Answer:
1.77 feet
Step-by-step explanation:
The circumference of the wheel is equal to the distance traveled in one revolution. Therefore, the total distance traveled is the circumference of the wheel multiplied by the number of revolutions. Given that the circumference of a circle with diameter (d) is πd, then we can express the distance traveled as:
[tex]\textsf{Distance traveled} = \pi d \times \textsf{Number of revolutions}[/tex]
Given that the wheel makes 9 revolutions and travels 50 feet, we can substitute these values into the equation along with π = 22/7 to create an equation in terms of d:
[tex]50=\dfrac{22}{7} \times d \times 9[/tex]
Now, solve for d:
[tex]\dfrac{50}{9}=\dfrac{22}{7} \times d\\\\\\d=\dfrac{50}{9} \div \dfrac{22}{7}\\\\\\d=\dfrac{50}{9} \times \dfrac{7}{22}\\\\\\d=\dfrac{350}{198}\\\\\\d=\dfrac{175}{9}\\\\\\d=1.77\; \sf ft\; (nearest\;hundredth)[/tex]
So, the estimated diameter of the wheel is approximately 1.77 feet.
Answer:
1.77 ft.
Step-by-step explanation:
To find the diameter of the wheel, we can use the formula for the circumference of a circle:
[tex] C = \pi \times d [/tex]
where
Given that the wheel makes 9 revolutions and travels 50 feet, we can find the total distance traveled by the wheel in one revolution by dividing the total distance by the number of revolutions:
[tex] \textsf{Distance per revolution} = \dfrac{50 \textsf{ feet}}{9 \textsf{ revolutions}} [/tex]
Now, the circumference of the wheel is equal to the distance traveled in one revolution.
Therefore:
[tex] C = \dfrac{\textsf{Distance}}{\textsf{ revolution}} [/tex]
So, we have:
[tex] \textsf{Distance per revolution} = \dfrac{50}{9} \textsf{ feet} [/tex]
Now, we know that:
[tex] C = \pi \times d [/tex]
Substituting the values:
[tex] \dfrac{50}{9} = \pi \times d [/tex]
To solve for [tex] d [/tex], we'll rearrange the equation:
[tex] d = \dfrac{\dfrac{50}{9}}{\pi} [/tex]
Now, we'll calculate [tex] d [/tex]:
[tex] d = \dfrac{50}{9 \pi} [/tex]
Approximating [tex] \pi [/tex] as [tex] \dfrac{22}{7} [/tex]:
[tex] d \approx \dfrac{50 \times 7}{9 \times 22} [/tex]
[tex] d \approx \dfrac{350}{198} [/tex]
Now, let's simplify the fraction:
[tex] d \approx 1.77 \textsf{ feet ( in 2 d.p.)} [/tex]
So, the diameter of the wheel is approximately 1.77 feet.
