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barxatty [35]
3 years ago
6

at the pet fair, Darlene's dog weighed 5 times as much as leah's dog. Together, the dogs weighed 84 pounds. how much did each do

g weigh? complete the bar model. write an equation and solve
Mathematics
1 answer:
Alex3 years ago
4 0
Let x = Darlene's dog's weightlet y = Leah's dog's weight x = 5yx + y = 84 so, 5y + y = 84, which means 6y = 84   so, y = 84/6 = 14x = 5(14) = 70    Darlene's dog = 70 lbs, Leah's dog = 14 lbs.<span> </span>
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Step-by-step explanation:

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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

#SPJ4

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