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AysviL [449]
3 years ago
5

Which equation is in point-slope form for the given point and slope?

Mathematics
1 answer:
grin007 [14]3 years ago
3 0
The equation of a straight line is y=mx+b, where m is the slope of the line.
By substituting the given information into this equation, we get:
7=(2/3×1)+b 
Now solve to find b, the y intercept:
7=2/3+b
b=19/3
So the equation of the line is y=2/3x+19/3
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C  is the central angle of the arc
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arc length = 2 π R (137/360) = 2.39R
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2 years ago
Write an expression the "difference of eighteen and two"
Bess [88]
(D): 18-2

Aside from its actual definition—“difference” in math terms talks about and/or means to subtract. The difference is the result/answer to a subtraction problem.
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2 years ago
Find the ratio of triangles to circles in the diagram below
Sedaia [141]

Answer:

Ratio should be equal so ratio is 3:3

<h3 /><h3>PLEASE MARK ME BRAINLIEST.</h3>
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2 years ago
Drag each shape to the correct category. Identify which shapes are similar to shape A and which are not.
Artyom0805 [142]
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3 years ago
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

3 0
1 year ago
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