NotThatSmart

3. 

Let's set up a system of equations to solve this problem. 

Let's let b be bagels

Let's let t be toast

And let's let m be muffins. 

From the problem, we have the system

\(2T + 1B = 3.15 \\ 1M + 1B = 3.50 \\ 1T + 2B + 3M = 8.15\)

Now, we want everything to be in terms of bagels. 

From the first two equations, we get the equations

\([3.15 - B ] / 2 = T \\ M = 3.50 - B\)

Now, plugging these values into the third equation, we get

\([ 3.15 - B ] / 2 + 2B + 3 [ 3.50 - B ] = 8.15 \\ [3.15 - B ] + 4B + 6 [ 3.50 -B ] = 16.30 \\ -3B + 3.15 + 21 = 16.30 \\ -3B = 16.30 - 3.15 - 21 \\ -3B = -7.85 \\ B = 7.85 / 3 ≈ $2.62\)

This is 262 cents or about 261 and 2/3 cents. 

Thanks! :)

Let's know that XZW will form a triangle

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

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

30 degrees is our answer. 

Thanks! :)

Let  \(BC = a, AC = b\) just to simplify our eyes a bit. 

Let's note that through the Pythagorean Theorem, we can find AM and BN in terms of a and b. 

We get \(\overline{AM} = \sqrt{b^2 + \left(\dfrac a2\right)^2} = \dfrac12 \sqrt{4b^2 + a^2}\)

and \(\overline{BN} = \sqrt{a^2+ \left(\dfrac{b}{2}\right)^2} = \dfrac12 \sqrt{4a^2 + b^2}\)

Because AM and BN both equal 1, we get the equations

\(4b^2 + a^2 = 1^2 \\4a^2 + b^2 = 1^2\)

Adding the two equations together and dividing, we get

\(a^2 + b^2 = \dfrac{1+ 1}{5} = 2/5\)

Since the hypotenuse is just the square root of a^2 + b^2, we get

\(\sqrt{a^2+ b^2} = \sqrt{2/5}\)

Thanks! :)

https://web2.0calc.com/questions/geometry_2526

We could multiply these numbers together, but we could also do this. 

\(40 \cdot 6 \cdot75\cdot 12 = \\ 40 \cdot6 \cdot[3 \cdot25] \cdot[ 3\cdot4 ] = \\ 40 \cdot 6 \cdot [ 3 \cdot 3 ] \cdot [ 25 \cdot4] = \\ 6 \cdot 40 \cdot 9 \cdot 100 = \\ 54 \cdot 4000 = \\ 54 \cdot 4 \cdot1000 = \\ 216 \cdot10^3\)

Thus, there are 3 terminal zeroes. 

Thanks! :)