Jason's part-time job pays him $105 a week
He got an amount of 105 dollars per week
Thhe cost price of the Dit bike = $900
The saving amount he had = $325
So, The amount he require more is the difference in the cost price of the bike to the saving amount
Require amount = 900-325
Require Amount = $575
Since he get the $105 per week
We need to find the number of weeks to get an amount of $575
So, divide the total required amount by the amount in one week he get
![\begin{gathered} \text{ Number of w}eek\text{ =}\frac{Total\text{ Amount}}{Amount\text{ he get p}er\text{ week}} \\ \text{Number of w}eek\text{ =}\frac{575}{105} \\ \text{Number of w}eeks\text{ =}5.47 \\ \text{Number of we}eks\text{ }\approx6 \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20%5Ctext%7B%20Number%20of%20w%7Deek%5Ctext%7B%20%3D%7D%5Cfrac%7BTotal%5Ctext%7B%20Amount%7D%7D%7BAmount%5Ctext%7B%20he%20get%20p%7Der%5Ctext%7B%20week%7D%7D%20%5C%5C%20%5Ctext%7BNumber%20of%20w%7Deek%5Ctext%7B%20%3D%7D%5Cfrac%7B575%7D%7B105%7D%20%5C%5C%20%5Ctext%7BNumber%20of%20w%7Deeks%5Ctext%7B%20%3D%7D5.47%20%5C%5C%20%5Ctext%7BNumber%20of%20we%7Deks%5Ctext%7B%20%7D%5Capprox6%20%5Cend%7Bgathered%7D)
So, he neeed to wokr for 6 weeks to pay for the dirt bike
Answer : 6 weeks
Answer:
y=4x
Step-by-step explanation:
Write the equation of linear function
Substitute:
Multiply the monomials:
Rearrange unknown terms to the left side of the equation:
Reduce the greatest common factor on both sides of the equation:
Rearrange unknown terms to the left side of the equation:
Divide both sides of the equation by the coefficient of variable:
Substitute one unknown quantity into the elimination:
Solve the equation: m=4
Write the solution set of the equations:
Substitute: y=4x
Answer: y=4x
<h2>
Answer: (−∞,∞)</h2>
Step-by-step explanation:
Answer:
a) ![S_{a}(t)=16Kg-0.16Kg*\frac{t}{min}](https://tex.z-dn.net/?f=S_%7Ba%7D%28t%29%3D16Kg-0.16Kg%2A%5Cfrac%7Bt%7D%7Bmin%7D)
b) ![S_{a}(20)=12.8Kg](https://tex.z-dn.net/?f=S_%7Ba%7D%2820%29%3D12.8Kg)
Step-by-step explanation:
It can be seen in the graph that the water velocity and solution velocity is the same, but the salt concentation will be lower
Water velocity ![V_{w} = 60\frac{L}{min}](https://tex.z-dn.net/?f=V_%7Bw%7D%20%3D%2060%5Cfrac%7BL%7D%7Bmin%7D)
Solution velocity ![V_{s} = 60\frac{L}{min}](https://tex.z-dn.net/?f=V_%7Bs%7D%20%3D%2060%5Cfrac%7BL%7D%7Bmin%7D)
Brine concentration = ![\frac{6,000L}{16Kg}=375\frac{L}{Kg}](https://tex.z-dn.net/?f=%5Cfrac%7B6%2C000L%7D%7B16Kg%7D%3D375%5Cfrac%7BL%7D%7BKg%7D)
a) Amount of salt as a funtion of time Sa(t)
![S_{a}(t)=16Kg-\frac{60Kg*L}{375L}*\frac{(t)}{min}=[tex]16Kg-0.16Kg*\frac{t}{min}](https://tex.z-dn.net/?f=S_%7Ba%7D%28t%29%3D16Kg-%5Cfrac%7B60Kg%2AL%7D%7B375L%7D%2A%5Cfrac%7B%28t%29%7D%7Bmin%7D%3D%5Btex%5D16Kg-0.16Kg%2A%5Cfrac%7Bt%7D%7Bmin%7D)
b) ![S_{a}(20)=16Kg-0.16\frac{Kg}{min}*(20min)=16Kg-3.2Kg=12.8Kg](https://tex.z-dn.net/?f=S_%7Ba%7D%2820%29%3D16Kg-0.16%5Cfrac%7BKg%7D%7Bmin%7D%2A%2820min%29%3D16Kg-3.2Kg%3D12.8Kg)
This value was to be expected since as the time passes the concentration will be lower due to the entrance to the pure water tank
34/3+12/3+2-6+9/6+4+7
Remember PEMDAS:
1. (34/3)+(12/3)+2-6+(9/6)+4+7
11.33+4+2-6+1.66+4+7
The rest can be added together into a calculator (or by hand if you are not supposed to use one)