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3 years ago

.The diameter of a bike wheel is 27 inches. If the wheel makes 15 complete rotations, how far does the bike travel?

Mathematics

1 answer:

kobusy [5.1K]3 years ago

5 0

Answer:

1271.7in

Step-by-step explanation:

A complete rotation is = circumference of the wheel.

So calculate the circumference which is pi times diameter.

27 times pi (which is 3.14) = 84.78in one rotation

so if you make 15 rotations, you are multiplying it by 15

15 x 84.78 = 1271.7in

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Bob won $800 at bingo last night.The first thing bob did was to take a cab to his mothers house.He gave the cab driver $20 and t

nirvana33 [79]

800-20 is 780. A third of 780 is 780(1/3) = 260. 780 - 260 = 520. 25% of $520 is 0.25(520) = 130. 520 - 130 = 390. One sixth of 390 is 390(1/6) = 65. 390 - 65 = 325. So, Bob kept $325 for himself.

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3 years ago

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Consider the differential equation: xy′(x2+7)y=cos(x)+e3xy. Put the differential equation into the form: y′+p(x)y=g(x), determin

icang [17]

Answer:

Linear and non-homogeneous.

Step-by-step explanation:

We are given that

$\frac{xy'}{(x^2+7)y}=cosx+\frac{e^{3x}}{y}$

We have to convert into y'+P(x)y=g(x) and determine P(x) and g(x).

We have also find type of differential equation.

$y'=\frac{(x^2+7)y}{x}(cosx+\frac{e^{3x}}{y}}$

$y'=\frac{(x^2+7)cosx}{x}y+\frac{(x^2+7)e^{3x}}{x}$

$y'-\frac{cosx(x^2+7)}{x}y=\frac{e^{3x}(x^2+7)}{x}$

It is linear differential equation because this equation is of the form

y'+P(x)y=g(x)

Compare it with first order first degree linear differential equation

$y'+P(x)y=g(x)$

$P(x)=-\frac{cosx (x^2+7)}{x},g(x)=\frac{e^{3x}(x^2+7)}{x}$

$\frac{dy}{dx}=\frac{(x^2+7)(ycosx+e^{3x})}{x}$

Homogeneous equation

$\frac{dy}{dx}=\frac{f(x,y)}{g(x,y)}$

Degree of f and g are same.

$f(x,y)=(x^2+7)(ycosx+e^{3x}),g(x,y)=x$

Degree of f and g are not same .

Therefore, it is non- homogeneous .

Linear and non-homogeneous.

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3 years ago

The length of a rectangle is 3 metern .<br> more than triple the width

mariarad [96]

It is 9 because you need to triple 3 so 3x3 is 9

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2 years ago

15 is 60% of what number

fiasKO [112]

Answer:

Step-by-step explanation:

If we call the number we're looking for n, then

n * 60/100 = 15

then n = 15/0.6 = 25

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3 years ago

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The coordinates of a particle in the metric xy-plane are differentiable functions of time t with:

frez [133]

Answer:

The rate of change of the distance $\frac{ds}{dt}$ when x = 9 and y = 12 is $-7.6\:\frac{m}{s}$ .

Step-by-step explanation:

This is an example of a related rate problem. A related rate problem is a problem in which we know one of the rates of change at a given instant $\frac{dx}{dt}$ and we want to find the other rate $\frac{dy}{dt}$ at that instant.

We know the rate of change of x-coordinate and y-coordinate:

$\frac{dx}{dt}=-2\:\frac{m}{s} \\\\\frac{dy}{dt}=-8\:\frac{m}{s}$

We want to find the rate of change of the distance $\frac{ds}{dt}$ when x = 9 and y = 12.

The distance of a point (x, y) and the origin is calculated by:

$s=\sqrt{x^2+y^2}$

We need to use the concept of implicit differentiation, we differentiate each side of an equation with two variables by treating one of the variables as a function of the other.

If we apply implicit differentiation in the formula of the distance we get

$s=\sqrt{x^2+y^2}\\\\s^2=x^2+y^2\\\\2s\frac{ds}{dt}= 2x\frac{dx}{dt}+2y\frac{dy}{dt}\\\\\frac{ds}{dt}=\frac{1}{s}(x\frac{dx}{dt}+y\frac{dy}{dt})$

Substituting the values we know into the above formula

$s=\sqrt{9^2+12^2}=15$

$\frac{ds}{dt}=\frac{1}{15}(9(-2)+12(-8))\\\\\frac{ds}{dt}=\frac{1}{15}\left(-18-96\right)\\\\\frac{ds}{dt}=\frac{1}{15}(-114)=-\frac{38}{5}=-7.6\:\frac{m}{s}$

The rate of change of the distance $\frac{ds}{dt}$ when x = 9 and y = 12 is $-7.6\:\frac{m}{s}$

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