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2 years ago

Given: Circle X with a Radius r and circle Y with radius s Prove: Circle x is similar to circle y

Mathematics

2 answers:

lana [24]2 years ago

8 0

Two figures are similar if one is the scaled version of the other.

This is always the case for circles, because their geometry is fixed, and you can't modify it in anyway, otherwise it wouldn't be a circle anymore.

To be more precise, you only need two steps to prove that every two circles are similar:

Translate one of the two circles so that they have the same center
Scale the inner circle (for example) unit it has the same radius of the outer one. You can obviously shrink the outer one as well

Now the two circles have the same center and the same radius, and thus they are the same. We just proved that any two circles can be reduced to be the same circle using only translations and scaling, which generate similar shapes.

Recapping, we have:

Start with circle X and radius r
Translate it so that it has the same center as circle Y. This new circle, say X', is similar to the first one, because you only translated it.
Scale the radius of circle X' until it becomes $s$ . This new circle, say X'', is similar to X' because you only scaled it

So, we passed from X to X' to X'', and they are all similar to each other, and in the end we have X''=Y, which ends the proof.

stealth61 [152]2 years ago

5 0

Answer:

Step-by-step explanation:

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Ahat [919]

Answer:

I'm not sure

Step-by-step explanation:

3 0

1 year ago

This background applies to the next several questions. Assume a 15 cm diameter wafer has a cost of 12, contains 84 dies, and has

nalin [4]

Answer:

1) $Yield_1= \frac{1}{(1+ 0.02 \frac{1}{2} 2.104)^2}=0.959$

$Yield_2= \frac{1}{(1+ 0.031 \frac{1}{2} 3.1415)^2}=0.909$

2) $Cost/die_1 = \frac{12}{84 x 0.959}=0.149$

$Cost/die_2 = \frac{15}{100 x 0.909}=0.165$

3) $Area_1 = \frac{1.1 \pi (7.5cm)^2}{84}=\frac{2.104 cm^2}{1.1}=1.913 cm^2$

$Area_2 = \frac{1.1 \pi (10cm)^2}{100}=\frac{3.1415 cm^2}{1.1}=2.856 cm^2$

And for the new yield we need to take in count the increase of 15% for the area and we got this:

$Yield_1= \frac{1}{(1+(1.15) 0.02 \frac{1}{2} 1.913)^2}=0.957$

$Yield_2= \frac{1}{(1+(1.15) 0.031 \frac{1}{2} 2.856)^2}=0.905$

4) $DR_{old}=\frac{1}{\sqrt{0.92}} -1$ =0.0426 defects/cm^2

$DR_{new}=\frac{1}{\sqrt{0.95}} -1$ =0.0260defects/cm^2

Step-by-step explanation:

Part 1

For this part first we need to find the die areas with the following formula:

$Area= \frac{W area}{Number count}$

$Area_1 = \frac{\pi (7.5cm)^2}{84}=2.104 cm^2$

$Area_2 = \frac{\pi (10cm)^2}{100}=3.1415 cm^2$

Now we can use the yield equation given by:

$Yield=\frac{1}{(1+ DR\frac{Area}{2})^2}$

And replacing we got:

$Yield_1= \frac{1}{(1+ 0.02 \frac{1}{2} 2.104)^2}=0.959$

$Yield_2= \frac{1}{(1+ 0.031 \frac{1}{2} 3.1415)^2}=0.909$

Part 2

For this part we can use the formula for cost per die like this:

$Cost/die = \frac{Cost per day_i}{Number count_i x Yield_i}$

And replacing we got:

$Cost/die_1 = \frac{12}{84 x 0.959}=0.149$

$Cost/die_2 = \frac{15}{100 x 0.909}=0.165$

Part 3

For this case we just need to calculate the new area and the new yield with the same formulas for part a, adn we got:

$Area_1 = \frac{1.1 \pi (7.5cm)^2}{84}=\frac{2.104 cm^2}{1.1}=1.913 cm^2$

$Area_2 = \frac{1.1 \pi (10cm)^2}{100}=\frac{3.1415 cm^2}{1.1}=2.856 cm^2$

And for the new yield we need to take in count the increase of 15% for the area and we got this:

$Yield_1= \frac{1}{(1+(1.15) 0.02 \frac{1}{2} 1.913)^2}=0.957$

$Yield_2= \frac{1}{(1+(1.15) 0.031 \frac{1}{2} 2.856)^2}=0.905$

Part 4

First we can convert the area to cm^2 and we got 2 cm^2 the yield would be on this case given by:

$Yield= \frac{1}{(1+DR\frac{2cm^2}{2})^2}=\frac{1}{1+(DR)^2}$

And if we solve for the Defect rate we got:

$DR= \frac{1}{\sqrt{Yield}}-1$

Now we can find the previous and new defect rate like this:

$DR_{old}=\frac{1}{\sqrt{0.92}} -1$ =0.0426 defects/cm^2

And for the new defect rate we got:

$DR_{new}=\frac{1}{\sqrt{0.95}} -1$ =0.0260defects/cm^2

5 0

2 years ago

The weather report says the temperature is 20°c and will drop 5°c per hour for the next 6 hours. Daryl plans to be gone for at l

umka21 [38]

The weather report says the temperature is 20°c and will drop 5°c per hour for the next 6 hours.

Temperature drops by 5°c per hour

So we find the temperature after 6 hours

In 1 hour temperature drops= 5°c

So in 6 hours = 6* 5 = 30°c

In 6 hours , temperature drops by 30°c

the actual temperature is 20°c. after 6 hours the temperature becomes

20°c - 30°c= -10°c

So, after 6 hours the temperature will be -10°c

Daryl plans to be gone for at least 6 hours. so its 6 hours or more than 6 hours

In 6 hours or more than 6 hours , the temperature will drop less than -10°c.

So, Daryl should move his plant to a warmer location before leaving

7 0

2 years ago

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A sapling, or young tree, is planted in a garden. After 3 years, it is 180 cm tall. After 7 years, it is 368 cm tall. How tall w

cestrela7 [59]

548cm tall because if you add 180 to 368 it is 548.

3 0

2 years ago

A student claims that the expression “9 times the sum of a number and 13” is translated to the algebraic expression 9n + 13. Is

Alex73 [517]

9*13+n
I think that is correct, try and split up the question.
9*
A sum of a number AND 13
So that means that 9 is multiplying by the SUM (adding) of 13 and a number (n)

8 0

2 years ago

Read 2 more answers