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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! :)