Answer:
B. The approximate length of EF is 4.47 units, and the approximate perimeter of triangle EFG is 12.94 units.
Step-by-step explanation:
First step is to determine the length of EF, since that will give us 2 sides of the triangle (since EG = EF).
From the diagram, we can easily make a rectangle triangle by dropping a vertical line from vertex E, let's name Z the meeting point of that line with the segment GF. Then we have a rectangle triangle EZF with a height of 4 and a base of 2, of which EF is the hypotenuse. So...
EF² = 4² + 2² = 16 + 4 = 20
EF = √20 = 4.47
Now that we have EF, we also have EG:
EF = 4.47
EG = 4.47
GF = 4 (visible on the graph)
Perimeter = 4.47 + 4.47 + 4 = 12.94 units.
<u>In Bar Graphs;</u>
- Bars have equal space
- One the y-axis, we have numbers & on the x-axis, we have data which can be anything.
<u> In Histograms;</u>
- Bars are fixed
- On the y-axis, we have numbers & and on the x-axis, we have data which in continuous & will always be number.
<u>An easy way you can remember the difference is looking at the spaces of the bars. </u>
<em>A bar graph has gaps</em>
<em>A histogram has no gaps.</em>
Answer:
$1170
Step-by-step explanation:
Let the sells for economy seats be =x
Let the sells for deluxe seats be=y
The inequalities that can be obtained are;
x≥1 --------------------at least 1 economy seats
y≥6 --------------------at least 6 deluxe seats
x+y=30-----------------maximum number of passengers allowed on each boat
Graph the inequalities
Use the graph tool to locate the point of maximum profit.The intersecting point for the three graphs
The point is (24,6)
Hence, x=24 and y=6
Profit for each
Economy seats 24×$40=$960
Deluxe seats 6×$35=$210
Maximum profit for one tour
$960+$210=$1170
What says odd integers???
Answer:
The value could replace q is 108 - n ⇒ the last answer
Step-by-step explanation:
* Lets study the values of quarters and nickles
- One quarter = 25 cents
- One nickle = 5 cents
- The number of coins is 108
- The coins are nickles or quarters only
- The coins worth $ 21
* Lets solve the problem
∵ The number of coins is 108 coins
- Let the quarter is q and the nickel is n
∴ q + n = 108 ⇒ (1)
∵ The coins worth $21
∵ $ 1 = 100 cents
∴ $21 = 21 × 100 = 2100 cents
∵ The quarter = 25 cents
∵ The nickels = 5 cents
∴ 25q + 5n = 2100 ⇒ (2)
- To find the value which replace q use equation (1)
∵ q + n = 108 ⇒ subtract n from both sides
∴ q = 108 - n
* The value could replace q is 108 - n