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just olya [345]
2 years ago
15

Two brothers, Paul and Julio, left the schoolyard and decided to have a very long footrace. Julio started running at exactly 4:0

0pm, and ran at a pace of 300 meters per minute. Paul started running at exactly 4:10pm and ran at a pace of 420 meters per minute. Write two equations where y = distance covered and x = the number of minutes after 4:00pm to represent Paul’s and Julio’s journey
Mathematics
1 answer:
alina1380 [7]2 years ago
6 0

Answer:

this is eassy

Step-by-step explanation:

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Answer:

20 feet below sea level

Step-by-step explanation:

6 0
2 years ago
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an inverted conical water tank with a height of 20 ft and a radius of 8 ft is drained through a hole in the vertex (bottom) at a
viktelen [127]

Answer:

the rate of change of the water depth when the water depth is 10 ft is;  \mathbf{\dfrac{dh}{dt}  = \dfrac{-25}{100  \pi} \  \ ft/s}

Step-by-step explanation:

Given that:

the inverted conical water tank with a height of 20 ft and a radius of 8 ft  is drained through a hole in the vertex (bottom) at a rate of 4 ft^3/sec.

We are meant to find the  rate of change of the water depth when the water depth is 10 ft.

The diagrammatic expression below clearly interprets the question.

From the image below, assuming h = the depth of the tank at  a time t and r = radius of the cone shaped at a time t

Then the similar triangles  ΔOCD and ΔOAB is as follows:

\dfrac{h}{r}= \dfrac{20}{8}    ( similar triangle property)

\dfrac{h}{r}= \dfrac{5}{2}

\dfrac{h}{r}= 2.5

h = 2.5r

r = \dfrac{h}{2.5}

The volume of the water in the tank is represented by the equation:

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

V = \dfrac{1}{3} \pi (\dfrac{h^2}{6.25}) h

V = \dfrac{1}{18.75} \pi \ h^3

The rate of change of the water depth  is :

\dfrac{dv}{dt}= \dfrac{\pi r^2}{6.25}\  \dfrac{dh}{dt}

Since the water is drained  through a hole in the vertex (bottom) at a rate of 4 ft^3/sec

Then,

\dfrac{dv}{dt}= - 4  \ ft^3/sec

Therefore,

-4 = \dfrac{\pi r^2}{6.25}\  \dfrac{dh}{dt}

the rate of change of the water at depth h = 10 ft is:

-4 = \dfrac{ 100 \ \pi }{6.25}\  \dfrac{dh}{dt}

100 \pi \dfrac{dh}{dt}  = -4 \times 6.25

100  \pi \dfrac{dh}{dt}  = -25

\dfrac{dh}{dt}  = \dfrac{-25}{100  \pi}

Thus, the rate of change of the water depth when the water depth is 10 ft is;  \mathtt{\dfrac{dh}{dt}  = \dfrac{-25}{100  \pi} \  \ ft/s}

4 0
3 years ago
Given the equation for this parabola : y=-(x-4)^2-3 does this parabola have a maximum or minimum value ?
noname [10]
The parabola has a minimum value of -3 due to the subtraction at the end of the equation.
8 0
2 years ago
You buy 4.5 pounds of bananas at $0.49 a pound. You hand the cashier a $20 bill. How much change will you get back
VARVARA [1.3K]

Answer:

You get 17.80 cents back

Step-by-step explanation:

4.5 times 0.49 is 2.20 cents

20 minus 2.20 is 17.80 dollars

4 0
2 years ago
Which is the side length of a cube with a volume of 1331 m3?
suter [353]
Let's remember the equation for the volume of a cube:
V= x^3 where x is the length of one side
Since a cube has equal length for length, width, and height.

Now, use what you're given
V= 1331

And put that in terms of x
x^3 = 1331

Now solve for x!
x= cube root (1331)

Think about it! cube root of 1331 * cube root of 1331 * cube root of 1331... will equal 1331 m^3!

Hope this helps
8 0
2 years ago
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