You can multiply 3 and 6 together then divide the product by 2
(6*3)/2=18/2=9
The answer is going to be 8 left over. hope that helped
Answer:
5 or 6
Step-by-step explanation:
So I did it in a different way than shown in rsm classes. I said that the speed was x-2 and x.
then I said the distance was 6 and 15. If the time taken was 1 hour more, then we can come up with this equation:
Since the time is 6/x-2 or 15/x
15/x-6/x-2=1
multiply everything to get a common denominator then cancel it out to get
(15x-30)-6x=x^2-2x
11x-30=x^2
x^2-11x+30=0
(x-5)(x-6) is factored form.
so x=5 or x=6. this means that the speed of the boat is either 5km/h or 6 km/h
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 :)
