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vovangra [49]
3 years ago
7

A pound of ground beef costs $5. At this rate, what is the cost of: 3 pounds

Mathematics
1 answer:
Vsevolod [243]3 years ago
6 0

Answer:

$15

Step-by-step explanation:

If one pound of beef costs 5 dollars then three would be 3 x 5 which is 15.

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Five angles of a hexagon measures 119, 129,104,139 and 95 degrees. What is the measure of the sixth angle?
Kisachek [45]

Answer:

x = 134

Step-by-step explanation:

Given that,

Five angles of a hexagon measures 119, 129,104,139 and 95 degrees.

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So,

119+129+104+139 +95+x = 720

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3 years ago
Tulio will invest​ $1,000,000 at​ 5% annual interest. Determine how much money will Tulio have after 10 years if the interest is
balu736 [363]

Answer:

$1,500,000

Step-by-step explanation:

Given

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Interest = 100000*5

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5 0
3 years ago
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}

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Answer: the 1 is > the one below it is < the one next to it unknown

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Step-by-step explanation:

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