Answer:
The answer is 112 families in a neighboring township take their dog for a walk each day.
Step-by-step explanation:
We know that 38%, or 0.38, of the customers take their dog for a walk every day. If we multiply 0.38 times the number of families in the neighboring township that own at least one dog, we get 111.72.
0.38 * 294 = 111.72.
Round that to the nearest whole number and we get 112.
The formula for a trapezoid relates the area A, the two bases, a and b, and the height, h.
<span>A=<span><span>(<span>a+b</span>)</span>2</span>h</span>
Solve for a.
<h3>
Answer: B) 1.0 year</h3>
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Explanation:
We have these two functions
- f(t) = 12(1.02)^(4t)
- g(t) = 12(1.09)t
which represent the amounts for his friend and William in that order. Strangely your teacher mentions William first, but then swaps the order when listing the exponential function as the first. This might be slightly confusing.
The table of values is shown below. We have t represent the number of years and t starts at 0.5. It increments by 0.1
The f(t) and g(t) columns represent the outputs for those mentioned values of t. For example, if t = 0.5 years (aka 6 months) then f(t) = 12.48 and that indicates his friend has 12,480 dollars in the account.
I've added a fourth column labeled |f - g| which represents the absolute value of the difference of the f and g columns. If f = g, then f-g = 0. The goal is to see if we get 0 in this column or try to get as close as possible. This occurs when we get 0.09 when t = 1.0
So we don't exactly get f(t) and g(t) perfectly equal, but they get very close when t = 1.0
It turns out that the more accurate solution is roughly t = 0.9925 which is close enough. I used a graphing calculator to find this approximate solution.
It takes about a year for the two accounts to have the same approximate amount of money.
Answer:

Step-by-step explanation:
There are two of them, so we will divide the lawn area mowed by 2.

-> This also can be simplified with seeing that 8 / 2 = 4, so
, but I wanted to show the whole mathematical process.