Live Fund receive 
it they sold half thier holding in Marks Brothers.
<u>Solution:</u>
Given: Sale price of Live Fund holding = 5000 dollar
To find: Amount that Live Fund will get if they sell half of their holding in Marks Brothers.
Assume the total number of shares held by Live Fund in Marks Brothers as A
Therfore, half the holdings (or half the number of shares) of Live Fund will be 
.
Thus, if each share is valued at 
, then the total value of the number of shares sold will be as follows,
Hence, Live Fund will receive 
Answer:
[2P-3.) ^2+(2P+3)^2[2P-3.) ^2+(2P+3)^2[2P-3.) ^2+(2P+3)^2
1.) The sum(addition) of 21 and 5 times(multiplication) a number f is(=) 61.
f = unknown number/variable [So 21 plus 5f(5 times f) equals 61]
21 + 5f = 61 [21(one-time) + 5f(number x variable) = 61(total)]
2.) Seventeen more(addition) than seven times(multiplication) a number j is(=) 87.
j = unknown number/variable [So 17 plus 7j(7 times j) equals 87]
17 + 7j = 87
3.) n = number of calls
18 + 0.05n = 50.50
[Company charges $18 plus five cents per call(n), and the total charge was $50.50]
4.) s = the number of students
40 + 30s = 220
[Tutor charges $40 plus $30 per student(s), and the total charge was $220]
Answer:
Infinite number of solutions.
Step-by-step explanation:
We are given system of equations
Firs we find determinant of system of equations
Let a matrix A=![\left[\begin{array}{ccc}5&4&5\\1&1&2\\2&1&-1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D5%264%265%5C%5C1%261%262%5C%5C2%261%26-1%5Cend%7Barray%7D%5Cright%5D)
and B=![\left[\begin{array}{ccc}-1\\1\\-3\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-1%5C%5C1%5C%5C-3%5Cend%7Barray%7D%5Cright%5D)
Determinant of given system of equation is zero therefore, the general solution of system of equation is many solution or no solution.
We are finding rank of matrix
Apply 
and 
![\left[\begin{array}{ccc}1&0&1\\1&1&2\\0&-1&-3\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%260%261%5C%5C1%261%262%5C%5C0%26-1%26-3%5Cend%7Barray%7D%5Cright%5D)
:![\left[\begin{array}{ccc}-5\\1\\-5\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-5%5C%5C1%5C%5C-5%5Cend%7Barray%7D%5Cright%5D)
Apply
![\left[\begin{array}{ccc}1&0&1\\0&1&1\\0&-1&-3\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%260%261%5C%5C0%261%261%5C%5C0%26-1%26-3%5Cend%7Barray%7D%5Cright%5D)
:![\left[\begin{array}{ccc}-5\\6\\-5\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-5%5C%5C6%5C%5C-5%5Cend%7Barray%7D%5Cright%5D)
Apply 
![\left[\begin{array}{ccc}1&0&1\\0&1&1\\0&0&-2\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%260%261%5C%5C0%261%261%5C%5C0%260%26-2%5Cend%7Barray%7D%5Cright%5D)
:![\left[\begin{array}{ccc}-5\\6\\1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-5%5C%5C6%5C%5C1%5Cend%7Barray%7D%5Cright%5D)
Apply 
and 
![\left[\begin{array}{ccc}1&0&1\\0&1&0\\0&0&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%260%261%5C%5C0%261%260%5C%5C0%260%261%5Cend%7Barray%7D%5Cright%5D)
:![\left[\begin{array}{ccc}-5\\\frac{13}{2}\\-\frac{1}{2}\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-5%5C%5C%5Cfrac%7B13%7D%7B2%7D%5C%5C-%5Cfrac%7B1%7D%7B2%7D%5Cend%7Barray%7D%5Cright%5D)
Apply 
![\left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%260%260%5C%5C0%261%260%5C%5C0%260%261%5Cend%7Barray%7D%5Cright%5D)
:![\left[\begin{array}{ccc}-\frac{9}{2}\\\frac{13}{2}\\-\frac{1}{2}\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-%5Cfrac%7B9%7D%7B2%7D%5C%5C%5Cfrac%7B13%7D%7B2%7D%5C%5C-%5Cfrac%7B1%7D%7B2%7D%5Cend%7Barray%7D%5Cright%5D)
Rank of matrix A and B are equal.Therefore, matrix A has infinite number of solutions.
Therefore, rank of matrix is equal to rank of B.
Answer:
The value of k is -11
Step-by-step explanation:
If (x+2) is a factor of x3 − 6x2 + kx + 10:
Then,
f(x)=x3 − 6x2 + kx + 10
f(-2)=0
f(-2)=(-2)³-6(-2)²+k(-2)+10=0
f(-2)= -8-6(4)-2k+10=0
f(-2)= -8-24-2k+10=0
Solve the like terms:
f(-2)=-2k-22=0
f(-2)=-2k=0+22
-2k=22
k=22/-2
k=-11
Hence the value of k is -11....
