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ikadub [295]
2 years ago
12

HELP WITH SOME GEOMETRY

Mathematics
1 answer:
Verizon [17]2 years ago
8 0

Answer:

9 is the <u>answer</u>

<h2><u>Hope</u><u> </u><u>it</u><u> </u><u>helps</u></h2>
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Evaluate 5 - 3(-2) + 1-3|​
marishachu [46]

Answer:

9

Step-by-step explanation:

5 - 3(-2) + 1-3

5 + 6 - 2

11 - 2

9

5 0
3 years ago
Read 2 more answers
What is the approximate circumference of the circle shown below?
Crank

Answer

Formula

Circumference\ of\ circle = 2\pi r

where r is the radius of a circle.

As shown in the image

Diameter of the circle = 16 .3 cm

Radius\ of\ the\ circle = \frac{16.3}{2}

                                   = 8.15 cm

Put in the formula

Circumference\ of\ circle = 2\pi\times 8.15

As

\pi = 3.14

Circumference\ of\ circle = 2\times 3.14\times 8.15

                                                 = 51.2 cm (approx)

Therefore the circumference of a circle is 51.2cm (approx) .

Option (D) is correct .

8 0
3 years ago
Read 2 more answers
WILL MARK BRAINLIEST HELP ASAP ASAP
Alina [70]

Answer:

B

Step-by-step explanation:

I got this answer correct on a test

5 0
2 years ago
A jewelry store is featuring a diamond in the shape of a square pyramid. The side length of
BARSIC [14]

Answer:

128√5/3 mm³

Step-by-step explanation:

Since we are not told what to find, we can as well look for the volume of the pyramid

Volume of a square pyramid: V = (1/3)a²h

a is the side length of the square

h is the height of the pyramid

Given

a = 8mm

l² = (a/2)² + h²

l² = (a/2)² + h²

6² = (8/2)² + h²

h² = 6² - 4²

h² = 36 - 16

h² = 20

h = √20

Volume of a square pyramid =  (1/3)*8²*√20

Volume of a square pyramid = 1/3 * 64 * 2√5

Volume of a square pyramid = 128√5/3 mm³

7 0
3 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

2\times 1.5\times (r')^2=0.321

r'^2=0.107009

r'=0.327 m

7 0
3 years ago
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