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solniwko [45]
3 years ago
10

Maya spent her allowance on playing an arcade game a few times and riding the Ferris wheel more than once.

Mathematics
1 answer:
faust18 [17]3 years ago
6 0

Answer:

Sample Response: To write a two-variable equation, I would first need to know how much Maya’s allowance was. Then, I would need the cost of playing the arcade game and of riding the Ferris wheel. I could let the equation be cost of playing the arcade games plus cost of riding the Ferris wheel equals the total allowance. My variables would represent the number of times Maya played the arcade game and the number of times she rode the Ferris wheel. With this equation I could solve for how many times she rode the Ferris wheel given the number of times she played the arcade game.

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Answer: Level of the liquid dropping at 28.28 inch/second when the liquid is 2 inches deep.

Step-by-step explanation:

Since we have given that

Height = 9 inches

Diameter = 6 inches

Radius = 3 inches

So, \dfrac{r}{h}=\dfrac{3}{9}=\dfrac{1}{3}\\\\r=\dfrac{1}{3}h

Volume of cone is given by

V=\dfrac{1}{3}\pi r^2h\\\\V=\dfrac{1}{3}\pi \dfrac{1}{9}h^2\times h\\\\V=\dfrac{1}{27}\pi h^3

By differentiating with respect to time t, we get that

\dfrac{dv}{dt}=\dfrac{1}{27}\pi \times 3\times h^2\dfrac{dh}{dt}=\dfrac{1}{9}\pi h^2\dfrac{dh}{dt}

Now,  the liquid drips out the bottom of the filter at the constant rate of 4 cubic inches per second, ie \dfrac{dv}{dt}=-4\ in^3

and h = 2 inches deep.

-4=\dfrac{1}{9}\times \pi\times (2)^2\dfrac{dh}{dt}\\\\-9\pi =\dfrac{dh}{dt}\\\\-28.28=\dfrac{dh}{dt}

Hence, level of the liquid dropping at 28.28 inch/second when the liquid is 2 inches deep.

7 0
3 years ago
The distance on a map between the entrance of the park in waterfall inside the park is 4 1/2 inches if 3 inches is 2 1/2 miles w
aleksandrvk [35]
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tankabanditka [31]

Answer:

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Step-by-step explanation:

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I need help putting correct answers in the yellow boxes.
olasank [31]

Answer:

<u>-5 ± √5² - 4 · 1 · 4</u>

         2 · 1

Step-by-step explanation:

ax²+bx+c=0 (quadratic equation)

x=<u> -b ± √b² - 4ac</u>

           2a

a= 1

b= 5

c= 4

-<u>5 ± √5² - 4 · 1 · 4</u>

            2 · 1

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