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:
<em>Good Luck!</em>
Step-by-step explanation:
Answer:
<h2>The answer is 15cm</h2>
Step-by-step explanation:
Let b be the length of the second parallel side
Area of a trapezium is given by

where
h is the height
a and b are the parallel sides
From the question
Area = 230 cm²
height = 20cm
one of parallel side / a = 8 cm
Substitute the values into the above formula and solve for b
That's

Reduce the fraction with 2
That's

Expand the terms
That's


Divide both sides by 10
We have the final answer as
<h3>b = 15cm</h3>
Hope this helps you