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Sunny_sXe [5.5K]
4 years ago
11

A tank, shaped like a cone has height 11 meter and base radius 4 meter. It is placed so that the circular part is upward. It is

full of water, and we have to pump it all out by a pipe that is always leveled at the surface of the water. Assume that a cubic meter of water weighs 10 000 N
Mathematics
1 answer:
aksik [14]4 years ago
3 0

Answer:

The tank is a cone, so the radius of the cross section increases linearly with h

At y=0, r=4 and at y=11, r=0

So

r(y)=4/11 (11−y)

A(y)=π[4(11−y)11]^2

Integral will be:

11∫0 (1000)(9.8)(4πy^2)(25)(11−y)dy

Step-by-step explanation:

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Step-by-step explanation:

v=√(pr)

v=√(32×0.5)

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v=4volts

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4 years ago
What addition expression is shown on the<br> number line?<br> 8<br> +<br> ?<br> 15
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3 years ago
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4 0
3 years ago
ANSWER ASAP!! Find the area of a triangle with legs that are: 12 m, 15 m, and 9 m.
Setler79 [48]

Answer:

<h2>54m²</h2>

Step-by-step explanation:

<h3>METHOD 1:</h3>

You can use the Heron's formula:

A=\sqrt{p(p-a)(p-b)(p-c)}

where

<em>p</em><em> - half of perimeter</em>

<em>a, b, c</em><em> - lengths of sides</em>

We have

a=12m;\ b=15m;\ c=9m

Calculate:

p=\dfrac{12+15+9}{2}=\dfrac{36}{2}=18\ (m)\\\\A=\sqrt{18(18-12)+(18-15)(18-9)}\\\\A=\sqrt{(18)(6)(3)(9)}\\\\A=\sqrt{2916}\\\\A=54\ (m^2)

<h3>METHOD 2:</h3>

Let's check that it is not a right triangle.

If the sum of the squares of the two shorter sides is equal to the square of the longest side, then this triangle is rectangular.

We have

9m < 12m

Check:

9^2+12^2=81+144=225\\15^2=225

This is a right trianglr wherew 9m and 12m are legs and 15m is a hypotenuse.

The formula of an area of a right triangle is:

A=\dfrac{ab}{2}

<em>a, b</em><em> - legs</em>

Substitute:

A=\dfrac{(9)(12)}{2}=\dfrac{108}{2}=54\ (m^2)

6 0
3 years ago
A school administrator will assign each student in a group of n students to one of m classrooms. If 3 &lt; m &lt; 13 &lt; n, is
tatiyna

Answer:

Hence to get same number of students in each classroom,the sufficient condition is that assign 13n students to each classroom.

Step-by-step explanation:

Given:

There are m classrooms and n be the students

3<m<13<n.

To Find:

Whether it is possible to assign each of n students to one of m classrooms with same no.of students.

Solution:

This problem is related to p/q form  has to be integer in order to get same no of students assigned to the classroom.

As similar as ,n/m ratio

So 1st condition is that,

If it is possible to assign the n/m must be integer and n should be multiple of m,

when we assign 3n students to m classrooms ,we cannot say that 3n/m= integer so that  n is greater than 13 i.e n=14 and m=6

hence they are not multiple of each other so they will not make same students in each classrooms.

Otherwise,n=14 and m=7 they will give same number but this condition is not sufficient condition to assign the student.

So 2nd condition is that ,

When we assign 13n students to m classrooms, as 13 is prime number and

3<m<13 which implies the 13n/m to be integer so n and m must be multiple of each other.

Suppose n=20 and m=5 classrooms

then 13*20=260 ,

260/5=52 students in each classroom,

4 0
3 years ago
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