The average hourly wage h(x) of workers in an industry is modeled by the function where \(h(x)=\frac{16.24x}{0.062x+39.42}\) x represents the number of years since 1970.
a) x represents the number of years since 1970.
b) h(x) represents [t]he average hourly wage ... of workers
c) This question asks for h(x) given an input x.
Since x represents the number of years since 1970, we will have to find the number of years that have elapsed by using subtraction.
\(x=1993-1970=23\text{ years}\)
Now, let's find h(x):
\(x=23\\ h(x)=\frac{16.24x}{0.062x+39.42}\) | This is the model of the hourly wage. |
\(h(23)=\frac{16.24*23}{0.062*23+39.42}\\\ h(23)=9.14459188...\approx\$9\) | I inputted the number of years and rounded appropriately to the nearest dollar |
d)
Since [t]he average hourly wage ... of workers is assumed to be $25, h(x) =25, and we are solving for x:
\(h(x)=25\\ h(x)=\frac{16.24x}{0.062x+39.42}\) | Substitute in 25 for h(x) |
\(25=\frac{16.24x}{0.062x+39.42}\\ 25(0.062x+39.42)=16.24x\\ 1.55x+985.5=16.24x\\ 985.5=14.69x\\ x=\frac{985.5}{14.69}\approx 67.09\text{ years} \) | Solve for x by multiplying by the LCD, 0.062x+39.42 |
67 years after 1970 certainly lands in 2037, but the extra 0.09 years would fall into the following year, so the hourly wage will hit $25 per hour in 2038.
60° and 240° are not the only solutions in the interval 0 < x < 360. 120º and 300º are both perfectly valid.
Let's isolate the trigonometric term and see where to go from there.
\(4\sin^2x=3\) | Divide by 4 on both sides. |
\(\sin^2x=\frac{3}{4}\) | Take the square root of both sides. |
\(|\sin x|=\sqrt{\frac{3}{4}}\) \(|\sin x|=\frac{\sqrt{3}}{2}\) | Use the definition of the absolute value to split this into two separate equations. |
\(\sin x=\frac{\sqrt{3}}{2}\\ \sin x=\frac{-\sqrt{3}}{2}\) | |
Here, you locate which angle yields an answer of \(\pm\frac{\sqrt{3}}{2}\) , which is \(\{60^{\circ},120^{\circ},240^{\circ},300^{\circ}\}\)
.For problems like these, I like to use a table. I like to fill in the table with the information given to me and then create an equation that I can solve.
For this table, I will use the equation d=rt as the structure for the table.
D | R | T | |
Large Snowplow | |||
Small Snowplow | |||
Large Snowplow and Small Snowplow |
We will use the equation d=rt as a metaphor for this particular question.
d is metaphorical for the total amount done; for example, in this problem, the large snowplow is clearing one parking lot of its snow. Therefore, d=1 for the large snowplow.
Although we do not know how long the small snowplow takes to clear the parking lot yet, we do know that it has to clear one parking lot by itself, so d=1 for the small snowplow, as well.
When the snowplows work together, they both only clear one parking lot in this problem, so d=1 for the large and small snowplow. Let's fill this information in now.
D | R | T | |
Large Snowplow | 1p | ||
Small Snowplow | 1p | ||
Large Snowplow and Small Snowplow | 1p |
Next, I will fill in the "T" column, which represents time.
The given information states that the large snowplow leaves the parking lot devoid of snow after 4 hours. Therefore, t=4.
We do not know how much time it took the small snowplow to clear this parking lot; after all, that is what we are solving for. I will assign a variable, x, for that slot.
Collectively, the snowplows unload all that snow in a matter of 3 hours, so t=3 in the final row.
This is what the table looks like now:
D | R | T | |
Large Snowplow | 1p | 4hr | |
Small Snowplow | 1p | xhr | |
Large Snowplow and Small Snowplow | 1p | 3hr |
To fill in the rates, I will refer back to that metaphorical formula. I can rearrange this formula to solve for r.
\(d=rt\) | divide by t to isolate r. |
\(r=\frac{d}{t}\) |
This rearrangement shows that all I have to do is divide the "D" values by the "T" values in order to fill in the elements of the "R" column. Here is an update of what the table looks like.
D | R | T | |
Large Snowplow | \(1p\) | \(\frac{1p}{4\text{hr}}\) | \(4\text{hr}\) |
Small Snowplow | \(1p\) | \(\frac{1p}{x\text{hr}}\) | \(x\text{hr}\) |
Large Snowplow and Small Snowplow | \(1p\) | \(\frac{1p}{3\text{hr}}\) | \(3\text{hr}\) |
Great! We have the table filled out. In order to generate an equation, I want you to focus on the "R" column, the one that represents the rate of the snowplows. When the snowplows are in cahoots, their combined rates equal that of the sum of the individual rates. This allows us to generate an equation.
\(\frac{1}{4}+\frac{1}{x}=\frac{1}{3}\)
\(\frac{1}{4}+\frac{1}{x}=\frac{1}{3}\) | There are many ways to approach this, but probably the simplest way is to multiply both sides by the LCD, 12x. |
\(3x+12=4x\) | Subtract 3x on both sides. |
\(x=12\text{hr.}\) | This is how long the small snowplow will take on its own. |