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erastovalidia [21]
3 years ago
11

Can an object be in pure translation as well as in pure rotation simultaneously?

Mathematics
1 answer:
Inessa [10]3 years ago
5 0
  <span>"Pure translational" motion means the object is moving from place to place but isn't rotating at all. Mathematically, that means every point on the object has the same velocity vector as every other point on the object. For example, say you have a box that's sliding along the ground, and say each corner on the box is moving at 15 feet per second at an angle of 26 degrees west of north. Since every point on the box has the same velocity vector, the motion is "pure translational." 

"Pure rotational" motion means the object is rotating, but its position in space isn't changing. Mathematically, that means that every point on the object moves in a circle around some axis (usually through the object's center of mass).

I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!
</span>
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Step 1: Find the Lowest Common Multiple between the denominators.

Step 2: Multiply the numerator and denominator of each fraction by a number (the one that will get them to the lcm) so that they have the LCM as their new denominator.

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Factor.<br><br> 3x(y−4)−2(y−4)<br><br><br><br> Enter your answer in the box.
NISA [10]

Answer: Use the distributive property to multiply 3 by y−4.

3y−12−2(y−4)

Use the distributive property to multiply −2 by y−4.

3y−12−2y+8

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Add −12 and 8 to get −4.

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Step-by-step explanation:

Hope this helps!

6 0
3 years ago
*In ∆ABC, on the extension of the side BC , draw a line segment CD ≅ CA . Draw the segment AD . The line segment CE is the angle
Leya [2.2K]

CE ⊥ CF because m∠ECA + m∠FCA = 90°

Step-by-step explanation:

In ∆ABC

  • On the extension of the side BC , draw a line segment CD ≅ CA
  • Draw the segment AD
  • The line segment CE is the angle bisector of ∠ACB
  • The line segment CF is the median towards AD in ∆ ACD

We want to prove that CF ⊥ CE

Look to the attached figure

In Δ ABC

∵ CE is the bisector of angle ACB

∴ ∠ACE ≅ ∠BCE

In Δ ACD

∵ CA = CD

∴ Δ ACD is an isosceles triangle

∵ AD is the median towards AD

- In any isosceles triangle the median from a vertex to its opposite

  side bisects this vertex

∴ AD bisects ∠ACD

∴ ∠ACF ≅ ∠DCF

∵ BCD is a straight segment

∵ CE , CA , CF are drawn from point C

∴ m∠BCE + m∠ACE + m∠ACF + m∠DCF = 180°

∵ m∠ACE ≅ m∠BCE

∵ m∠ACF ≅ m∠DCF

- Replace m∠BCE by m∠ACE and m∠DCF by m∠ACF

∴ m∠ACE + m∠ACE + m∠ACF + m∠ACF = 180°

∴ 2 m∠ACE + 2 m∠ACF = 180°

- Divide all terms by 2

∴ m∠ACE + m∠ACF = 90°

∴ EC ⊥ CF

CE ⊥ CF because m∠ECA + m∠FCA = 90°

Learn more:

You can learn more about perpendicular lines in brainly.com/question/11223427

#LearnwithBrainly

7 0
3 years ago
A bank says you can double your money in 10 years if you put $1,000 in a simple interest account. What annual interest rate does
katrin2010 [14]

The simple interest formula allows us to calculate I, which is the interest earned or charged on a loan. According to this formula, the amount of interest is given by I = Prt, where P is the principal, r is the annual interest rate in decimal form, and t is the loan period expressed in years. The rate r must be converted from a percentage into decimal form.

Then, 2,000 = 1,000 * r * 10 ;

 Finally, r = 2 ÷ 10 = 20 ÷ 100 = 0.2

hope this helps you

8 0
3 years ago
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