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USPshnik [31]
3 years ago
14

Two gears are connected and are rotating simultaneously. The smaller gear has a radius of 2 inches, and the larger gear has a ra

dius of 7 inches.
Part 1: What is the angle measure, in degrees and rounded to the nearest tenth, through which the larger gear has rotated when the smaller gear has made one complete rotation?
Part 2: How many rotations will the smaller gear make during one complete rotation of the larger gear?
Show all work.
Mathematics
2 answers:
sleet_krkn [62]3 years ago
5 0

Answer:

Let r represent the radius of the smaller circle and R the radius of the larger circle.

Apply ratios: the radius of smaller circle to radius of larger circle, i.e.

r: R = 3 : 7.

I complete rotation = 360 degrees.

Part 1:

For one complete rotation of the smaller circle, the larger circle is rotated through: (3/7)*(360) = 154.3 degrees

Part 2:

For one complete rotation of the larger circle, the larger circle is rotated through: (7/3)*(360) = 840.0 degrees

This is equivalent to (840/360) = 2.3 rotations

Alternatively, use the ratios:

The number of rotations = R/r = 7/3 = 2.3 rotations  

BRAINLIEST PLEASE??

Step-by-step explanation:

Anika [276]3 years ago
3 0

Answer:

Step-by-step explanation:

for smaller gear circumference=2π×2=4π inches

for larger gear

l=4π

θ=l/r

θ=(4 ×180)/7=720/7=102 .9°

for larger gear circumference=2π×7=14π

number of rotations=(14π)/4π=7/2=3.5 rotations or 3 1/2 rotations.

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Consider the lengths of consecutive 1-2 blocks.

block 1 - 1, 2 - length 2

block 2 - 1, 2, 2 - length 3

block 3 - 1, 2, 2, 2 - length 4

block 4 - 1, 2, 2, 2, 2 - length 5

and so on.


Recall the formula for the sum of consecutive positive integers,

\displaystyle \sum_{i=1}^j i = 1 + 2 + 3 + \cdots + j = \frac{j(j+1)}2 \implies \sum_{i=2}^j = \frac{j(j+1) - 2}2

Now,

1234 = \dfrac{j(j+1)-2}2 \implies 2470 = j(j+1) \implies j\approx49.2016

which means that the 1234th term in the sequence occurs somewhere about 1/5 of the way through the 49th 1-2 block.

In the first 48 blocks, the sequence contains 48 copies of 1 and 1 + 2 + 3 + ... + 47 copies of 2, hence they make up a total of

\displaystyle \sum_{i=1}^48 1 + \sum_{i=1}^{48} i = 48+\frac{48(48+1)}2 = 1224

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\displaystyle \sum_{i=1}^{48} 1 + \sum_{i=1}^{48} 2i = 48 + 48(48+1) = 48\times50 = 2400

This leaves us with the contribution of the first 10 terms in the 49th block, which consist of one 1 and nine 2s with a sum of 1+9\times2=19.

So, the sum of the first 1234 terms in the sequence is 2419.

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