According to the statement, the variable to be modelled is the depth of the hole, which decreases whereas is filled. Under the assumption that the hole is filled at constant rate, we obtain the following expression:

$h(t) = h_{o}-\frac{\Delta h}{\Delta t}\cdot t$ (1)

Where:

$h(t)$ - Current depth of the hole, measured in inches.

$h_{o}$ - Initial depth of the hole, measured in inches.

$\Delta h$ - Filled level, measured in inches.

$\Delta t$ - Filling time, measured in hours.

$t$ - Time, measured in hours.

If we know that $h_{o} = 24\,in$ , $\Delta h = 6\,in$ and $\Delta t = 0.5\,h$ , then the function that models the depth of the hole is:

$h(t) = 24 - 12\cdot t$

The function that models the depth of the hole in feet over time in hours is $h(t) = 24 - 12\cdot t$ .

6 0

2 years ago

Drag the operator to the correct location on the image.

timama [110]

Answer:

Thanks you

Step-by-step explanation:

8 0

3 years ago

Find the exact solution for the following system of equations. 2x+3y=48 x+6y=11

TEA [102]

Question 2x+3y=48
Answer x = 24 + -1.5y

Question x+6y=11
Answer x = 11 + -6y

4 0

3 years ago

Please answer these questions using BODMAS: 1) - 4 - 3 x ( - 2 ) 2) 4 + 8 ÷ ( - 2 ) - 3

Paul [167]

Um I think this is it hope this helps

3 0

2 years ago