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4 years ago

25 POINTS to answer question!

Mathematics

2 answers:

wel4 years ago

6 0

The answer is A.

Hope I helped!

Let me know if you need anything else!

~ Zoe

Amiraneli [1.4K]4 years ago

5 0

The answer would be "A".

This is because the line would have to cross through -4 on the y-axis on the graph. "A" is the only answer that has a line through -4.

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When a figure is translated, what is true, compared to the preimage of the figure

Charra [1.4K]

When you translate something in geometry, you're simply moving it around. You don't distort it in any way. If you translate a segment, it remains a segment, and its length doesn't change. Similarly, if you translate an angle, the measure of the angle doesn't change.

6 0

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

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3 years ago

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2 years ago

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Otrada [13]

Answer:

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Step-by-step explanation:

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