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 :)
Your answer is 2’3’3!! (2*3*3 = 18)
hopefully this helps, good luck bud! :)
Answer:
a. 144 cubic foot
b. 12 ft
c. 18 ft
Step-by-step explanation:
Let the volume of prism A be V_a
and that of prism b be V_b
ATQ, V_a + V_b= 432
also V_a= 0.5 V_b
⇒1.5 V_b= 432
= V_b= 432/1.5= 288 cubic feet
therefore V_a= 144 cubic feet
volume of prism= area of base×height = V_a
24×h = 144
⇒h= 12 ft
A_b= 2/3×24
⇒base area of prism B= 16 sq.ft
now 16×h_b= 288⇒h_b= 288/16= 18 ft
Top right I’m pretty sure