Answer:
The ship is located at (3,5)
Explanation:
In the first test, the equation of the position was:
5x² - y² = 20 ...........> equation I
In the second test, the equation of the position was:
y² - 2x² = 7 ..............> equation II
This equation can be rewritten as:
y² = 2x² + 7 ............> equation III
Since the ship did not move in the duration between the two tests, therefore, the position of the ship is the same in the two tests which means that:
equation I = equation II
To get the position of the ship, we will simply need to solve equation I and equation II simultaneously and get their solution.
Substitute with equation III in equation I to solve for x as follows:
5x²-y² = 20
5x² - (2x²+7) = 20
5x² - 2y² - 7 = 20
3x² = 27
x² = 9
x = <span>± </span>√9
We are given that the ship lies in the first quadrant. This means that both its x and y coordinates are positive. This means that:
x = √9 = 3
Substitute with x in equation III to get y as follows:
y² = 2x² + 7
y² = 2(3)² + 7
y = 18 + 7
y = 25
y = +√25
y = 5
Based on the above, the position of the ship is (3,5).
Hope this helps :)
Answer:
627200LJ
Step-by-step explanation:
Height of box=5 m
Side of square base=4 m
Volume of rectangular box=
Using the formula
Volume of rectangular box=
Density of material=
We know that
Using the formula
Mass of rectangular box=
Gravity=
Weight of rectangular box,F=
Let L be the vertical distance traveled by box
Total work done =
Therefore,total work done against gravity =
=627200 L J
Hence, the box requires work against gravity=627200LJ
7 cm: 35000 cm (7*5000)
2.8 cm: 14000 cm (2.8*5000)
2 cm: 10000 cm (2*5000)
2a+4(7+5a)
Distribute 4 into 7 and 5a
2a+28+20a
Combine like terms
Final Answer: 22a+28
