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2 years ago

10

WILL MARK BRANLIEST PLEASE ANSWER Guys what is the perimeter? -PHOTO INCLUDED-

2 answers:

My name is Ann [436]2 years ago

5 0

Perimeter = 9+8+14
=31

Black_prince [1.1K]2 years ago

3 0

$\qquad \qquad\huge \underline{\boxed{\sf Answer}}$

Perimeter of a triangle = sum of measures of each side ;

$\qquad \rightarrow \sf p = 9 \: ft + 8y + 14 \: ft$

$\qquad \rightarrow \sf p =23 \: ft + 8y$

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snow_lady [41]

Answer:

3x + 2 = 59

Step-by-step explanation:

because the total is 59 so is definetly =59

and no of oranges in bag so there are 3 bags it means its 3x so the equation is 3x + 2 = 59

Hope it will help you

Lots of love from nepal <3 :)

5 0

3 years ago

If ∠A > ∠B, then what is the relation between AC and BC?<br>

prohojiy [21]

Answer:

ab and ac are congruent with each other

8 0

2 years ago

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Write the equation of the center (2,5) and passes through the point (8,5)

BARSIC [14]

Answer: varibles coffeeefishents exponets study then eat

Step-by-step explanation:

3 0

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

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3 0

2 years ago

Help ill Give brainliest

patriot [66]

A= 16777216
B= 279936
C= 32768
D= 49

3 0

3 years ago

Read 2 more answers