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1 year ago

Brainliest if its right

Mathematics

2 answers:

Lorico [155]1 year ago

7 0

Answer:

given ,

radius of circle = 7 inches.

( value of pi = 22/7 )

to calculate - circumference of the circle.

$circumference = 2\pi \: r \\ \\ \dashrightarrow 2 \times \frac{22}{\cancel7} \times \cancel7 \\ \\ \dashrightarrow \: 2 \times 22 \\ \\ \dashrightarrow44 \: inches {}^{2}$

hope helpful :D

ira [324]1 year ago

4 0

Formula : C=2πr

43.98

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2 years ago

If u={1,2,3,4,5},A={2,4} and Beta {2,5,5}find n(AUB)

slavikrds [6]

u={1,2,3,4,5},A={2,4} and Beta {2,5,5}

now, (AUB)={1,3,3,4,5}

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n(AUB)=5

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3 years ago

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Shalnov [3]

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2 years ago

A hoop, a uniform solid cylinder, a spherical shell, and a uniform solid sphere are released from rest at the top of an incline.

Serjik [45]

Answer:

Step-by-step explanation:

Given

Hoop, Uniform Solid Cylinder, Spherical shell and a uniform Solid sphere released from Rest from same height

Suppose they have same mass and radius

time Period is given by

$t=\sqrt{\frac{2h}{a}}$ ,where h=height of release

a=acceleration

$a=\frac{g\sin \theta }{1+\frac{I}{mr^2}}$

Where I=moment of inertia

a for hoop

$a=\frac{g\sin \theta }{1+\frac{mr^2}{mr^2}}$

$a=\frac{g\sin \theta }{2}$

a for Uniform solid cylinder

$a=\frac{g\sin \theta }{1+\frac{mr^2}{2mr^2}}$

$a=\frac{2g\sin \theta }{3}$

a for spherical shell

$a=\frac{g\sin \theta }{1+\frac{2mr^2}{3mr^2}}$

$a=\frac{3g\sin \theta }{5}$

a for Uniform Solid

$a=\frac{g\sin \theta }{1+\frac{2mr^2}{5mr^2}}$

$a=\frac{5g\sin \theta }{7}$

time taken will be inversely proportional to the square root of acceleration

$t_1=k\sqrt{2}=1.414k$

$t_2=k\sqrt{\frac{3}{2}}=1.224k$

$t_3=k\sqrt{\frac{5}{3}}=1.2909k$

$t_4=k\sqrt{\frac{7}{5}}=1.183k$

thus first one to reach is Solid Sphere

second is Uniform solid cylinder

third is Spherical Shell

Fourth is hoop

3 0

3 years ago

Look at this ................. HI

OLEGan [10]

Answer: Yes

Step-by-step explanation:

You do yes + yes = yes

Yes is the answer xD

8 0

2 years ago