All Answers 356464 Answers

You had a typo in 39+3x=31+7x it should be 21 instead of 31. so x=9/2 so TP=21/5. Thank you for all your help

15 * 45  = 20 * x

[15 * 45 ] / 20  = x

x = 33.75 min =  33 min 45 sec

https://web2.0calc.com/questions/geometry_43278

See a very similar problem  here :

https://web2.0calc.com/questions/help-pls_105

https://web2.0calc.com/questions/geometry_81445

WX * XY =  2 * 3    =  6  =  [ WXYZ]

Let O be the center of the larger circle and let N be the center of the smaller circle

Extend   BA   and ON  to meet at  M

Let MN = 3 + x      MT = 6 + x      OM =  13 + x    NA =  3    BO = 7

First relationship

MN / NA  = MT / TP

NA / MN  = TP  / MT

3 / (3 + x)  = TP /  (6 + x)

TP  = 3(6 + x) / (3 + x)

Second relationship

TP / MT =  BO / OM

3 (6 + x) /  (3 + x)         7

______________  = ________

        (6 + x)               (13 + x)

3 / ( 3 + x)  = 7 / (13 + x)

3(13 + x) = 7(3 + x)

39 + 3x  = 21 + 7x

18  = 4x

x = 18/4  = 9/2

TP  = 3(6 + 9/2) / (3 + 9/2)  =  21 / 5 =  4.2

CORRECTED  !!!!

If you are willing to show that you indeed can solve this problem, do it. 

There's no point in insulting the moderators. They're volunteers graciously spending their time on this forum. 

Like I said, if you have a problem with this website, you don't have to be on it. 

However, there are people who DO like this website, and you can't change their minds. 

With these comments, you aren't helping anybody. 

Yes, I know this problem is not impossible. I know it involves cos, tan, theta, and creativity. 

I'm simply chalenging those on the forum. 

I don't care if no one answered. 

In fact, I do know the moderators gace it an attempt. I saw them on this promblem, trying to do it.

That's all I asked for, and that's what I got. 

I got sqrt57/2

thank you!

Let's plug everything in and go from there. 

Plugging in 4, 2, and -5,

We have \(\sqrt[3]{4(4^4)(2^5)}+\frac{4-(-5)}{(2-5)^2}\)

Simplfying, we get

\(\sqrt[3]{256*32*4}+\frac{9}{(-3)^2}\\ = \sqrt[3]{32768}+1\\ =32+1\\ =33\)

So our final answer is 33. 

Thanks! :)

Let's note something really important first. 

Note that XZW forms a triangle!

Thus, we have

\(\angle ZXW = 90\\ \angle XWZ = 60 \)

Thus, \(\angle XZW = 180 - 90 - 60 = 30\)

So our answer is 30. 

Thanks! :)

Let's note that median of the trapezoid is the average of the two bases. 

12 has to be smaller base, since it's impossible to have negative sidelengths. 

Thus, we can write the equation

\(\frac{12+(12+16)}{2}=40/2 = 20\)

So our answer is 20. 

Thanks! :)

In the first 1000 numbers, there are \(31\) perfect squares, \(10\) perfect squares, and 3 of both. Thus there are

\(1000-(31+10) + 3 = 1000 - 38 = 962 \)

So we have 38 numbers to go! 

We have \(32^2=1024\), so before that we tack on another 23 numbers. We have 15 more numbers. 

The next number we worry about is \(33^2=1089\)

After 1024, the next 15th number is 1039. 

So 1039 is our answer. 

I might be wrong though...

Thanks! :)

so i would start this problem by starting with a number like 1300 then getting rid of the squares and cubes less that 1300 and see how close i am. so 1300 is about 1296 or 36^2. 10^3 is the greatest cube less than 1300. so there would be 36+10 or 46 squares and cubes. but we overcounted the 6th powers. 1, 2 and 3 are in the range so we must substract 3. 43 in total. which means our initial guess is way too high. so ill try 1050. there are still 10 cubes but only 32 squares. so there are 39 values that work. so still too high. ill try 1030. still 10 cubes and 32 squares. now ill try \(\boxed{1039}\), and i think it works!!!

in my previous solution i got 5*4*3 /6 but now i think its 5*4*3/3 since you can rotate the equilateral triangle to only three sides. so im pretty sure its \(\boxed{20}\)?

Hopefully this is helpful, https://web2.0calc.com/questions/hard-hard-hard. It might be wrong, and it might not answer them all, but i hope this helps

Thank you! :)

~NTS

Good job, NTS !!!

We can set variables to complete this problem quite efiiciently. 

First, let's set the center of the large circle as M and the center of the smaller circle as N. 

Now, let's extend AB and and the line connecting the centers until they meet each other at a pont, O. 

Let the distance of the left edge of the smaller circle and O be x. 

Now, let's note that triangle BMO and ANO are similar. This is quite important. 

From this, we can write

\(BM / MO = AN / NO \\ 4 / (4 + 2 + x) = 1 / ( 1 + x) \\ 4 / ( 6 + x) = 1 /(1 +x) \\ 4(1 + x) = 1 (6 + x) \\ 4 + 4x = 6 + x \\ 3x = 2 \\ x = 2/3\)

Now that we have the value for x, we can easily solve for AB. We get that

\(OA = \sqrt{ NO^2 - NA^2 } = \sqrt{ (1 + 2/3)^2 - 1^2} = \sqrt{ 25/9 -1} = \sqrt{16/9} = 4/3 \\ OB = \sqrt{ MO^2 - MB^2} = \sqrt{ (6 + 2/3)^2 - 4^2 } = \sqrt{ 400/9 -16} = \sqrt{256/9} = 16/3 \\ AB = OB = OA = 16/3 - 4/3 = 12/3 = 4 \)

So our answer is 4

Feel free to let me know if I messed up!

Thanks! :)

https://web2.0calc.com/questions/circles_146

https://web2.0calc.com/questions/circles_147

We can use an arithmetic sequence to solve this problem. 

The first number in this list is \(1043\), scine 1040 is divisble by 65. 

We just add 65 from there, until we get to 1953, since 1950 is divisble by 65. 

Thus, we have the sequence

\(1043, 1108, 1173...1953\)

Subtracting every term 3 and dividing by 65. we get

\(16,17,18,...30\)

Next, we subtract 15 from every term. 

We get

\(1,2,3...15\)

So our answer is 15...just like what Shmewy said!

Thanks! :)