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3 years ago

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what are the matching the functions with their ranges y = 3sin(x − π), (-∞, ∞) ,y = 1 − sin(x) ,[-1, 7] , y = 3 + 4cos(x − π) ,[

-3, 3] , y = 2 + cot(x), [0, 2]

1 answer:

BabaBlast [244]3 years ago

7 0

You need to know:

Tangent function has range of <span>(-∞, ∞)

Sine function has range of [-1,1], which is the same as cosine function

So [-3,3] matches the 1st, [0,2] matches the 2nd, [-1,7] matches the 3rd, the 4th will be the rest </span>

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What is the solution to the following system of equations? 4x + 2y = 18 x − y = 3

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For this case we have the following system of equations:

$4x + 2y = 18\\x-y = 3$

To solve, we clear "x" from the second equation:

$x = 3 + y$

We substitute "x" in the first equation:

$4 (3 + y) + 2y = 18\\12 + 4y + 2y = 18\\12 + 6y = 18$

We clear the value of the variable "y":

$6y = 18-12\\6y = 6\\y = \frac {6} {6}\\y = 1$

We look for the value of the variable "x":

$x = 3 + y\\x = 3 + 1\\x = 4$

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$(x, y) :( 4,1)$

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$(x, y) :( 4,1)$

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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

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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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