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2 years ago

Whats the perimeter of the figure below?

Mathematics

2 answers:

baherus [9]2 years ago

7 0

Answer:

Step-by-step explanation:

2+9+11+2+9+11

sp2606 [1]2 years ago

3 0

Answer:

Perimeter: 40 in

Step-by-step explanation:

You already know that perimeter refers to the lengths of the SIDES of a shape, so let's apply it here.

As we can see this is a hexagon, a shape with 6 SIDES

You are given the numbers 2, 9, 11, 2 already as 4 out of the 6 SIDES needed to solve this problem.

You need two more sides.

As you can see the sides that we're missing here has no numbers, but we can look at other side lengths and see if they can help.

The other side lengths that can help are 9 in and 2 in. Why? because when subtracted, they equal the missing length.

If you took away the 2 from the 9, you get a length of 7, which is one of the missing sides.

The NEXT missing side, you can determine by subtracting 11-2, because then you are left with the extra missing space. 11-2= 9

Now that we have 6/6 SIDES, let's now add.

7+9+2+2+9+11=40

Therefore your answer is 40 in.

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2 years ago

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2 years ago

A student on a piano stool rotates freely with an angular speed of 2.85 rev/s . The student holds a 1.50 kg mass in each outstre

Vlad1618 [11]

Answer:r'=0.327 m

Step-by-step explanation:

Given

$N=2.85 rev/s$

angular velocity $\omega =2\pi N=17.90 rad/s$

mass of objects $m=1.5 kg$

distance of objects from stool $r_1=0.789 m$

Combined moment of inertia of stool and student $=5.53 kg.m^2$

Now student pull off his hands so as to increase its speed to 3.60 rev/s

$\omega _2=2\pi N_2$

$\omega _2=2\pi 3.6=22.62 rad/s$

Initial moment of inertia of two masses $I_0=2mr_^2$

$I_0=2\times 1.5\times (0.789)^2=1.867$

After Pulling off hands so that r' is the distance of masses from stool

$I_0'=2\times 1.5\times (r')^2$

Conserving angular momentum

$I_1\omega =I_2\omega _2$

$(5.53+1.867)\cdot 17.90=(5.53+I_o')\cdot 22.62$

$I_0'=1.397\times 0.791$

$I_0'=5.851$

$5.53+2\times 1.5\times (r')^2=5.851$

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$r'^2=0.107009$

$r'=0.327 m$

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3 years ago

The circumference of the equator of a sphere was measured to be 82 82 cm with a possible error of 0.5 0.5 cm. Use linear approxi

True [87]

Answer:

The maximum error in the calculated surface area is approximately 8.3083 square centimeters.

Step-by-step explanation:

The circumference ( $s$ ), in centimeters, and the surface area ( $A_{s}$ ), in square centimeters, of a sphere are represented by following formulas:

$A_{s} = 4\pi\cdot r^{2}$ (1)

$s = 2\pi\cdot r$ (2)

Where $r$ is the radius of the sphere, in centimeters.

By applying (2) in (1), we derive this expression:

$A_{s} = 4\pi\cdot \left(\frac{s}{2\pi} \right)^{2}$

$A_{s} = \frac{s^{2}}{\pi^{2}}$ (3)

By definition of Total Differential, which is equivalent to definition of Linear Approximation in this case, we determine an expression for the maximum error in the calculated surface area ( $\Delta A_{s}$ ), in square centimeters: