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3 years ago

James is x years old now and the sum of his age is and his brothers age is 10.

Mathematics

1 answer:

Vesnalui [34]3 years ago

5 0

Answer:

1) James brother is 10-x years old

2) James was 3 years ago x-3 years old

3) James brother was 3 years ago 7-x years old

4) 3(x-3)=7-x

5) James is 4 years old, his brother is 6 years old

Step-by-step explanation:

3) 10 - x - 3 = 7 - x

5) 3* James age 3 years ago = James brothers age 3 years ago

<=> 3*(x-3) = 7-x

<=> 3x - 9 = 7 - x

<=> 4x = 16

<=> x = 4

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3 years ago

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Step-by-step explanation:

For a

the solution is x>-17

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3 years ago

Find the radius r of the sphere.

AleksandrR [38]

Given:

The volume of the sphere = 12348π in³

To find the radius of the sphere.

Formula

The volume of a sphere of radius r is

$V = \frac{4}{3} \pi r^{3}$

According to the problem,

$\frac{4}{3} \pi r^{3} = 12348\pi$

Eliminating π from both the side.

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or, $r^{3}=9641$

or, $r=\sqrt[3]{9261}$

or, $r=21$

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3 0

3 years ago

Read 2 more answers

Are the following triangles similar, Explain your answer

Pie

Yes the have the same 78° angle . The smaller one should be able to fit inside the larger one.

3 0

3 years ago

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

Ilya [14]

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.

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

$x^2 + y^2 = 1\\\\z = 4$

- 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: ( $y = - \sqrt{1 - x^2}$ , $y = \sqrt{1 - x^2}$ ). 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: ( $y = - \sqrt{1 - x^2}$ , $y = \sqrt{1 - x^2}$ )

$V_s = \int [ {4y - x^2.y - \frac{4}{3} y^3} ] \, | .dx\\\\V_s = \int [ {8\sqrt{( 1 - x^2 )} - 2x^2*\sqrt{( 1 - x^2 )} - \frac{8}{3} ( 1 - x^2 )^\frac{3}{2} } ] . dx$

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

$V_s = [ 4. ( arcsin ( x ) + x\sqrt{1 - x^2} ) - \frac{arcsin ( x ) - 2x ( 1 -x^2 )^\frac{3}{2} + x\sqrt{1 - x^2} }{2} - \frac{ 3*arcsin ( x ) + 2x ( 1 -x^2 )^\frac{3}{2} + 3x\sqrt{1 - x^2} }{3} ] | \limits^1_-_1\\\\V_s = [ \frac{5}{2} *arcsin ( x ) + \frac{5}{3}*x ( 1 -x^2 )^\frac{3}{2} + \frac{5}{2} *x\sqrt{1 - x^2} ) ] | \limits^1_-_1\\\\V_s = [ \frac{5\pi }{2} + 0 + 0 ] \\\\V_s = \frac{5\pi }{2}$

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

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3 years ago