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3 years ago

2/11 , 8/15, 7/13, 5/12 Which fraction does not belong with the other three?

Mathematics

1 answer:

ololo11 [35]3 years ago

3 0

It's 2/11 because all the others round to 1/2.

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Use cylindrical coordinates to evaluate the triple integral ∭ where E is the solid bounded by the circular paraboloid z = 9 - 16

4vir4ik [10]

Answer:

$\mathbf{\iiint_E E \sqrt{x^2+y^2} \ dV =\dfrac{81 \ \pi}{80}}$

Step-by-step explanation:

The Cylindrical coordinates are:

x = rcosθ, y = rsinθ and z = z

From the question, on the xy-plane;

$9 -16 (x^2 + y^2) = 0 \\ \\ 16 (x^2 + y^2) = 9 \\ \\ x^2+y^2 = \dfrac{9}{16}$

$x^2+y^2 = (\dfrac{3}{4})^2$

where:

0 ≤ r ≤ $\dfrac{3}{4}$ and 0 ≤ θ ≤ 2π

∴

$\iiint_E E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} \int ^{\dfrac{3}{4}}_{0} \int ^{9-16r^2}_{0} \ r \times rdzdrd \theta$

$\iiint_E E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} \int ^{\dfrac{3}{4}}_{0} r^2 z|^{z= 9-16r^2}_{z=0} \ \ \ drd \theta$

$\iiint_E E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} \int ^{\dfrac{3}{4}}_{0} r^2 ( 9-16r^2}) \ drd \theta$

$\iiint_E E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} \int ^{\dfrac{3}{4}}_{0} ( 9r^2-16r^4}) \ drd \theta$

$\iiint_E E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} ( \dfrac{9r^3}{3}-\dfrac{16r^5}{5}})|^{\dfrac{3}{4}}_{0} \ drd \theta$

$\iiint_E E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} ( 3r^3}-\dfrac{16r^5}{5}})|^{\dfrac{3}{4}}_{0} \ drd \theta$

$\iiint_E E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} ( 3(\dfrac{3}{4})^3}-\dfrac{16(\dfrac{3}{4})^5}{5}}) d \theta$

$\iiint_E E \sqrt{x^2+y^2} \ dV =( 3(\dfrac{3}{4})^3}-\dfrac{16(\dfrac{3}{4})^5}{5}}) \theta |^{2 \pi}_{0}$

$\iiint_E E \sqrt{x^2+y^2} \ dV =( 3(\dfrac{3}{4})^3}-\dfrac{16(\dfrac{3}{4})^5}{5}})2 \pi$

$\iiint_E E \sqrt{x^2+y^2} \ dV =(\dfrac{81}{64}}-\dfrac{243}{320}})2 \pi$

$\iiint_E E \sqrt{x^2+y^2} \ dV =(\dfrac{81}{160}})2 \pi$

$\mathbf{\iiint_E E \sqrt{x^2+y^2} \ dV =\dfrac{81 \ \pi}{80}}$

4 0

3 years ago

The top tree broke and fell over.the remaining tree teunk is 3 feet tall.the tip of the tree rests on the ground 14 feet from th

Bas_tet [7]

Answer:

14.3 feet.

14.3 feet

Step-by-step explanation:

The problem forms a right triangle in which:

The Vertical Leg of the Right Triangle = 3 feet

The Horizontal Leg of the Right Triangle =14 feet

We are to determine the length of the broken part of the tree. This is the Hypotenuse of the Right Triangle,

Using Pythagoras Theorem

$Hypotenuse^2=Opposite^2+Adjacent^2\\Hypotenuse^2=14^2+3^2\\Hypotenuse^2=205\\Hypotenuse=\sqrt{205}\\Hypotenuse=14.32\\ \approx 14.3 feet $(to the nearest tenth of a foot).\\Therefore, the lenght of the broken part of the tree to the nearest tenth of a foot is 14.3 feet.$

5 0

3 years ago

Suppose you are given a square with an inscribed circle. If the area of the square is 36 m^2. What is the area of the circle?

loris [4]

If the square has sides of 6 the circle has a radius of 3 so it's area is 9pi

4 0

3 years ago

Read 2 more answers

I need help quick please help me thnak you

quester [9]

Answer:

28 numbers

Step-by-step explanation:

There is a distance of 81 between -0.5 and 80.5. Divide this number by 3, since you are only counting every third number and you will get 27, but you need to count the first number, which has been excluded. Add one more to get the answer, 28.

3 0

3 years ago

Each small cube is 1 ft³. The length of the large cube is 15 ft. What is the volume of the large cube?

Dmitriy789 [7]

The answer your question is A

3 0

3 years ago