Answer:
2x + 3y ≥ 5
Step-by-step explanation:
See the graph attached.
The bold straight line passes through the points (1,1) and (4,-1).
Therefore, the equation of the straight line will be
⇒ 3(y + 1) = - 2(x - 4)
⇒ 3y + 3 = - 2x + 8
⇒ 2x + 3y = 5 ............. (1)
Now, the shaded region i.e. the solution to the inequality does not include the origin(0,0).
So, putting x = 0 and y = 0 in the equation (1) we get, 0 < 5
Therefore, the inequality equation is 2x + 3y ≥ 5 (Answer)
A) zeroes
P(n) = -250 n^2 + 2500n - 5250
Extract common factor:
P(n)= -250 (n^2 - 10n + 21)
Factor (find two numbers that sum -10 and its product is 21)
P(n) = -250(n - 3)(n - 7)
Zeroes ==> n - 3 = 0 or n -7 = 0
Then n = 3 and n = 7 are the zeros.
They rerpesent that if the promoter sells tickets at 3 or 7 dollars the profit is zero.
B) Maximum profit
Completion of squares
n^2 - 10n + 21 = n^2 - 10n + 25 - 4 = (n^2 - 10n+ 25) - 4 = (n - 5)^2 - 4
P(n) = - 250[(n-5)^2 -4] = -250(n-5)^2 + 1000
Maximum ==> - 250 (n - 5)^2 = 0 ==> n = 5 and P(5) = 1000
Maximum profit =1000 at n = 5
C) Axis of symmetry
Vertex = (h,k) when the equation is in the form A(n-h)^2 + k
Comparing A(n-h)^2 + k with - 250(n - 5)^2 + 1000
Vertex = (5, 1000) and the symmetry axis is n = 5.
Answer:
x=1.4 or 21/15
Step-by-step explanation:
Answer: A. ![\left[\begin{array}{ccc}29&13\\13&10\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D29%2613%5C%5C13%2610%5C%5C%5Cend%7Barray%7D%5Cright%5D)
Step-by-step explanation:
The question is asking us to find the product of the matrices. The key difference is the second A has a little <em>T</em> in the exponent. This <em>T</em> means transpose. You multiply A by the transpose of A. To find the transpose, you turn the rows into columns.
![A^T=\left[\begin{array}{ccc}5&3\\2&-1\\\end{array}\right]](https://tex.z-dn.net/?f=A%5ET%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D5%263%5C%5C2%26-1%5C%5C%5Cend%7Barray%7D%5Cright%5D)
Now that we have our transpose, we can multiply the matrices.
![\left[\begin{array}{ccc}5&2\\3&-1\\\end{array}\right] \left[\begin{array}{ccc}5&3\\2&-1\\\end{array}\right] =\left[\begin{array}{ccc}5*5+2*2&5*3+2(-1)\\3*5+2(-1)&3*3+(-1)(-1)\\\end{array}\right] =\left[\begin{array}{ccc}29&13\\13&10\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D5%262%5C%5C3%26-1%5C%5C%5Cend%7Barray%7D%5Cright%5D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D5%263%5C%5C2%26-1%5C%5C%5Cend%7Barray%7D%5Cright%5D%20%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D5%2A5%2B2%2A2%265%2A3%2B2%28-1%29%5C%5C3%2A5%2B2%28-1%29%263%2A3%2B%28-1%29%28-1%29%5C%5C%5Cend%7Barray%7D%5Cright%5D%20%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D29%2613%5C%5C13%2610%5C%5C%5Cend%7Barray%7D%5Cright%5D)
