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statuscvo [17]
3 years ago
10

Su has 3 times as many dolls as Bertha. When Su gives Bertha 4 dolls, they now have the same amount. How many dolls did they EAC

H have at the begining? How many did they EACH have at the end?
Mathematics
2 answers:
inna [77]3 years ago
7 0

I also think it would be 8
Usimov [2.4K]3 years ago
5 0

Answer: i believe its 8

Step-by-step explanation:

i say 8 because 3x4 is 12 and so if bertha orignally had 4 and was given 4 she would have 8 and if su had 12 and gave away four she would also have 8

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Answer:

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Step-by-step explanation:

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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

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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

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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
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Sedbober [7]
She made an error in the STEP 1
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4 0
2 years ago
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Alecsey [184]

Answer:

I think the answer is B= 62  

Step-by-step explanation:

Please mark brainliest and have a great day!

8 0
2 years ago
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