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3 years ago

Helpppppppppppp!!!!!!!!!!!!!!!!!!

Mathematics

1 answer:

serious [3.7K]3 years ago

4 0

Answer: 2

Step-by-step explanation: angle measures don't change when changing size

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25 7/5 tons = ib ......

Fittoniya [83]

Answer:

c. 51,400

Step-by-step explanation:

25 x 2,000=50,000

2,000 x .7= 1,400

50,000+1,400=51,400

5 0

3 years ago

Write an equation of the line below in the picture?

ad-work [718]

I hope my explanation helps

6 0

3 years ago

Read 2 more answers

Using 6 as a denominator, what fraction completes the following

mixas84 [53]

Answer:

Step-by-step explanation:

We can start by converting 1 into 6/6 and 1/3 into 2/6. This makes subtraction much easier, and by subtracting the numerators, (6/6 - 2/6) you should get 4/6.

5 0

3 years ago

How to solve for b = -1 -7a when the input,a, is 6

azamat

Answer:

-43

Step-by-step explanation:

Input 6 where a is so then do -7 x 6 which would be -42

the equation is now b = -1 - 42

-1 -42 = -43

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

4 0

4 years ago