Answer: hello your question lacks some data hence I will be making an assumption to help resolve the problem within the scope of the question
answer:
≈ 95 units ( output level )
Step-by-step explanation:
Given data :
P = 2000 - Q/10
TC = 2Q^2 + 10Q + 200 ( assumed value )
<u>The output level where a purely monopolistic market will maximize profit</u>
<u>at MR = MC </u>
P = 2000 - Q/10 ------ ( 1 )
PQ = 2000Q - Q^2 / 10 ( aka TR )
MR = d (TR ) / dQ = 2000 - 2Q/10 = 2000 - Q/5
TC = 2Q^2 + 10Q + 200 ---- ( 2 )
MC = d (TC) / dQ = 4Q + 10
equating MR = MC
2000 - Q/5 = 4Q + 10
2000 - 10 = 4Q + Q/5
1990 = 20Q + Q
∴ Q = 1990 / 21 = 94.76 ≈ 95 units ( output level )
Answer:
a = -5
Step-by-step explanation:
(5 - 3)/(5 - a)
2/(5 - a)
perpendicular of 2/(5 - a) = -(5 - a)/2
(a - 0)/(1 - 0) = -(5 - a)/2
a/1 = -(5 - a)/2
2a = -5 + a
a = -5
Answer:
![\:c_{11}=-2,\:\:\:c_{12}=-2](https://tex.z-dn.net/?f=%5C%3Ac_%7B11%7D%3D-2%2C%5C%3A%5C%3A%5C%3Ac_%7B12%7D%3D-2)
![\:c_{21}=5,\:\:\:c_{22}=8](https://tex.z-dn.net/?f=%5C%3Ac_%7B21%7D%3D5%2C%5C%3A%5C%3A%5C%3Ac_%7B22%7D%3D8)
Step-by-step explanation:
Given the matrices
![A=\begin{pmatrix}1&-1\\ 2&1\end{pmatrix}](https://tex.z-dn.net/?f=A%3D%5Cbegin%7Bpmatrix%7D1%26-1%5C%5C%202%261%5Cend%7Bpmatrix%7D)
![B=\begin{pmatrix}1&2\\ \:3&4\end{pmatrix}](https://tex.z-dn.net/?f=B%3D%5Cbegin%7Bpmatrix%7D1%262%5C%5C%20%5C%3A3%264%5Cend%7Bpmatrix%7D)
Calculating AB:
![\begin{pmatrix}1&-1\\ \:\:2&1\end{pmatrix}\times \:\begin{pmatrix}1&2\\ \:\:3&4\end{pmatrix}=\begin{pmatrix}c_{11}&c_{12}\\ \:\:\:c_{21}&c_{22}\end{pmatrix}](https://tex.z-dn.net/?f=%5Cbegin%7Bpmatrix%7D1%26-1%5C%5C%20%5C%3A%5C%3A2%261%5Cend%7Bpmatrix%7D%5Ctimes%20%5C%3A%5Cbegin%7Bpmatrix%7D1%262%5C%5C%20%5C%3A%5C%3A3%264%5Cend%7Bpmatrix%7D%3D%5Cbegin%7Bpmatrix%7Dc_%7B11%7D%26c_%7B12%7D%5C%5C%20%5C%3A%5C%3A%5C%3Ac_%7B21%7D%26c_%7B22%7D%5Cend%7Bpmatrix%7D)
Multiply the rows of the first matrix by the columns of the second matrix
![=\begin{pmatrix}1\cdot \:1+\left(-1\right)\cdot \:3&1\cdot \:2+\left(-1\right)\cdot \:4\\ 2\cdot \:1+1\cdot \:3&2\cdot \:2+1\cdot \:4\end{pmatrix}](https://tex.z-dn.net/?f=%3D%5Cbegin%7Bpmatrix%7D1%5Ccdot%20%5C%3A1%2B%5Cleft%28-1%5Cright%29%5Ccdot%20%5C%3A3%261%5Ccdot%20%5C%3A2%2B%5Cleft%28-1%5Cright%29%5Ccdot%20%5C%3A4%5C%5C%202%5Ccdot%20%5C%3A1%2B1%5Ccdot%20%5C%3A3%262%5Ccdot%20%5C%3A2%2B1%5Ccdot%20%5C%3A4%5Cend%7Bpmatrix%7D)
![=\begin{pmatrix}-2&-2\\ 5&8\end{pmatrix}](https://tex.z-dn.net/?f=%3D%5Cbegin%7Bpmatrix%7D-2%26-2%5C%5C%205%268%5Cend%7Bpmatrix%7D)
Hence,
![\begin{pmatrix}c_{11}&c_{12}\\ \:\:\:c_{21}&c_{22}\end{pmatrix}=\begin{pmatrix}-2&-2\\ \:5&8\end{pmatrix}](https://tex.z-dn.net/?f=%5Cbegin%7Bpmatrix%7Dc_%7B11%7D%26c_%7B12%7D%5C%5C%20%5C%3A%5C%3A%5C%3Ac_%7B21%7D%26c_%7B22%7D%5Cend%7Bpmatrix%7D%3D%5Cbegin%7Bpmatrix%7D-2%26-2%5C%5C%20%5C%3A5%268%5Cend%7Bpmatrix%7D)
Therefore,
![\:c_{11}=-2,\:\:\:c_{12}=-2](https://tex.z-dn.net/?f=%5C%3Ac_%7B11%7D%3D-2%2C%5C%3A%5C%3A%5C%3Ac_%7B12%7D%3D-2)
![\:c_{21}=5,\:\:\:c_{22}=8](https://tex.z-dn.net/?f=%5C%3Ac_%7B21%7D%3D5%2C%5C%3A%5C%3A%5C%3Ac_%7B22%7D%3D8)
Answer: See explanation
Step-by-step explanation:
The perimeter of a pentagon is gotten through the summation of its five sides. Let the first side be represented by x. Since each side of a pentagon is 10 cm greater than the previous side, then the sides will be:
First side = x
Second side = x + 10
Third side = x + 10 + 10 = x + 20
Forth side = x + 30
Fifty side = x + 40
Therefore,
x + (x + 10) + (x + 20) + (x + 30) + (x + 40) = 500
5x + 100 = 500
5x = 500 - 100
5x = 400
x = 400/5
x = 80
Therefore, the lengths will be:
First side = x = 80cm
Second side = x + 10 = 80 + 10 = 90cm
Third side = x + 20 = 80 + 20 = 100cm
Forth side = x + 30 = 80 + 30 = 110cm
Fifty side = x + 40 = 80 + 40 = 120cm