Math homework Help please? - diagram of sailboat types
A boat is located 5 km east of buoy O. A sailboat is located 3 miles west and 6 miles north of the buoy. Draw a diagram with a coordinate system with origin O. To what extent the two boats?
How do I know how the ships?
Saturday, December 5, 2009
Diagram Of Sailboat Types Math Homework Help Please?
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is a triangle ...
the buoy at 0
put the boat on the line 5 places on the right side
presented the vessel 3 squares left and 6 seats
A 6
B 5 plus 3 .. Na 8
6 ^ 2 + 8 ^ 2 = c ^ 2
Whoever ends up 9.16
I hope that helps
is a triangle ...
the buoy at 0
put the boat on the line 5 places on the right side
presented the vessel 3 squares left and 6 seats
A 6
B 5 plus 3 .. Na 8
6 ^ 2 + 8 ^ 2 = c ^ 2
Whoever ends up 9.16
I hope that helps
when plotted on the coordinate system has essentially two sides of a triangle, and they want to find the hypotenuse. You can use the Pythagorean theorum --
a ^ 2 + b ^ 2 = c ^ 2
* A * represents a leg
* B * represents the other leg
* C represents the hypotenuse.
Would be 8 * a * b * 6, then only solve for c * to *.
a ^ 2 + b ^ 2 = c ^ 2
8 ^ 2 + 6 ^ 2 = c ^ 2
64 + 36 = 100
Therefore, as equal to 100 * c ^ 2 comes from the square root of 100th is and the boats are 10 miles away.
The treatment of the buoy to the origin of a plot of X and Y. And in the north is positive and negative is south-east of the positive x and x is negative west. The first would be rowing (5.0) and the second boat (-3.6). Use the form distance D = [(x2-x1) ^ 2 + (y2-y1) ^ 2] ^ (1 / 2) to find the distance between them.
Country of origin and the two ships in the diagram. Then you will be able to see that the exception is 8 miles from east to west and 6 miles from north to south. with the Pythagorean theorem, the square of the distance plus the square of the distance EW suare NS equal to the distance between the vessels. SOOO
8 ^ 2 + 6 ^ 2 = D ^ 2
D ^ 2 = 64 +36 = 100 to the square root of both sides of d = 10
Under the coordinate system with the buoy that the origin and east as positive X and Y. North as a positive
Bounding boot = (5, 0)
Bounding candle = (-3, 6)
= √ (distance (x ₂ - x ₁) ² + (y ₂ - and ₁) ²)
............. = √ ((-3 - 5) ² + (6 - 0) ²)
............. = √ ((-8) ² + (6) ²)
............. = 10 mi
With coordinates.
The boat is at (-5, 0) in the case of the X-axis negative for the East.
And the ship is (3, 6) the axis is of course the west and north is positive.
Combine the two and use the distance formula to determine their distance.
First, draw the pattern, because the problem, "he said. Graph paper can help, but it is really necessary. There is a right triangle (the two boats and have a point 3 miles west of the buoy) and the rate of the Pythagoras (or solve triple) for you.
put them in a point similar to Graph # well, but much more. Then put in the middle of the buoy to buoy as the source. 5 Then go right to the boat ... and comtinue. North is up, down south, right, east, west, on the left. Place a compass on the map in the case of JST.
First, the buoy to 0 in a graph. Then place the boat launch at 5.0 and -3, 6 Then take the elevator. (x1 - x2) + (Y1 - Y2)
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