Q # 1 solve the problem
2 answers:
The solution is shown in the graph attached.
Explanation:
I am goind to teach you how to get that solution.
1) Restricctions: both x and y cannot be negative, i.e. x ≥ 0 and y ≥ 0
⇒ the solution is on the first quadrant.
2) Using the prices of the tickets, $ 8 for adults, and $ 6 for children, the linear equation for the costs is: cost = 8x + 6y.
3) Since the radio station is willing to spend a maximum of $ 172 you have the final restriction:
⇒ 8x + 6y ≤ 172.
4) Then the solutions that meet the three restrictions (x ≥0, y ≥ 0, and 8x + 6y ≤ 172) is found graphically by drawing the line 8 x + 6y = 172
5) To draw the line 8x + 6y = 172, use the axis intercepts:
x = 0 ⇒ y = 172/6 = 86/3 ⇒ point (0, 86/3)
y = 0 ⇒ x = 172/8 = 86/4 ⇒ point (86/4, 0)
6) Once you have the line you shade the region that is between the line and the two axis. That region contain of the possible solutions.
Explanation:
I am goind to teach you how to get that solution.
1) Restricctions: both x and y cannot be negative, i.e. x ≥ 0 and y ≥ 0
⇒ the solution is on the first quadrant.
2) Using the prices of the tickets, $ 8 for adults, and $ 6 for children, the linear equation for the costs is: cost = 8x + 6y.
3) Since the radio station is willing to spend a maximum of $ 172 you have the final restriction:
⇒ 8x + 6y ≤ 172.
4) Then the solutions that meet the three restrictions (x ≥0, y ≥ 0, and 8x + 6y ≤ 172) is found graphically by drawing the line 8 x + 6y = 172
5) To draw the line 8x + 6y = 172, use the axis intercepts:
x = 0 ⇒ y = 172/6 = 86/3 ⇒ point (0, 86/3)
y = 0 ⇒ x = 172/8 = 86/4 ⇒ point (86/4, 0)
6) Once you have the line you shade the region that is between the line and the two axis. That region contain of the possible solutions.
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The imaginary unit belongs to the set of complex numbers, denoted by
. These numbers take the form
, where
are any real numbers.
The set of real numbers, , is a subset of
, where each number in
can be obtained by taking
and letting
be any real number.
But any number in with non-zero imaginary part is not a real number. This includes
.
- "is it possible that i can use an imaginary number for a real number"
I'm not sure what you mean by this part of your question. It is possible to represent any real number as a complex number, but not a purely imaginary one. All real numbers are complex, but not all complex numbers are real. For example, 2 is real and complex because .
There are some operations that you can carry out on purely imaginary numbers to get a purely real number. A famous example is raising to the
-th power. Since
, we have