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mmm...let's note something really quickly. 

In the problem, it said that

\(QR + PR < PQ \)

However, it said that PQR is a triangle, but this violates the Triangle Inequality Theorem.

So this problem is invalid. 

thanks! :)

mmm...take a look at the points given. 

\(P_1 P_2 + P_2 P_3 + P_3 P_4 + \dots + P_9 P_{10} + P_{10} P_1\)

This is essentially just the perimeter of the decagon \(P_1P_2P_3P_4P_5...P_{10}\)

The lenngth of one side is essentially

\(radius/2 ( -1 +\sqrt 5)\)

Now that we know one side, we can find the perimeter through the equation

\(10 (1/2) ( -1 + \sqrt 5) = 5 ( -1 + \sqrt 5) ≈ 6.18\)

Thus, the answer is 6.18, 

Thanks! :)

Look for the largest prime number from 100-200 

199  would be the largest class size 

     2003,  2011,  2017,  2027,  2029       

_.

lol, yeah. "just know your primes" 

I love your solution, Umeko1.0 😂😂😂

Nice work!

This is pretty simple. NotThatSmart already showed one way. If you know your primes, you'll find 2, 3, 5, and 7 to be prime. 

A prime number is a number that cannot be divisble except for 1 and itself. 

However, 1 does not qualify, since it technically only 1 factor. 

Now, the numbers \(2,3,5,7\) are all prime numbers. 

However, \(4=2*2, 6=2*3, 8=2*4, 9=3*3, 10=2*5\)

So the answer is 2,3,5,7

Thanks! :)

Let's simplify the two equations a bit first. We get that

\(8a + b = -3 \\ 9a + (2 +m)b = k \)

Now, 

This will have infinte solutions when the equations are the  same...so....multiply the first equation by 9 and the second by 8 and we get

\(72a + 9b = -27 \\\ 72a + 8 (2 + m)b = 8k\)

Thus, we can no easily solve for m and k. Solving for m, we get

\(9 = 8(2 + m) \\ 9 = 16 + 8m \\ -7 = 8m \\ m = -7/8\)

Solving for k, we get

\(8 k = -27 \\ k = -27/8 \)

Thanks! :)

oops, sorry. just realized I sent the wrong link...silly me

https://web2.0calc.com/questions/algebra_5878

First, let's simplify both equations to setup an interesting way to solve the problem. 

\((s/2) + 5t = 3 - 7t + 8s → s + 10t = 6 - 14t + 16s → -15s + 24t = 6 \space\space\space \text{(1st equation)} \\ 3t - 6s = 9 -2t → 6s + 5t = 9 \space\space\space \text{(2nd equation)} \)

Now, let's mulitply the 1st equation by 6 and the 2nd equation by 15

These numbers seem random, we when we write it out, we can see why it's so important. We get

\(-90s + 144t = 36 \\ 90s + 75t = 135 \)

Now we have the opposite s coefficients. Adding these equations, we get

\(219 t = 171 \)

\(t = 171 / 219 = 57 / 73 \)

Now, we can find s through the value of t. We get

\(6s + 5(57/73) = 9 \\ 6s + 285/73 = 9 \\ 6s = 9 - 285/73 \\ 6s = 372/73 \\ s = 62 / 73\)

Thus, our ordered pair (s, t) is \((57/73, 62/73)\)

Thanks! :)

First, let's find the common rate of change through the arithmetic sequence. Let's let this be d.

From the problem, we can write the equation

\(1/4 + 10d = -1/5 \\ 10d = -1/5 - 1/4 = -9/20 \\ d = -9/200\)

Now, we can find the 43rd term of the sequence. We have

\(43rd term = -1/5 + 10d = -1/5 + 10 (-9/200) = -1/5 - 9/20 = -65/100 = -13 / 20 \)

So our final answer is -13/20. 

Thanks! :) 

The answer is  \(2\)

The answer is \(2\)

The answer is \(4\)

Question nicely answered here

mmm...not sure about my solution, but here it is. 

First, let's combine all the like terms. Isolating 115, we get

\(-8x+xy=115\)

Now factoring out x from the left side, and isolating x, we find

\(x\left(y-8\right)=115\\ x=\frac{115}{y-8}\)

Now, I guess we want integers, so let's just see what y can be to make x an integer. 

115 only has 2 factors other and 1 and itself: 5 and 23. 

So x + y has 2 cases. 

First, y could be 13, and x is 23. 

The other option is for y to be 31, and x is 5. 

So we have \(13+23=31+5=36\)

So the answer is 36. 

Thanks! :)