A wheel has a radius of 28 cm. If it does 50 revolutions, what distance will it travel?

Mathematics

1 answer:

LuckyWell [14K]3 years ago

5 0

Answer:

87.96 meters to the nearest cm.

Step-by-step explanation:

First find the circumference of the wheel

= 2 π r

= 2 π * 28

= 175.929 cm.

So one revolution is equivalent to travelling 175.929 cm.

50 revolutions = 175.929 * 50

= 8796.46 cm.

= 8796.46 / 100

= 87.96 m (answer).

You might be interested in

A. x-4=0 b. x+4=0 c. y-4=0 d. y+4=0

Radda [10]

I thinks it’s c but cover your name so people don’t see it

5 0

2 years ago

Write an equation for the horizontal line that contains point E(-3, -1)

ryzh [129]

A horizontal line is represented by y = a number...with that number being the y coordinate of ur point (-3,-1).

Therefore, ur equation is : y = -1

5 0

3 years ago

Part 4: Identify the vertex, focus and directrix of each. Then sketch the graph.

Nezavi [6.7K]

Answer: 1) Vertex: (6, -2) Focus: (6, -7/4) Directrix: y = -9/4

2) Vertex: (-2, -1) Focus: (-7/4, -1) Directrix: x = -9/4

Step-by-step explanation:

Rewrite the equation in vertex format y = a(x - h)² + k or x = a(y - k)² + h by completing the square. Divide the b-value by 2 and square it - add that value to both sides of the equation.

(h, k) is the vertex
p is the distance from the vertex to the focus
-p is the distance from the vertex to the directrix

$\bullet\quad a=\dfrac{1}{4p}$

1) y = x² - 12x + 34

$y-34=x^2-12x\\\\y-34+\bigg(\dfrac{-12}{2}\bigg)^2=x^2-12x+\bigg(\dfrac{-12}{2}\bigg)^2\\\\\\y-34+36=(x-6)^2\\\\y+2=(x-6)^2\\\\y=(x-6)^2-2\qquad \rightarrow \qquad a=1\quad (h,k)=(6,-2)\\$

$a=\dfrac{1}{4p}\quad \rightarrow \quad 1=\dfrac{1}{4p}\quad \rightarrow \quad p=\dfrac{1}{4}\\\\\\\text{Focus = Vertex + p}\\.\qquad \quad =\dfrac{-8}{4}+\dfrac{1}{4}\\\\.\qquad \quad =-\dfrac{7}{4}\quad \rightarrow \quad\text{Focus}=\bigg(6,-\dfrac{7}{4}\bigg)\\\\\\\text{Directrix: y= Vertex - p}\\.\qquad \qquad y=\dfrac{-8}{4}-\dfrac{1}{4}\\\\.\qquad \qquad y=-\dfrac{9}{4}$

*******************************************************************************************

2) x = y² + 2y - 1

$x+1=y^2+2y\\\\x+1+\bigg(\dfrac{2}{2}\bigg)^2=y^2+2y+\bigg(\dfrac{2}{2}\bigg)^2\\\\\\x+1+1=(y+1)^2\\\\x+2=(y+1)^2\\\\x=(y+1)^2-2\qquad \rightarrow \qquad a=1\quad (h,k)=(-2,-1)\\$

$a=\dfrac{1}{4p}\quad \rightarrow \quad 1=\dfrac{1}{4p}\quad \rightarrow \quad p=\dfrac{1}{4}\\\\\\\text{Focus = Vertex + p}\\.\qquad \quad =\dfrac{-8}{4}+\dfrac{1}{4}\\\\.\qquad \quad =-\dfrac{7}{4}\quad \rightarrow \quad\text{Focus}=\bigg(-\dfrac{7}{4},-1\bigg)\\\\\\\text{Directrix: x= Vertex - p}\\.\qquad \qquad y=\dfrac{-8}{4}-\dfrac{1}{4}\\\\.\qquad \qquad x=-\dfrac{9}{4}$