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3 years ago

15

A wheel has ten chairs and people are seated two per chair. Every minute a chair passes the exit platform. The wheel opens at 11

am for thirty minutes – how many people took a ride in that time?

1 answer:

Anit [1.1K]3 years ago

6 0

I’m not really sure,but I got 60

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QUESTION : WHICH LINES ARE PREPENDICULAR ? HOW DO YOU KNOW ? WILL REPORT THE ONES THAT DONT HELP !!!!

Minchanka [31]

Answer: Lines R and B are perpendicular

Step-by-step explanation: Slope of R is -1 and the slope of B is 1. Perpendicular lines have opposite slopes

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3 years ago

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the width of a piece of rectangular land is 5m shorter rhan 1/3 of its length .find the width of the land if the length is 60m,1

aniked [119]

Answer:

If width = 60 m, length = 195 m.

If width = 150 m, length = 465 m.

Step-by-step explanation:

The area of a rectangle is l * w or length times width.

So we can set an equation (letting w = width and x = length)

w = $1/3x - 5$

Part 1: width = 60m:

60 = $1/3x - 5$

65 = $1/3x$

x = 195 m

Part 2: width = 150 m:

150 = $1/3x - 5$

155 = $1/3x$

x = 465 m

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3 years ago

The cost of oranges varies directly as the number of oranges sold. If 8 oranges cost $10.40, how many oranges can be bought for

alexdok [17]

Answer:

26

Step-by-step explanation:

if you can get 8 oranges for 10.40 then divide 10.40 by 8. you can get each orange for 1.3. you need to know how many oranges you can get for 33.8 so divide that by 1.3 and you get 26

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3 years ago

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What is the slope of this equation please help me

galina1969 [7]

The slope is always change in y value over change in x value so if you pick any 2 points on the line you’ll see that the slope is 30/40 and that simplified is 3/4 so the slop is 3/4

4 0

3 years ago

Find the solution of the differential equation dy/dt = ky, k a constant, that satisfies the given conditions. y(0) = 50, y(5) =

irga5000 [103]

Answer: The required solution is $y=50e^{0.1386t}.$

Step-by-step explanation:

We are given to solve the following differential equation :

$\dfrac{dy}{dt}=ky~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)$

where k is a constant and the equation satisfies the conditions y(0) = 50, y(5) = 100.

From equation (i), we have

$\dfrac{dy}{y}=kdt.$

Integrating both sides, we get

$\int\dfrac{dy}{y}=\int kdt\\\\\Rightarrow \log y=kt+c~~~~~~[\textup{c is a constant of integration}]\\\\\Rightarrow y=e^{kt+c}\\\\\Rightarrow y=ae^{kt}~~~~[\textup{where }a=e^c\textup{ is another constant}]$

Also, the conditions are

$y(0)=50\\\\\Rightarrow ae^0=50\\\\\Rightarrow a=50$

and

$y(5)=100\\\\\Rightarrow 50e^{5k}=100\\\\\Rightarrow e^{5k}=2\\\\\Rightarrow 5k=\log_e2\\\\\Rightarrow 5k=0.6931\\\\\Rightarrow k=0.1386.$

Thus, the required solution is $y=50e^{0.1386t}.$

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3 years ago

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