How would I write these problems?

21) Total number of cups: 26
6) Solution of the equation: x = -12
7) Father's age: 40
4) Distance driven: 400 miles
Step-by-step explanation:
21)
Let's call:
[tex]h=10 cm[/tex] the height of each cup
[tex]H=30 cm[/tex] the total height available in the dispenser
We are told that each cup added to the stack add a height of
[tex]\Delta h = 0.8 cm[/tex]
We can therefore solve the problem by writing the following equation:
[tex]H=h+N\Delta h[/tex]
which means that the total height available (H) is filled by the height of one cup (h) + the number of additional cup multiplied by 0.8.
Solving for N:
[tex]N=\frac{H-h}{\Delta h}=\frac{30-10}{0.8}=25[/tex]
And including the first cup, the total number of cups is
[tex]N+1=25+1=26[/tex]
6)
The equation that we have to solve is
[tex]\frac{5}{2}x-1=-31[/tex]
We proceed as follows:
First of all, we add +1 on both sides of the equation,
[tex]\frac{5}{2}x-1+1=-31+1\\\frac{5}{2}x=-30[/tex]
Now we multiply by 2 on both sides:
[tex]\frac{5}{2}x\cdot 2=-30\cdot 2\\5x=-60[/tex]
And now we divide both sides by 5:
[tex]\frac{5x}{5}=\frac{-60}{5}\\x=-12[/tex]
7)
Let's call:
k = Karma's age
f = father's age
We are told that Karma is 13 years old, so
k = 13
Also, Karma's age is 2 years less than 3/8 of the father's age, so we can write
[tex]k=\frac{3}{8}f-2[/tex]
By combining the two equations,
[tex]13=\frac{3}{8}f-2[/tex]
And solving for f, we find:
[tex]\frac{3}{8}f=13+2\\\frac{3}{8}f=15\\3f=15\cdot 8 =120\\f=\frac{120}{3}=40[/tex]
So, the father's age is 40.
4)
Let's call:
T = 16 the total volume of the tank (16 gallons)
We also know that the car burns 0.03 gallons per mile, so the amount of fuel consumed after x miles is
[tex]0.03 x[/tex]
This means that the amount of fuel left after x miles is
[tex]16-0.03x[/tex]
We also know that the amount of fuel left is 4 gallons, therefore:
[tex]16-0.03x=4\\0.03x=16-4\\0.03x=12\\x=\frac{12}{0.03}=400[/tex]
So, 400 miles.
Learn more about equations:
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