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2 years ago

Simply the expression by combining like terms 20c+8d+10c-5d

Mathematics

1 answer:

valentinak56 [21]2 years ago

6 0

First, combine 20c and 10c because they are like terms then combine 8d and -5d.

It would look like this:

20c+8d+10c-5d

30c+3d

Hope that helps

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Use polar coordinates to find the volume of the given solid. Inside both the cylinder x2 y2 = 1 and the ellipsoid 4x2 4y2 z2 = 6

Anton [14]

The Volume of the given solid using polar coordinate is: $\frac{-1}{6} \int\limits^{2\pi}_ {0} [(60) ^{3/2} \; -(64) ^{3/2} ] d\theta$

V= $\frac{-1}{6} \int\limits^{2\pi}_ {0} [(60) ^{3/2} \; -(64) ^{3/2} ] d\theta$

<h3>What is Volume of Solid in polar coordinates?</h3>

To find the volume in polar coordinates bounded above by a surface z=f(r,θ) over a region on the xy-plane, use a double integral in polar coordinates.

Consider the cylinder, $x^{2}+y^{2} =1$ and the ellipsoid, $4x^{2}+ 4y^{2} + z^{2} =64$

In polar coordinates, we know that

$x^{2}+y^{2} =r^{2}$

So, the ellipsoid gives

$4{(x^{2}+ y^{2)} + z^{2} =64$

4( $r^{2}$ ) + $z^{2}$ = 64

$z^{2}$ = 64- 4( $r^{2}$ )

z=± $\sqrt{64-4r^{2} }$

So, the volume of the solid is given by:

V= $\int\limits^{2\pi}_ 0 \int\limits^1_0{} \, [\sqrt{64-4r^{2} }- (-\sqrt{64-4r^{2} })] r dr d\theta$

= $2\int\limits^{2\pi}_ 0 \int\limits^1_0 \, r\sqrt{64-4r^{2} } r dr d\theta$

To solve the integral take, $64-4r^{2}$ = t

dt= -8rdr

rdr = $\frac{-1}{8} dt$

So, the integral $\int\ r\sqrt{64-4r^{2} } rdr$ become

= $\int\ \sqrt{t } \frac{-1}{8} dt$

= $\frac{-1}{12} t^{3/2}$

= $\frac{-1}{12} (64-4r^{2}) ^{3/2}$

so on applying the limit, the volume becomes

V= $2\int\limits^{2\pi}_ {0} \int\limits^1_0{} \, \frac{-1}{12} (64-4r^{2}) ^{3/2} d\theta$

= $\frac{-1}{6} \int\limits^{2\pi}_ {0} [(64-4(1)^{2}) ^{3/2} \; -(64-4(2)^{0}) ^{3/2} ] d\theta$

V = $\frac{-1}{6} \int\limits^{2\pi}_ {0} [(60) ^{3/2} \; -(64) ^{3/2} ] d\theta$

Since, further the integral isn't having any term of $\theta$ .

we will end here.

The Volume of the given solid using polar coordinate is: $\frac{-1}{6} \int\limits^{2\pi}_ {0} [(60) ^{3/2} \; -(64) ^{3/2} ] d\theta$

Learn more about Volume in polar coordinate here:

brainly.com/question/25172004

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3 0

1 year ago

Which expressions are equivalent to 6 (–x) 2x (–7) 2x? Check all that apply. X x 6 – 7 x 2x 2 x 3 – x 2x – 4 2x x – 1 x 1.

qaws [65]

Equivalent expression is the expression which is equal to the original equation but the way of representation is different. The equivalent expression of the given expression are,

$x+x+6-7+x$

$3-x+2x-4+2x=3x-1$

Given information

The given expression in the problem is,

$6+(-x)+2x+(-7)+2x$

<h3>Equivalent expression</h3>

Equivalent expression is the expression which is equal to the original equation but the way of representation is different.

The given equation is,

$6+(-x)+2x+(-7)+2x$

As the product of the negative and positive is negative. Thus solve the equation by opening the bracket.

$6+(-x)+2x+(-7)+2x=6-x+2x-7+2x\\
6+(-x)+2x+(-7)+2x=-x+2x+2x+6-7\\
6+(-x)+2x+(-7)+2x=3x-1$

Check all the option,

1) The expression given in the option 1 is,

$x+x+6-7+x$

Solve it further,

$x+x+6-7+x=x+x+x+6-7\\
x+x+6-7+x=3x-1\\$

The result is equal to the result of the given expression. Thus the expression of the option 1 is equivalent to the expression given in the problem.

2) The expression given in the option 2 is,

$2x+2+x$

Solve it further,

$2x+2+x=2x+x+2\\2x+2+x=3x+2$

The result is not equal to the result of the given expression. Thus the expression of the option 2 is not equivalent to the expression given in the problem.

3) The expression given in the option 3 is,

$3-x+2x-4+2x$

Solve it further,

$3-x+2x-4+2x=3x-1$

The result is equal to the result of the given expression. Thus the expression of the option 3 is equivalent to the expression given in the problem.