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whenever i go to the link all i see is black

That means 5228X is divisible by 2 and 3.

So X must be an even number, and \(5 + 2 + 2 + 8 + X\) is divisible by 3.

But that means \(2 + 2 + 2 + 2 + X \) is divisible by 3. Taking mod 3 again on both sides,

\(2 + X\) is divisible by 3, and don't forget that X is an even digit.

\(X = 4\) is the only possible solution.

OA, OB, OC divide the circle into three congruent parts \(\Longleftrightarrow\) \(\angle AOB = \angle BOC = \angle COA = 120^\circ\)

D is a point on arc AB such that arc AD has half the measure of arc DB \(\Longleftrightarrow\) \(m\stackrel{\mbox{$\large\frown$}}{AD} = 40^\circ\), \(m\stackrel{\mbox{$\large\frown$}}{DB} = 80^\circ\)

\(\text{reflex }\angle COD = m\stackrel{\mbox{$\large\frown$}}{DB} + m\stackrel{\mbox{$\large\frown$}}{BC} = 200^\circ\)

\(m\angle COD = 360^\circ - 200^\circ = 160^\circ\)

Because OC = OD,

\(m\angle OCD = m\angle ODC\)

So \(160^\circ + 2 \left(m\angle OCD\right) = 180^\circ\)

I believe you can continue from here. It's just one step away from the answer.

There are 15 numbers, and we are choosing 5 digits. 

__  __  __  __  __

Permutation means order matters, so there are 15 options for the first digit, 14 for the second, 13 for the third, etc.

15  14  13  12  11 = 15*14*13*12*11

But there are repeating digits in the set. We have to divide by the number of ways the 2s, 3s, 4s, and 5s can repeat. Since there are two repeating 2s, there are 2! different ways they can repeat. Since there are three repeating 3s there are 3! different ways they can repeat (using the same logic as when we did 15 options, 14 options, 13 options, etc.

(15*14*13*12*11) / (2!*3!*4!*5!) 

However, this doesn't come out as an integer :( ... I'm not sure what I did wrong. Maybe you can find my mistake. Hopefully this helps anyways!

isn't this a math problem?

then why is it off-topic?

https://www.google.com/search?safe=active&rlz=1CAHJUL_enUS905&sxsrf=ALeKk00yxR0t7ZSFLeSwwRLLVF-LFmR5XQ%3A1592070821375&ei=pRLlXuKxFq6awbkP9_WxiAs&q=google+translate&oq=goo&gs_lcp=CgZwc3ktYWIQARgAMgQIIxAnMgcIABCDARBDMgQIABBDMgUIABCxAzIHCAAQsQMQQzIHCAAQsQMQQzIHCAAQsQMQQzIHCAAQsQMQQzIHCAAQsQMQQzIFCAAQsQM6BQgAEJECOgYIABAHEB46BQgAEIMBOgIIADoHCCMQ6gIQJ1D8Flj1RmDcVGgBcAB4AIABdYgBrQSSAQMwLjWYAQCgAQGqAQdnd3Mtd2l6sAEK&sclient=psy-ab

try going to this link

https://www.khanacademy.org/math/geometry-home/transformations

hope it helps

hi jimk

if u don't need help then just "delete" the question

since u can't exactly delete it just edit it then write "deleted" so people won't waste their time clicking on the question

We can obtain a larger value if we add it like this:

3*2(+8/2 + 2) = 6(6) = 36

Treat the addition symbol as a positive sign.

No, it isn't. The rotation center is (0, 2).

If you translate the entire diagram 2 unit downwards, then you would realise it is rotated 180 degrees about (0, 0). Then you can translate the entire diagram upwards again, including that rotation center, and you will realise the rotation center is (0, 2). 

Because \((a,b )\in \mathbb Z^2\), it is optimal to minimize b and maximize a.

Because of that reason, if we choose (a, b) = (50, -50), then we get our minimum value, which is -2500.

(If a was larger, then b would be smaller. Same otherwise. It is actually optimal to choose |a| = |b|, or make |a| as close as |b|.)

i got its a rotation from the centre 3,-2 ?

The closer two numbers with the same sum are, the bigger their product.

In this case, the problem did not specify whether positive or negative so the minimum value is: 50-(-50)=100

Therefore, a*b= -2500

the minimum value of b = 1 so the minimum value of a = 101      1*101= 101

Oh I feel sorry for you! :(

What's your attempt on this question?

if negatives are fine, then its -2500

if it's only positive answers, then I'm pretty sure its 101

"deleted" realized my answer is wrong

yes what? 

nice solution thanks too!

thanks

We set up an equation first: 0.25q + 0.10d = 2.65

Multiplying by 100 and dividing by 5, we get 5q + 2d = 53

At this point, we can just use trial and error, noting that q can only be an odd number because when it's even, 3-0 = 3, which is not divisible by 2.

After testing out a few cases, the only pair that satisfies q > d is q=9 and d=4.