1/2 and u = 0, the third entries don't match. So there can be no solution.

(I've assumed all your prior calculations were correct, didn't check them.)

leehuan · May 21, 2016

1008 said:
Thanks, it is indeed easier!

I found my mistake, so when I calculate DE = (1,1,1) and EF = (1,1,2), I find cos(DEF) = 4/sqrt(18). Previously, I found 4/(sqrt(12). I feel so dumb....

Thanks!

Also, there isn't an answer provided at the back of the problems book for this, could you please explain if I am right or not? Here is the question:

The answer I get is:
Equation of altitude through A (l1) = (0,1,2) + t(2,-6,-14)
Equation of altitude through B (l2) = (-1,4,1) + u(0,3,3)
I found u and t by the cross product of BC and CD as well as AD and AC respectively.
Then, to determine if the lines intersect or not:

$\begin{pmatrix}0\\1\\2\end{pmatrix} + t \begin{pmatrix}2\\-6\\-14\end{pmatrix} = \begin{pmatrix}-1\\4\\1\end{pmatrix} + u \begin{pmatrix}0\\3\\3\end{pmatrix} \\ \Longrightarrow t\begin{pmatrix}-2\\6\\14\end{pmatrix} +u \begin{pmatrix}0\\3\\3\end{pmatrix}= \begin{pmatrix}1\\-3\\1\end{pmatrix} \\\text{Writing this as an augmented matrix Ax=b: }\\\begin{bmatrix}\left.\begin{matrix} -2& 0& \\ 6& 3& \\ 14& 3& \end{matrix}\right|\begin{matrix}1& \\ -3& \\ 1& \end{matrix}\end{bmatrix}$

As evident, b has to be a leading column, and the leading element in b doesn't equal 0 as lines aren't parallel. So altitudes don't intersect as there is no solution to this matrix.

Pretty much like InteGrand said, don't forget the methods that Josef Dick explained

1. Use Gaussian elimination to bring out a R-E form.

Via matlab, assuming that your altitude through A is correct (the one through B definitely is correct) the reduced row-echelon form is

1 0 | 0
0 1 | 0
0 0 | 1

(Obviously, there's no need to actually bring out the reduced row echelon form on paper)

2. So since the right hand column is indeed a leading column like you said, there exists no solution. Hence the conclusion is also correct.

Edit: Also, this is me being pedantic now but technically Ax=b is this

$\left[\begin{matrix}-2&0\\6&3\\14&3\end{matrix}\right]\left(\begin{matrix}t\\u\end{matrix}\right)=\left(\begin{matrix}1\\-3\\1\end{matrix}\right)$

Augmented matrix notation is technically [A | b]

1008 · May 24, 2016

leehuan said:
Pretty much like InteGrand said, don't forget the methods that Josef Dick explained

1. Use Gaussian elimination to bring out a R-E form.

Via matlab, assuming that your altitude through A is correct (the one through B definitely is correct) the reduced row-echelon form is

1 0 | 0
0 1 | 0
0 0 | 1

(Obviously, there's no need to actually bring out the reduced row echelon form on paper)

2. So since the right hand column is indeed a leading column like you said, there exists no solution. Hence the conclusion is also correct.

Edit: Also, this is me being pedantic now but technically Ax=b is this

$\left[\begin{matrix}-2&0\\6&3\\14&3\end{matrix}\right]\left(\begin{matrix}t\\u\end{matrix}\right)=\left(\begin{matrix}1\\-3\\1\end{matrix}\right)$

Augmented matrix notation is technically [A | b]

Thanks leehuan! And yeah you're right, my notation is wrong for the augmented matrix, appreciate that.

I have another question:
$\noindent 1. Prove that $|Ax-y|$ is least when $y-Ax$ is orthogonal to $\sigma =$span(columns of $A).$

$\text{Given a set of vectors} \{\textbf{v}_{1},\textbf{v}_{2},...,\textbf{v}_{n}\} \text{such that none of}\\\text{ the vectors is a linear combination of ones earlier in the list,}\\\text{ prove that the following procedure contructs an orthonormal}\\\text{ set} \{\textbf{u}_{1},\textbf{u}_{2},...,\textbf{u}_{n}\}\text{ with the property that for each }k,\\ 1\leq k \leq n,\\\text{span} \{\textbf{u}_{1},\textbf{u}_{2},...,\textbf{u}_{k}\} = \text{span}\{\textbf{v}_{1},\textbf{v}_{2},...,\textbf{v}_{k}\} = \sigma_{k}.\\\text{Set} \textbf{u}_{1} = \frac{\textbf{v}_{1}}{|\textbf{v}_{1}|}.\\\text{For }k=2,...,n\text{ in turn set }\\\textbf{w}_{k}=\textbf{v}_{k}-\text{proj}_{\sigma_{k-1}}\textbf{v}_{k},\\\textbf{u}_{k} = \frac{\textbf{w}_{1}}{|\textbf{w}_{1}|}.$

InteGrand · May 24, 2016

1008 said:
Thanks leehuan! And yeah you're right, my notation is wrong for the augmented matrix, appreciate that.

I have a few more questions:
$\noindent 1. Prove that $|Ax-y|$ is least when $y-Ax$ is orthogonal to $\sigma =$span(columns of $A).$

$$\noindent 2. Given a set of vectors ${ \textbf{v_{1},v_{2},...,v_{n}}$ such that none of the vectors is a linear combination of ones earlier in the list, prove that the following procedure constructs an orthonormal set ${ \textbf{u_{1},u_{2},...,u_{n}}$ with the property that for each $ k, 1\leq k \leq n, $ span ${\textbf{u_{1},u_{2},...,u_{k}}} = $ span ${\textbf{v_{1},v_{2},...,v_{k}}} = \sigma_{k}.\\ $ Set $ u_{1}=\frac{v_{1}}{|v_{1}|}\\$ For k=2,...,n in turn set $\\ \textbf{w_{k}} = \textbf{v_{k}} - $proj$_{\sigma_{k-1}}\textbf{v_{k}},\\ u_{k}= \frac{w_{k}}{|w_{k}|}.$

A proof of 1) may be found here: http://community.boredofstudies.org/238/extracurricular-topics/350230/least-squares.html .

For 2), it is known as the Gram-Schmidt orthonormalisation process, and the proof is typically done by induction.

1008 · May 24, 2016

InteGrand said:
A proof of 1) may be found here: http://community.boredofstudies.org/238/extracurricular-topics/350230/least-squares.html .

For 2), it is known as the Gram-Schmidt orthonormalisation process, and the proof is typically done by induction.

Thanks, I'll leave the latter for later!

I'm pretty frustrated by these...What would be the shortest way of doing them?
Prove the following properties of matrices:
1) If the product AB exists, then A(λB) = λ(AB) = (λA)B
2) Associative law of matrix multiplication. If products AB and BC exist, then A(BC) = (AB)C
3) AI = A and IA = A where I represents identity matrices of the appropriate (possibly different) sizes
4) Left distributive law. If A+B and AC exist, then (A+B)C = AC+BC
5) Right distributive law If B+C and AB exist, then A(B+C) = AB+AC

I know you could do these writing each of them out as an nxn matrix...but it takes way to long to do so test-wise (+ its frustrating).

leehuan · May 24, 2016

It's actually possible to prove both distributive laws by considering component wise what goes on and then using some rules over the real numbers.

Associativity is much harder (because you can't assume distribution to prove association obviously). If I remember to I'll go grab my solution when my break ends

1008 · May 24, 2016

leehuan said:
It's actually possible to prove both distributive laws by considering component wise what goes on and then using some rules over the real numbers.

Associativity is much harder (because you can't assume distribution to prove association obviously). If I remember to I'll go grab my solution when my break ends

Thanks

But is there a way to prove the distributive laws WITHOUT considering the matrices component wise?

InteGrand · May 24, 2016

1008 said:
Thanks, I'll leave the latter for later!

I'm pretty frustrated by these...What would be the shortest way of doing them?
Prove the following properties of matrices:
1) If the product AB exists, then A(λB) = λ(AB) = (λA)B
2) Associative law of matrix multiplication. If products AB and BC exist, then A(BC) = (AB)C
3) AI = A and IA = A where I represents identity matrices of the appropriate (possibly different) sizes
4) Left distributive law. If A+B and AC exist, then (A+B)C = AC+BC
5) Right distributive law If B+C and AB exist, then A(B+C) = AB+AC

I know you could do these writing each of them out...but it takes way to long to do so test-wise (+ its frustrating).

$\noindent These come down to properties of sums and the definition of the $i,j$ entry of a matrix product (since this involves a sum). For the second one, the following fact will be handy:$

$\noindent If $a_1,\ldots a_n$ are real numbers and $b_1,\ldots, b_m$ are complex numbers (and hence this result also holds for real numbers), then $\sum _{i=1} ^n \sum _{j=1} ^{m} a_i b_j = \sum _{j=1}^{m} \sum _{i=1} ^n a_i b_j$. In fact, both are just equal to $\left(\sum _{i=1} ^n a_i \right) \left( \sum _{j=1} ^{m}b_j\right)$. To see this, note that$

$\begin{align*}\left(\sum _{i=1} ^n a_i \right) \left( \sum _{j=1} ^{m}b_j\right) &= \left(a_1 + a_2 + \cdots + a_n\right)\left( \sum _{j=1} ^{m}b_j\right) \\ &= a_1 \left( \sum _{j=1} ^{m}b_j\right) + a_2 \left( \sum _{j=1} ^{m}b_j\right) + \cdots + a_n \left( \sum _{j=1} ^{m}b_j\right) \quad (\text{distributive law}) \\ &= \sum _{i=1}^{n} a_i\left( \sum _{j=1} ^{m}b_j\right) \\ &= \sum _{i=1}^{n} \left( \sum _{j=1} ^{m} a_i b_j\right) \quad (\text{the }a_i \text{ can be moved inside, by distributivity}) \\ &\equiv \sum _{i=1}^{n} \sum _{j=1} ^{m} a_i b_j .\end{align*}$

$\noindent In a similar way, we can show that $\left(\sum _{i=1} ^n a_i \right) \left( \sum _{j=1} ^{m}b_j\right)= \sum _{j=1} ^{m} \sum _{i=1} ^n a_i b_j$. To do this, expand the $b$ sum instead of the $a$ sum that we did above. Alternatively, note that $\left(\sum _{i=1} ^n a_i \right) \left( \sum _{j=1} ^{m}b_j\right)=\left( \sum _{j=1} ^{m}b_j\right) \left(\sum _{i=1} ^n a_i \right) $ by commutativity, and then by the result proved above, $\left( \sum _{j=1} ^{m}b_j\right) \left(\sum _{i=1} ^n a_i \right)=\sum _{j=1} ^{m} \sum _{i=1} ^n a_i b_j$, so $\sum _{j=1} ^{m} \sum _{i=1} ^n a_i b_j=\left(\sum _{i=1} ^n a_i \right) \left( \sum _{j=1} ^{m}b_j\right)$ too.$

InteGrand · May 24, 2016

1008 said:
Thanks

But is there a way to prove the distributive laws WITHOUT considering the matrices component wise?

You don't need to write out an actual nxn matrix. You just need to use the definition of the i,j entry of a matrix product as a sum (essentially a dot product) and use properties of summations, denoting the i,j entry of a matrix A by say a_ij.

1008 · May 24, 2016

InteGrand said:
You don't need to write out an actual nxn matrix. You just need to use the definition of the i,j entry of a matrix product as a sum (essentially a dot product) and use properties of summations, denoting the i,j entry of a matrix A by say a_ij.

Is this right?

This is for the first question, btw

InteGrand · May 24, 2016

1008 said:
Is this right?

This is for the first question, btw

$\noindent The second line there is invalid (that'd be saying the sum of product of terms is equal to the product of the sums, which is not true, e.g. $1\times 5 + 2\times 6 \neq (1+2)\times (5+6)$). It's rather simple in fact though, here's how to do it.$

$\noindent Let $A$ have entries denoted $a_{ij}$, and similarly $b_{ij}$ for $B$ ($A$ an $m \times n$ and $B$ an $n \times p$ matrix over some field, e.g. the reals or complex numbers). Let $\lambda$ be a scalar from this field. Then we have for any $i$ from $1$ to $m$ and $j$ from $1$ to $p$ that the $i-j$ entry of $AB$ is$

$\begin{align*}\left[A\left(\lambda B\right)\right] _{ij} &= \sum _{k=1}^{n} a_{ik}\left(\lambda b_{kj}\right) \quad (\text{since the }i-j \text{ entry of }\lambda B\text{ is }\lambda b_{jk} \text{ by definition of scalar multiplication of a matrix}) \\ &= \sum _{k=1}^{n} \left(\lambda a_{ik}\right) b_{kj}= \left[\left(\lambda A\right)B\right]_{ij} \quad (\text{due to associativity and commutative of multiplication in a field}) \\ &= \lambda \sum _{k=1}^{n} a_{ik} b_{kj}= \lambda \left[ AB\right]_{ij} \quad (\text{due to the distributive law and associative law of multiplication in a field (the latter lets us firstly write }\left(\lambda a_{ik}\right) b_{kj} = \lambda a_{ik} b_{kj} \text{ and the former then lets us take the }\lambda \text{ outside the sum.} ) . \end{align*}$

In other words, the first part of the question follows due to the familiar properties of arithmetic in fields like the real/complex numbers.

(Also, your sums have been referring to the i-j components of matrices rather than the matrices themselves, so it wouldn't be right to say A(lambda B) equals those, for instance.)

Need help, URGENT maths question: (1 Viewer)

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