Respuesta :
For the answer to the question above, f x is the number of days she works, she'll earn $90x
After buying the laptop, she'll have $90x - $700 left over, which will pay for ($90x - $700) / $150 days of travel. So we have y = ($90x - $700) / $150 = (9x - 70) / 15 = 0.6x - (14/3)
Note that y can't be negative. Also, if y = 0, then Emma doesn't get to travel at all, so we should avoid that. So we have:
0.6x - (14/3) > 0
0.6x > 14/3
x > (14/3) / 0.6
x > 70/9
The question says that x can be up to 40, so the domain is 70/9 < x <= 40
That's approximately 7.777... < x <= 40
Multiply those numbers by 0.6 and then subtract 700 to get the range:
0 < y <= 58/3
That's approximately 0 < y <= 19.333
After buying the laptop, she'll have $90x - $700 left over, which will pay for ($90x - $700) / $150 days of travel. So we have y = ($90x - $700) / $150 = (9x - 70) / 15 = 0.6x - (14/3)
Note that y can't be negative. Also, if y = 0, then Emma doesn't get to travel at all, so we should avoid that. So we have:
0.6x - (14/3) > 0
0.6x > 14/3
x > (14/3) / 0.6
x > 70/9
The question says that x can be up to 40, so the domain is 70/9 < x <= 40
That's approximately 7.777... < x <= 40
Multiply those numbers by 0.6 and then subtract 700 to get the range:
0 < y <= 58/3
That's approximately 0 < y <= 19.333
Answer:
(a) The required equation is [tex]90x-150y=700[/tex].
(b) The graph of given relation is shown below.
(c) [tex]Domain=\{x|x\in Z,0\leq x\leq 40\}[/tex] and [tex]Range=\{y|y\in Z,0\leq x\leq \frac{58}{3}\}[/tex].
(d) She needs to work for 18 days.
Step-by-step explanation:
Let x be the number of days she can work and y is the number of days she can travel.
She can earn $90 each day. She expects each day of travel will cost her $150 and the laptop she hopes to buy costs $700.
[tex]90x-150y=700[/tex]
Therefore the required equation is 90x-150y=700.
(b)
The given relation is
[tex]90x-150y=700[/tex] .... (1)
Rewrite the above equation in slope intercept form.
[tex]150y=90x-700[/tex]
Divide both sides by 150.
[tex]y=\frac{90x-700}{150}[/tex]
[tex]y=\frac{3}{5}x-\frac{14}{3}[/tex]
It is a straight line with slope 3/5 and y-intercept (14)/3. The graph of given relation is shown below.
(c)
The relation is
[tex]y=\frac{3}{5}x-\frac{14}{3}[/tex]
Here, x is the number of days she can work and she can work a maximum of 40 days. So the domain of the function is
[tex]Domain=\{x|x\in Z,0\leq x\leq 40\}[/tex]
At x=0,
[tex]y=\frac{3}{5}(0)-\frac{14}{3}=-\frac{14}{3}[/tex]
At x=40,
[tex]y=\frac{3}{5}(40)-\frac{14}{3}=\frac{58}{3}[/tex]
y is the number of days she can travel, so y cannot be negative.
The range of the relation is
[tex]Range=\{y|y\in Z,0\leq x\leq \frac{58}{3}\}[/tex]
(d)
We need to find the number of days she need to work if she plans to travel for 6 days.
Substitute y=6 in equation (1) to find the value of x.
[tex]90x-150(6)=700[/tex]
[tex]90x-900=700[/tex]
[tex]90x=700+900[/tex]
[tex]90x=1600[/tex]
[tex]90x=\frac{1600}{90}[/tex]
[tex]x=17.77\approx 18[/tex]
Therefore she needs to work for 18 days.
