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Mama L [17]
2 years ago
7

Once again help, please!!!

Mathematics
1 answer:
inessss [21]2 years ago
6 0

Answer: 5.29pi m^2

Step-by-step explanation:

The circumference = 2 * pi * radius

Therefore, the radius = circumference/(2pi)

Radius = 4.6pi/2/pi

Radius = 2.3meters

Area = pi*r^2

Area = pi*(2.3)^2

Area = 5.29pi m^2

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Hi guys, Can anyone help me with this tripple integral? Thank you:)
OleMash [197]

I don't usually do calculus on Brainly and I'm pretty rusty but this looked interesting.

We have to turn K into the limits of integration on our integrals.

Clearly 0 is the lower limit for all three of x, y and z.

Now we have to incorporate

x+y+z ≤ 1

Let's do the outer integral over x.  It can go the full range from 0 to 1 without violating the constraint.  So the upper limit on the outer integral is 1.

Next integral is over y.  y ≤ 1-x-z.   We haven't worried about z yet; we have to conservatively consider it zero here for the full range of y.  So the upper limit on the middle integration is 1-x, the maximum possible value of y given x.

Similarly the inner integral goes from z=0 to z=1-x-y

We've transformed our integral into the more tractable

\displaystyle \int_0^1 \int_0^{1-x} \int _0^{1-x-y} (x^2-z^2)dz \; dy \; dx

For the inner integral we get to treat x like a constant.

\displaystyle \int _0^{1-x-y} (x^2-z^2)dz = (x^2z - z^3/3)\bigg|_{z=0}^{z= 1-x-y}=x^2(1-x-y) - (1-x-y)^3/3

Let's expand that as a polynomial in y for the next integration,

= y^3/3 +(x-1) y^2 + (2x+1)y -(2x^3+1)/3

The middle integration is

\displaystyle \int_0^{1-x} ( y^3/3 +(x-1) y^2 + (2x+1)y -(2x^3+1)/3)dy

= y^4/12 + (x-1)y^3/3+ (2x+1)y^2/2- (2x^3+1)y/3 \bigg|_{y=0}^{y=1-x}

= (1-x)^4/12 + (x-1)(1-x)^3/3+ (2x+1)(1-x)^2/2- (2x^3+1)(1-x)/3

Expanding, that's

=\frac{1}{12}(5 x^4 + 16 x^3 - 36 x^2 + 16 x - 1)

so our outer integral is

\displaystyle \int_0^1 \frac{1}{12}(5 x^4 + 16 x^3 - 36 x^2 + 16 x - 1) dx

That one's easy enough that we can skip some steps; we'll integrate and plug in x=1 at the same time for our answer (the x=0 part doesn't contribute).

= (5/5 + 16/4 - 36/3 + 16/2 - 1)/12

=0

That's a surprise. You might want to check it.

Answer: 0

6 0
3 years ago
The solution for a system of two linear equations is (0, 0). Which equation below could be one of the equations in the system? A
earnstyle [38]

Answer:

y = -3x

Step-by-step explanation:

For the equation to be a part of a system, the point (0,0) must be a part of the line. Since y = -3x is proportional meaning b = 0, its y-intercept is (0,0). This means it must cross through and include (0,0). This is a part of the system.

4 0
4 years ago
10 pts
Simora [160]
Area = pi * radius^2
radius = diameter/2 = 26/2 = 13 feet

Area = 3.14 * (13^2)
Area = 3.14 * 169 = 530.66 ft^2
4 0
3 years ago
I really need help!!!
antiseptic1488 [7]
(a)

A right angle sums to 90 

6x + 4x + 10 = 90

10 x = 80

x = 8

(b) straight line = 180

5x + 13  + 3x + 7 = 180

x = 20

(c)  Sum of triangle angles = 180

3x + 5  + 2x + 18 + 2x + 17 = 180

x = 20

(d) Sum of two angles in right angle = 90

90 = x + 30

x = 60


5 0
3 years ago
Is surface area 3-dimensional?
masha68 [24]
Considered as it only taking the area of flat surfaces of a 3D object, no
6 0
3 years ago
Read 2 more answers
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