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goldfiish [28.3K]
3 years ago
10

If the volume if a cylindrical solid is 1320 and base area is 264 . find its height​

Mathematics
1 answer:
Mandarinka [93]3 years ago
5 0

Answer:

5

Step-by-step explanation:

V = h*A where h is the height and A is the base area

1320 = h*264

Therefore, h = 5

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The difference between two numbers is 40. The first number and four times the second number is 55. Find the two numbers and show
posledela
The numbers are 43 and 3. The difference between these two is 40. 43+3*4 equals 55.
5 0
4 years ago
Let C be the boundary of the region in the first quadrant bounded by the x-axis, a quarter-circle with radius 9, and the y-axis,
rewona [7]

Solution :

Along the edge $C_1$

The parametric equation for $C_1$ is given :

$x_1(t) = 9t ,  y_2(t) = 0   \ \ for \ \ 0 \leq t \leq 1$

Along edge $C_2$

The curve here is a quarter circle with the radius 9. Therefore, the parametric equation with the domain $0 \leq t \leq 1 $ is then given by :

$x_2(t) = 9 \cos \left(\frac{\pi }{2}t\right)$

$y_2(t) = 9 \sin \left(\frac{\pi }{2}t\right)$

Along edge $C_3$

The parametric equation for $C_3$ is :

$x_1(t) = 0, \ \ \ y_2(t) = 9t  \ \ \ for \ 0 \leq t \leq 1$

Now,

x = 9t, ⇒ dx = 9 dt

y = 0, ⇒ dy = 0

$\int_{C_{1}}y^2 x dx + x^2 y dy = \int_0^1 (0)(0)+(0)(0) = 0$

And

$x(t) = 9 \cos \left(\frac{\pi}{2}t\right) \Rightarrow dx = -\frac{7 \pi}{2} \sin \left(\frac{\pi}{2}t\right)$

$y(t) = 9 \sin \left(\frac{\pi}{2}t\right) \Rightarrow dy = -\frac{7 \pi}{2} \cos \left(\frac{\pi}{2}t\right)$

Then :

$\int_{C_1} y^2 x dx + x^2 y dy$

$=\int_0^1 \left[\left( 9 \sin \frac{\pi}{2}t\right)^2\left(9 \cos \frac{\pi}{2}t\right)\left(-\frac{7 \pi}{2} \sin \frac{\pi}{2}t dt\right) + \left( 9 \cos \frac{\pi}{2}t\right)^2\left(9 \sin \frac{\pi}{2}t\right)\left(\frac{7 \pi}{2} \cos \frac{\pi}{2}t dt\right) \right]$

$=\left[-9^4\ \frac{\cos^4\left(\frac{\pi}{2}t\right)}{\frac{\pi}{2}} -9^4\ \frac{\sin^4\left(\frac{\pi}{2}t\right)}{\frac{\pi}{2}} \right]_0^1$

= 0

And

x = 0,  ⇒ dx = 0

y = 9 t,  ⇒ dy = 9 dt

$\int_{C_3} y^2 x dx + x^2 y dy = \int_0^1 (0)(0)+(0)(0) = 0$

Therefore,

$ \oint y^2xdx +x^2ydy = \int_{C_1} y^2 x dx + x^2 x dx+ \int_{C_2} y^2 x dx + x^2 x dx+ \int_{C_3} y^2 x dx + x^2 x dx  $

                        = 0 + 0 + 0

Applying the Green's theorem

$x^2 +y^2 = 81 \Rightarrow x \pm \sqrt{81-y^2}$

$\int_C P dx + Q dy = \int \int_R\left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) dx dy $

Here,

$P(x,y) = y^2x \Rightarrow \frac{\partial P}{\partial y} = 2xy$

$Q(x,y) = x^2y \Rightarrow \frac{\partial Q}{\partial x} = 2xy$

$\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y} \right) = 2xy - 2xy = 0$

Therefore,

$\oint_Cy^2xdx+x^2ydy = \int_0^9 \int_0^{\sqrt{81-y^2}}0 \ dx dy$

                            $= \int_0^9 0\ dy = 0$

The vector field F is = $y^2 x \hat i+x^2 y \hat j$  is conservative.

5 0
3 years ago
Given: circle k(O), m∠P=95°,
Sergeu [11.5K]

The answer is 75 degrees

6 0
4 years ago
What is the product?
Maurinko [17]

Answer:

option C

  15         14

  -1           9

Step-by-step explanation:

Given in the question a 2x3 and 3x2 matrix

Since, the column of first matrix and row of second matrix is of equal integer, 3, so we can multiply them.

The resultant matrix will be of 2x2

We will use dot product

  1(2) + 3(3) + 1(4)              1(-2) + 3(5) + 1(1)      

-2(2) + 1(3) + 0(4)             -2(-2) + 1(5) + 0(1)  

      15         14

      -1          9

 

5 0
3 years ago
Read 2 more answers
7 out of 28 is what percent?
jeka57 [31]
Here's what I did and I checked it So it is correct!!

\frac{7}{28} = \frac{1}{4}

\frac{1}{4} = 0.25

25% is your answer.

28 \times 25\% = 7

--------------------------------------------------------

I hope that helps you!! Any more questions please feel free to ask!!
8 0
3 years ago
Read 2 more answers
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