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2 years ago

Divide.

2 answers:

8 0

Hello!

Mixed number is one of the three forms I am able to calculate.

The answer is: -2 3/157.

Happy to help and have a great night.

Sladkaya [172]2 years ago

3 0

- 634 ÷ 314 can be written as -->

- 317 ÷ 157

On dividing , quotient is 2 while remainder is 3

The fraction can be written as -->
- 2 3/157.

Hope This Helps You!

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2 years ago

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2 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; $\mathtt{\dfrac{dh}{dt} = \dfrac{-25}{100 \pi} \ \ ft/s}$

4 0

3 years ago

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Answer:

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Step-by-step explanation:

7 0

2 years ago

Construct 3 linear equation starting with qiven solution z = 1/3

Andreyy89

Answer:

(a) $9z+2=5$

(b) $21z-11=-4$

Step-by-step explanation:

We are required to construct 3 linear equations starting with the given solution z = 1/3.

Equation 1

$z=\frac{1}{3}$

Multiply both sides by 9

$9z=\frac{1}{3}\times 9\\9z=3$

Rewrite 3 as 5-2

9z=5-2

Add 2 to both sides

Our first equation is: $9z+2=5$

Equation 2

$z=\frac{1}{3}$

Multiply both sides by 21

$21z=\frac{1}{3}\times 21\\21z=7$

Rewrite 7 as 11-4

21z=11-4

Subtract 11 from both sides

Our second equation is: $21z-11=-4$

Equation 3

$z=\frac{1}{3}$

Multiply both sides by 6

$6z=\frac{1}{3}\times 6\\6z=2$

Rewrite 6z as 4z+2z

4z+2z=2

Subtract 2z from both sides

Our third equation is: $4z=2-2z$

4 0

2 years ago