Answer:
Step-by-step explanation:
The height of the marshmallow (in feet) is represented by the equation h=−16t2+8t+48,
where t is the time (in seconds) after he throws the marshmallow.
The equation is a quadratic equation. When the marshmallow hits the ground, its height from the ground will be zero. Therefore
−16t^2+8t+48 = 0
- 2t^2 + t + 6 = 0
Multiplying through by the negative sign. It becomes
2t^2 - t - 6 = 0
2t^2 + 3t - 4t - 6 = 0
t(2t + 3) - 2(2t + 3) = 0
(t - 2)(2t + 3) = 0
t - 2 = 0 or 2t + 3 = 0
t = 2 or t = - 3/2
The time cannot be negative
So it will take 2 seconds to hit the ground
Answer:
-1 or 1/-1
Step-by-step explanation:
24/3=8
Therefore one third of the can is 8 liters.
24-8=16.
Peter has to fill 16 liters.
Given the equation:

(a) You can identify that the student applied the Subtraction Property of Equality by subtraction 3 from both sides of the equation:

However, the student made a mistake when adding the numbers on the right side.
Since you have two numbers with the same sign on the right side of the equation, you must add them, not subtract them and use the same sign in the result. Then, the steps to add them are:
- Add their Absolute values (their values without the negative sign).
- Write the sum with the negative sign.
Then:

(b) The correct procedure is:
1. Apply the Subtraction Property of Equality by subtracting 3 from both sides (as you did in the previous part):

2. Apply the Multiplication Property of Equality by multiplying both sides of the equation by 6:

Hence, the answers are:
(a) The student made a mistake by adding the numbers -18 and -3:

(b) The value of "x" should be:
Answer:
130.826 kilojoules
Step-by-step explanation:
The Work required to pump water = pgV
Where, p = density of water = 1000 kg/m³
g = acceleration due to gravity = 9.81ms-¹
V = volume of water
Since, radius r = 5 ft and height h = 9ft
Volume of water in the cylindrical tank = (2/3)πr²h
V = (2/3)π *5²*9 = 471.24 ft³ = 471.24 * 0.0283m³ = 13.336m³
Work required = 1000*9.81*13.336
W = 130.826 KJ
Therefore, the work required to pump water 2/3 of the volume of the cylindrical tank = 130.826 kilojoules.