The perimeter of a playing card is 30 centimeters. The playing card is 5 centimeters wide. How tall is it?
1 answer:
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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.
Answer:
True
Step-by-step explanation:
The integral
in the integral we have a product of a monomial:
and an exponential function:
In general case, when we have a combination of these two things you can use the integration by parts, where will be
and
will be
.
The statement is true