MATHEMATICS COLLEGE

I NEED YOUR HELP 22 POINTSA cone has a base of 30 in^2 and a height of 8.8 what is the volume of the cone?
Plz show how to do it I'm stuck

Answers

Answer 1
Answer:

Hello from MrBillDoesMath!

Answer:

88 in^3

Discussion:

The volume of a cone of height h and base radius r is

          Pi * R^2 * (h/3)

But Pi * r^2 ( = 30) is the area of the base so the volume is:

         (30) * (8.8)/3 =

         (30/3) ( 8.8) =

         10 * 8.8 =

         88 in^3


Thank you,

MrB
Answer 2
Answer:

Answer:

8293.8in

Step-by-step explanation:

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4. Find the volume of the given solid bounded by the elliptic paraboloid z = 4 - x^2 - 4y^2, the cylinder x^2 + y^2 = 1 and the plane z = 0.5. Sketch the region of integration and change the order of integration.

Answers

Answer:

2.5π units^3

Step-by-step explanation:

Solution:-

- We will evaluate the solid formed by a function defined as an elliptical paraboloid as follows:-

                                  z = 4 - x^2 -4y^2

- To sketch the elliptical paraboloid we need to know the two things first is the intersection point on the z-axis and the orientation of the paraboloid ( upward / downward cup ).

- To determine the intersection point on the z-axis. We will substitute the following x = y = 0 into the given function. We get:

                                 z = 4 - 0 -4*0 = 4

- The intersection point of surface is z = 4. To determine the orientation of the paraboloid we see the linear term in the equation. The independent coordinates ( x^2 and y^2 ) are non-linear while ( z ) is linear. Hence, the paraboloid is directed along the z-axis.

- To determine the cup upward or downwards we will look at the signs of both non-linear terms ( x^2 and y^2 ). Both non-linear terms are accompanied by the negative sign ( - ). Hence, the surface is cup downwards. The sketch is shown in the attachment.

- Theboundary conditions are expressed in the form of a cylinder and a plane expressed as:

                                x^2 + y^2 = 1\n\nz = 4

- To cylinder is basically an extension of the circle that lies in the ( x - y ) plane out to the missing coordinate direction. Hence, the circle ( x^2 + y^2 = 1 ) of radius = 1 unit is extended along the z - axis ( coordinate missing in the equation ).

- The cylinder bounds the paraboloid in the x-y plane and the plane z = 0 and the intersection coordinate z = 4 of the paraboloid bounds the required solid in the z-direction. ( See the complete sketch in the attachment )

- To determine the volume of solid defined by the elliptical paraboloid bounded by a cylinder and plane we will employ the use of tripple integrals.

- We will first integrate the solid in 3-dimension along the z-direction. With limits: ( z = 0 , z = 4 - x^2 -4y^2 ). Then we will integrate the projection of the solid on the x-y plane bounded by a circle ( cylinder ) along the y-direction. With limits: ( y = - √(1 - x^2) , y = √(1 - x^2) ). Finally evaluate along the x-direction represented by a 1-dimensional line with end points ( -1 , 1 ).

- We set up our integral as follows:

                            V_s = \int\int\int {} \, dz.dy.dx

- Integrate with respect to ( dz ) with limits: ( z = 0 , z = 4 - x^2 -4y^2 ):

                           V_s = \int\int [ {4 - x^2 - 4y^2} ] \, dy.dx

- Integrate with respect to ( dy ) with limits: ( y = - √(1 - x^2) , y = √(1 - x^2) )

                        V_s = \int [ {4y - x^2.y - (4)/(3) y^3} ] \, | .dx\n\nV_s = \int [ {8√(( 1 - x^2 )) - 2x^2*√(( 1 - x^2 )) - (8)/(3) ( 1 - x^2 )^(3)/(2) } ] . dx

- Integrate with respect to ( dx ) with limits: ( -1 , 1 )

                       V_s = [ 4. ( arcsin ( x ) + x√(1 - x^2) ) - (arcsin ( x ) - 2x ( 1 -x^2 )^(3)/(2) + x√(1 - x^2) )/(2) - ( 3*arcsin ( x ) + 2x ( 1 -x^2 )^(3)/(2) + 3x√(1 - x^2) )/(3) ] | \limits^1_-_1\n\nV_s = [ (5)/(2) *arcsin ( x ) + (5)/(3)*x ( 1 -x^2 )^(3)/(2) + (5)/(2) *x√(1 - x^2) ) ] | \limits^1_-_1\n\nV_s = [ (5\pi )/(2) + 0 + 0 ] \n\nV_s = (5\pi )/(2)

Answer: The volume of the solid bounded by the curves is ( 5π/2 ) units^3.

Final answer:

The volume of the bounded region is found by setting up a triple integral, changing to cylindrical coordinates, and integrating to get 3.5π. The region of integration is a solid capped by an elliptic paraboloid, lying inside the unit circle above the xy-plane. Changing the order of integration doesn't apply here as the given order is already the most ideal.

Explanation:

The subject of this question is

Calculating Volume

in integral calculus, specifically dealing with triple integrals. Given the equations z = 4 - x^2 - 4y^2, x^2 + y^2 = 1, and z = 0, we find the volume by setting up a triple integral. In cylindrical coordinates, this is  ∫ ∫ (4 - x^2 - 4y^2) rdrdθ from θ=0 to 2π and r=0 to 1. Changing to cylindrical coordinates, x = rcosθ and y = rsinθ, gives  ∫ ∫ (4 - r^2) rdrdθ. This evaluates to π(4r - (r^2)/2) evaluated from 0 to 1, which simplifies to π(4 - 0.5) = 3.5π.

Sketching the Region of Integration

, the integrand and bounds describe a solid capped by the elliptic paraboloid and lying above the xy-plane inside the unit circle. The request to 'change the order of integration' would apply if this were an improper triple integral being evaluated in Cartesian coordinates. Here, the order of integration (r, then θ) is itself the most simple and meaningful approach.

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A person invests 4000 dollars in a bank. The bank pays 6.25% interest compoundedannually. To the nearest tenth of a year, how long must the person leave the money
in the bank until it reaches 9600 dollars?

Answers

Answer:

14.4

Step-by-step explanation:

Given that :

Principal = 4000

Interest (r) = 6.25% compounded annually

Calculate time, t, if final amount A = 9600

Using the compound interest formula :

A = P(1 + r/n)^n*t

A = final amount

n = number of times interest is applied per period

9600 = 4000(1 + 0.0625)^t

9600 = 4000(1.0625)^t

9600/4000 = 1.0625^t

2.4 = 1.0625^t

Take the log of both sides

0.3802112 = 0.0263289t

t = 0.3802112/0.0263289

t = 14.440811

t = 14.4 ( nearest tenth)

B(n)=2^n A binary code word of length n is a string of 0's and 1's with n digits. For example, 1001 is a binary code word of length 4. The number of binary code words, B(n), of length n, is shown above. If the length is increased from n to n+1, how many more binary code words will there be? The answer is 2^n, but I don't get how they got that answer. I would think 2^n+1 minus 2^n would be 2. Please help me! Thank you!

Answers

Answer:

More number of words that can be made: \bold{2^n}

Please refer to below proof.

Step-by-step explanation:

Given that:

The number of binary code words that can be made:

B(n) =2^n

where n is the length of binary numbers.

Binary numbers means 2 possibilities either 0 or 1.

Here, suppose if we have 5 as the length of binary number.

And there are 2 possibilities for each digit.

So, total number of possibilities will be 2* 2* 2* 2* 2 = 2^5

If the length of binary number is 2.

The total words possible are 2^2.

These numbers are:

{00, 01, 10, 11}

If the length of binary number is 3. (increasing the 'n' by 1)

The total words possible are 2^3.

These words are:

{000, 001, 010, 100, 011, 101, 110, 111}

So, number of More binary words = 8 - 4 = 4 or 2^2 or 2^n.

So, the answer is 2^n.

Let us try to prove in generic terms:

B(n) = 2^n

Increasing the n by 1:

B(n+1) = 2^(n+1)

Number of more words made by increasing n by 1:

B(n+1) -B(n)= 2^(n+1) -2^n\n\Rightarrow 2* 2^(n) -2^n\n\Rightarrow 2^n(2-1)\n\Rightarrow \bold{2^n}

Hence, proved that:

More number of words that can be made: \bold{2^n}

Final answer:

When the length of a binary code word increases from n to n+1, the number of additional binary code words is equal to the number of binary code words of length n, which is 2^n.

Explanation:

When the length is increased from n to n+1, the number of binary code words of length n+1 is equal to the number of binary code words of length n multiplied by 2. This is because for each binary code word of length n, we can append a 0 or a 1 to create two new binary code words of length n+1. Therefore, the number of additional binary code words is equal to the number of binary code words of length n, which is2^n.

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Please help!! Need to know ASAP

Answers

Answer:

Quadratic function

Step-by-step explanation:

Hope this helps you. Have a great day and I hope to get a brainliest.

What are the following proof triangle LMN equals triangle OPQ

Answers

Answer:

D. SSS

Step-by-step explanation:

Was given to us that the corresponding sides are congruent so is SSS.

Side Side Side Theorem tells us that if am the sides of a triangle are having the same measurement with the corresponding sides of another triangle then the two triangles are congruent.

WILL GIVE BRAINLIEST AND 40 POINTS TO CORRECT ANSWER (URGENT)chef daniels places her favorite recipes in a bag for 4 pasta dishes, 5 casseroles, 3 types of chili, and 8 desserts. If Chef Daniel chooses two recipes at random (but replaces the first one before drawing the second), what is the probability that the first recipe she selects is a casserole and the second recipe she selects is a dessert?

Answers

Answer:

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