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Sedaia [141]
3 years ago
15

-12×4÷6-4= how much is it​

Mathematics
2 answers:
Nat2105 [25]3 years ago
8 0

Answer:

-12

Step-by-step explanation:

Hope this helps :D

Vera_Pavlovna [14]3 years ago
4 0
-24


hope that helps
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Expand the fraction 72/125 .so that it has a denominator of 1000.
Dmitrij [34]
X/1000=72/125  multiply both sides by 1000

x=72000/125

x=576

So the fraction is 576/1000
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3 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
Find the area for all of these numbers
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answer : 29
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2 years ago
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Answer:

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Step-by-step explanation:

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3 years ago
Help me with this please
KiRa [710]
No pongas foto por que no puedo ver y no te puedo responder
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3 years ago
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