NotThatSmart

For this problem, we have to go throug every function and find when f(x) = x. 

First off, we have $f(x)=2x+8$. 

We have to solve $2x+8=x$, in which we get $x=-8$. However, -8 isn't in the range $1 \le x \le 2$, so it isn't a valid solution. 

Next, we have $f(x) = 13 - 5x$

Solving for $13-5x=x$, we get $x=13/6$. Indeed, 13/6 is in the range $2 < x \le 3$, so it is a valid solution. 

After, we have $f(x) = 20 - 14x$

Solving for $20-14x=x$, we get $x=4/3$, which is not a valid solution since it's not in the range $3 < x \le 4$. 

Lastly, we have $f(x)= 40 + 5x$. 

We have $40+5x=x$, in which we get $x=-10$, which is not a valid solution. 

So, we only have x = 13/6

Thanks! :)

We basically just have to go thorugh every function and find when f(x) = x. 

First, we have f(x) = 2x + 8. 

We basically have $2x+8=x$. Solving for this, we get $x=-8$. Unfortunately, that is not in the range $1 \le x \le 2$, so that wouldn't work. 

Second, we have f(x) = 13 - 5x. 

We have $13-5x=x$. Solving this, we get $x=13/6$. This is in the range $2 < x \le 3$, so 13/6 is a valid solution

Third, we have f(x) = 20 - 14x. 

We have $20-14x=x$. Solving this, we get $x=4/3$. This is not in the range of $3 < x \le 4$, so it is not a valid solution. 

Lastly, we have f(x) = 40 + 5x. 

We have $40+5x=x$. Solving this, we get $x = -10$. This is not in the range $4 < x \le 5$, so it is not a valid solution. 

So finally, we have $x=13/6$. 

Thanks! :)

First, let's note somethings. 

Notice that right triangles ABC, ABE, ABD, BHD, BEC are 30-60-90 triangles.

Right triangles ABe and ADC are 45-45-90 triangles. 

Now, let's draw the altitude for triangle ABC by connecting point C to line segment AB.

This line intersects point H on the way. 

We know that BAC is is 60 degrees, so that leaves HCA would be 30 degrees. 

So 30 degrees is our answer!

Thanks! :)

We could use casework for this problem. 

First, if x, y, and z are all postive, we have 

$\frac{x}{|x|}+\frac{y}{|y|}+\frac{z}{|z|}+\frac{xyz}{|xyz|}\,=\,\frac{x}{x}+\frac{y}{y}+\frac{z}{z}+\frac{xyz}{xyz}\,=\,1+1+1+1\,=\,4$

Now, if we have only x is negative, is 0, we have

$\frac{x}{|x|}+\frac{y}{|y|}+\frac{z}{|z|}+\frac{xyz}{|xyz|}\,=\,\frac{-|x|}{|x|}+\frac{y}{y}+\frac{z}{z}+\frac{-|x|yz}{|xyz|}\,=\,-1+1+1-1\,=\,0$

In the same way, having only y and z being negative will still be 0.

If only z is postive, we have

$\frac{x}{|x|}+\frac{y}{|y|}+\frac{z}{|z|}+\frac{xyz}{|xyz|}\,=\,\frac{-|x|}{|x|}+\frac{-|y|}{|y|}+\frac{z}{z}+\frac{(-|x|)(-|y|)z}{|xyz|}\,=\,-1-1+1+1\,=\,0$

In  the same way, if only x and y are positive, the result will also be 0.

If all three x, y, and z are negative, we have

$\frac{x}{|x|}+\frac{y}{|y|}+\frac{z}{|z|}+\frac{xyz}{|xyz|}\,=\,\frac{-|x|}{|x|}+\frac{-|y|}{|y|}+\frac{-|z|}{|z|}+\frac{(-|x|)(-|y|)(-|z|)}{|xyz|}\,=\,-1-1-1-1\,=\,-4$

So, the three possible values are 0, -4, and 4. 

Thanks! :) 

First off, let's let $DC = 3BD$

We can write two equations off the problem

$AD = \sqrt{AC^2 - (3BD)^2 } = \sqrt{15^2 - 9BD^2 } \\ AD = \sqrt{AB^2 - BD^2} = \sqrt{10^2 - BD^2 }$

Notice that we have AD on both equations, so we can equate the two to get

$\sqrt{15^2 - 9BD^2} = \sqrt {10^2 -BD^2 }$

Now, we just have to solve for BD^2. 

$15^2 - 9BD^2 = 10^2 - BD^2 \\ 225 - 9BD^2 = 100 - BD^2 \\ 125 = 8BD^2 \\ 125/8 = BD^2$

We know that 

$AD =\sqrt{10^2 - BD^2}$ so we can just plug in BD^2 to find AD. 

$AD = \sqrt{100 - 125/8} = \sqrt { 675/8} = \frac{15\sqrt3 }{2 \sqrt2}$

This rounds to about 9.19. 

Thanks! :)

From the problem, we can write the system

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

 where t is toast, b is bagel and m is muffin. 

From the first two equations, we have

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

Subbing this in to the third equation, we have 

$[ 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 about 262 cents. 

Thanks! :)

Not too bad of a problem. 

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

Thanks! :)

We knowt hat AK is the altitude, so we have

$AK = \sqrt{AC^2 - CK^2} = \sqrt{8^2 - 3^2} = \sqrt{55}$ from the pythagorean theorem. 

We can now find AB, with

$AB = \sqrt{AK^2 + BK^2} = \sqrt{ 55 + 4 } = \sqrt { 59 }$

so AB is $\sqrt{59}$

Thanks! :)

Well, unfortunately, this triangle can't ever exist. 

Using the Law of Cosines, we have

$AC^2 = CM^2 + AM^2 - 2 (AM * MC)cos (AMC) \\ AB^2 = BM^2 + AM^2 - 2 (AM * BM) (-cos (AMC) \\ 17^2 = x^2 + 12^2 - 2 (AM * x) cos (AMC) \\ 7^2 = x^2 + 12^2 + 2(AM * x) cos (AMC)$

Adding these two equations together, we get 

$338 = 2x^2 + 288 \\ 50 = 2x^2 \\ 25 = x^2 \\ 5 = x \\$

Alright. 

Now, we have $BM + CM = 5 + 5 = 10 = BC$. 

However, according to the Triangle Inequality Theorem, we have

$AB+BC>AC \\ AB +BC > 17$

However, we have

$AB + BC = 7 + 10 = 17$

Meaning this is an invalid triangle. 

Thanks! :)