All categories

3 years ago

Helpppppppppppppppppp

Mathematics

1 answer:

Wittaler [7]3 years ago

7 0

8 ÷ 2 x 4 = 16

I hope I helped. If I got the right answer, Brainliest would be appreciated.

You might be interested in

A tank in the form of a right-circular cylinder standing on end is leaking water through a circular hole in its bottom. As we sa

sashaice [31]

The question is incomplete so, here is the complete question.

A tank in form of a right-circular cylinder standing on end is leaking water through a circular hole in its bottom. When friction and contraction of water at the hole are ignored, the height h of water in the tank is described by

$\frac{dh}{dt} = - \frac{A_{h} }{A_{w} } \sqrt{2gh}$ , where $A_{h}$ and $A_{w}$ are cross-sectional areas of the hole and the water, respectively. (a) Solve for h(t) if the initial height of the water is H. By hand, sketch the graph of h(t) and give its interval I of definition in terms of the symbols Aw, Ah and H. Use g = 32ft/s². (b) Suppose the tank is 12 feet high and has radius 4 feet and the circular hole har radius 1/2 inch. If the tank is initially full, how long will it take to empty?

Answer: a) h(t) = $(\sqrt{H} - 4 \frac{A_{h} }{A_{w} }t )^{2}$ , with interval 0≤t≤ $\frac{A_{w}\sqrt{H} }{4A_{h} }$

b) It takes 133 minutes.

Step-by-step explanation: a) The height per time is expressed as

$\frac{dh}{dt} = - \frac{A_{h} }{A_{w} } \sqrt{2gh}$

Using g=32ft/s²: $\sqrt{2gh}$ = $\sqrt{2.32.h}$ $= 8\sqrt{h}$

$\frac{dh}{dt} = - \frac{A_{h} }{A_{w} } \sqrt{2gh}$

$\frac{dh}{\sqrt{h} }$ = - 8. $\frac{A_{h} }{A_{w} }$ dt

$\int\limits^a_b {\frac{dh}{\sqrt{h} } } = - \int\limits^a_b {8.\frac{A_{h} }{A_{w} } } \, dt$

Calculating the indefinite integrals:

2 $\sqrt{h}$ = - 8 $\frac{A_{h} }{A_{w} }$ .t + c

According to the question, when t=0 h₀ for water is H. so

2 $\sqrt{H}$ = - 8 $\frac{A_{h} }{A_{w} }$ .0 + c

c = 2 $\sqrt{H}$

2 $\sqrt{h}$ = - 8 $\frac{A_{h} }{A_{w} }$ .t + 2 $\sqrt{H}$

Dividing each term by 2, then, the equation h(t) is

h(t) = ( $\sqrt{H} -$ $4\frac{A_{h} }{A_{w} } .t$ )²

Now, to find the interval, when the tank is empty, there is not water height so:

0 = ( $\sqrt{H} -$ $4\frac{A_{h} }{A_{w} } .t$ )²

0 = ( $\sqrt{H} -$ $4\frac{A_{h} }{A_{w} } .t$ )

4t $\frac{A_{h} }{A_{w} }$ = $\sqrt{H}$

t = $\frac{A_{w} \sqrt{H} }{4.A_{h} }$

Thus, the interval will be : 0 ≤ t ≤ $\frac{A_{w}\sqrt{H} }{4A_{h} }$

b) $r_{w}$ = 4 ft $h_{w}$ = 12 ft and $r_{h}$ = 1/2 in

Area of the water:

$A_{w} = \pi r^{2}$

$A_{w} = \pi .4^{2}$

$A_{w}$ = 16πft²

Area of the hole:

1 in = $\frac{1}{12}$ ft

so, $r_{h}$ = $\frac{1}{2}.\frac{1}{12}$ = $\frac{1}{24}$

$A_{h} = \pi .r^{2}$

$A_{h} = \pi .\frac{1}{24} ^{2}$

$A_{h}$ = $\frac{\pi }{576}$ ft²

As H=10 and having t = $\frac{A_{w} \sqrt{H} }{4.A_{h} }$ :

t = $\frac{16.\pi .\sqrt{12} }{4.\frac{\pi }{576} }$

t = 4.576. $\sqrt{12}$

t = 7981.3 seconds

t = $\frac{7981.3}{60}$ = 133 minutes

It takes 133 minutes to empty.

8 0

3 years ago

Please please help or i fail math

bixtya [17]

The answer is D.

I hope this helps you! Good luck :)

4 0

4 years ago

Read 2 more answers

Which equation equivalent to logx36=2

Anna71 [15]

First you convert the expression to an exponential equation
36=x²
6²=x²
therefore x=6

4 0

3 years ago

What is the slope of a line parallel to y=(1\2)x+3

Lina20 [59]

The slope is 1/2............

3 0

4 years ago

If the spinner below is spun twice, find the probability that it lands on N, then a vowel.<br><br>

gogolik [260]

Answer:

0.29 or 29%

Step-by-step explanation:

There are 7 options per spin. Of which 3 vowels.

For 2 spins, total spins is 14.

conditions:

First spin is 1/7

Second spin is 3/7

so,

4/14 = 0.29 or 29% I think

5 0

2 years ago