NotThatSmart

We can set up an equation to help us solve this problem. Let's define a variable. 

Let n be the smallest  number. We have \(n + n+1 + n+2 ....... n+6 = 7n + 21\)

Now, this sequence adds up to 28/5 times the largest number, n + 6, so we can write the equation

\(7n+21=28/5(n+6)\)

From here, we just solve for n. We have

\(7n+21=\frac{28}{5}n + 168/5\\ 7/5n=63/5\)

Now, multiplying both sides by the reciprocol of 7/5, we get

\(n=\frac{63}{5} \cdot \frac{5}{7}=9\)

So 9 is our final answer!

Thanks! :)

First, we have to find the values of PA, PQ, and AQ.

Using the pythagorean theorem for all 3 of them, we get

 \(AP = \sqrt{4^2+5^2}=\sqrt{41}\)

\(PQ = \sqrt{1^2+3^2}=\sqrt{10}\)

\(AQ=\sqrt{1^2+5^2} = \sqrt{26}\)

Now, we apply the Law of Cosines. According to the law, 

\(PQ^2 = AQ^2 + PA^2 - 2(AQ * PA) * cos (PAQ)\)

Plugging in all the values of AP, PQ, and AQ we figured out, we get

\(10=26+41-2(\sqrt{26}*\sqrt{41})*cos(PAQ)\\ 10=67-2\sqrt{1066}*cos(PAQ)\\ cos(PAQ) = \frac{-57}{-2\sqrt{1066}}\)

Now, sin(PAQ) is just 1 - cos(PAQ), so we get

\(sin(PAQ)=1- \frac{-57}{-2\sqrt{1066}}\\ sin(PAQ)=1+ \frac{57}{-2\sqrt{1066}}\\ sin(PAQ) = \frac{-2\sqrt{1066}+57}{-2\sqrt{1066}}\)

This is approxaimtely \(0.12709645414\)

So our answer is \(\frac{-2\sqrt{1066}+57}{-2\sqrt{1066}}\)

Thanks! :)

Let's first convert both numbers into base 10 before adding them together. We have

\((A710)_{16} = (10 × 16^3) + (7 × 16^2) + (1 × 16^1) + (0 × 16^0) = (42768)_{10}\)

\((51)_{16} = (5 × 16^1) + (1 × 16^0) = (81)_{10}\)

Now we have to add up 81 and 42768. We get \(42768+81=42849\)

Now we want to convert 42849 into base 16. 

We go in order, so we have

\(42849/16 = 2678 R1\)

\(2678/16 =167 R6\)

\(167/16 =10R 7 \)

\(10/16 =0R 10 \)

So our final answer is \(A761\)

Thanks! :)

First, let's set a variable. Let's set x as angle PQR. 

From the problem, we can write that 

\(cos(x) = \frac{8^2 + 11² - 5²}{2*8*11}=10/11\)

Now, we can sue the law of cosines to write

\( (PM)² = 5.5² + 8² - 2*5.5*8 * cos(x)\\ \)

Now, we can find PM. 

\(PM^2 = 10/11*6.25\\ PM \approx 2.38365647311\)

I'm not exactly sure if this is correct. I may have made a mistake. 

Thanks! :)

First, let's note that the median of the trapezoid is half the sum of the two bases. 

The area is irrelevant to the problem.

If one of the bases has length \(12\), then the other base must have length \(12+16=28\)

Thus, the median is \(\frac{28+12}{2} = 40/2 = 20\)

So 20 is our ansswer. 

Thanks! :)

We can use a clever trick to make \( 2.\overline{57}\) a fraction. 

First, let's set \(x= 2.\overline{57}\)

If we multiply both sides by 100, we get \( 100x=257.\overline{57}\)

Subtracting x from 100x, we get \(100x-x=257.\overline{57}- 2.\overline{57}\)

\(99x= 257.\overline{57}-2.\overline{57}\\ 99x=257\\ x=257/99\\ x=2 \frac{19}{33}\)

So our final answer is \(2 \frac{19}{33}\)

Thanks! :)

We know that the slope of AB is 2, which means we can write a formula. 

The slope of a line is \(\frac{y_2-y_1}{x_2-x_1}\). Plugging in the points \((a, a^2)\) and \((b, b^2)\), we can write the equation

\(\text{Slope} = \frac{b^2 - a^2 }{b - a}\)

\(\frac{b^2-a^2}{b-a}=2\)

Knowing that \(b^2-a^2=(b-a)(b+a)\), we have

\(\frac{ (b -a) (b + a) }{(b -a)} = b + a = 2\)

So our final answer is just 2. 

Thanks! :)

We want to convert \( 0.\overline{2123} \) into a fraction, and we can do this with a very handy trick. 

First, let's set x as \( 0.\overline{2123} \). 

This means that \(10000x=2123.\overline{2123}\)

Doing subtracting x from 10000x, we have

 \(10000x-x= 2123.\overline{2123}-0.\overline{2123}\\ 9999x=2123 x=2123/9999\\ x=193/909\)

So our final answer is \(\frac{192}{909}\)

Thanks! :)

All the points 5 units away from (2, 7) form a nice little circle. 

The formula for the circle is \((x-2)^2+(x-7)^2=25\). 

We now have a system of equations with

\((x-2)^2+(x-7)^2=25\\ x+y=13\)

Let's focus on the first equation. Factoring it, we get

\(2x^2-18x+53=25\\ 2x^2-18x+28=0\\ x^2-9x+14=0\\ (x-7)(x-2)=0\\ x=7,2\)

Plugging these two values into the second equation, we get 

\(y=13-2=11\\ y=13-7=6\)

So our two solutions are

\(\begin{pmatrix}x=2,\:&y=11\\ x=7,\:&y=6\end{pmatrix}\)

So (2,11) and (7,6) are our answers!

Thanks! :)