NotThatSmart

If angle XWZ = 60°, angle WXY = 180° - 60° = 120° because XY and WZ are parallels.

However, this means that Angle ZXW must be less than 120°.

In the problem though, it clearly defines ZXW as 120, meaning this problem is invalid. 

Thanks! :)

Let's note that If AD  = AB  then $\angle  ABD  =  \angle ADB$

This means that $\angle ABD = (180 - 17) / 2 = 81.5°$

So 81.5 is the answer. 

Thanks! :)

Alright. Let's first put y in terms of x. 

We get $y=25-x$. 

Subbing this value back into the second equation, we get 

. $3x+75$

Since x can't be negative, the smallest possible value of x is 0. 

When x is 0, we have $3(0)+75=75$

So 75 is our final answer. 

Thanks! :)

Might have the wrong question, learnmgcat ;)

Let's make some observations of the problem. 

First, note that triangle ABC is a 45-45-90 right triangle. This means that $AB = BC = 12/\sqrt 2 = 6\sqrt 2$

Also, we have that

$AD = AD\\ BD = BD \\ AB = BC$

This means that triangles  ABD  and CBD  are congruent

Thus, we have $[ ABD ] = (1/2) [ABC] = (1/2) ( 1/2) ( 6\sqrt 2)^2 = (1/4) (72) = 18$

So our aswer is 18. 

Thanks! :)

We can complete this problem in two different ways. 

The first tactic is to essentially compare this root to the quadratic equation of $5x^2+21x+v = 0$

From this quadratic, we can identify that a = 5, b = 21, c = v. 

We have the equation

$\frac{-21-\sqrt{201}}{10} = {-b \pm \sqrt{b^2-4ac} \over 2a}\\ \frac{-21-\sqrt{201}}{10} = {-21 \pm \sqrt{21^2-4(5)(v)} \over 2(5)}\\ \frac{-21-\sqrt{201}}{10} = {-21 - \sqrt{441-20v} \over 10}\\ \sqrt{201} = \sqrt{441-20v}\\ 201 = 441-20v\\ v=12$

This is a bit complicated and takes a lot of computations, but it does give us the correct answer. 

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The second tactic is to use conjugations of square roots. 

This is because the conjugate root theorem states that if a root of a polynomial is a square root $a+\sqrt b$, then its conjugate,  $a-\sqrt b$is also a root

We can apply that to this problem. If  $\frac{-21-\sqrt{201}}{10}$ is a root, then $\frac{-21+\sqrt{201}}{10}$ is also a root. 

The product of the roots is $[ (-21)^2 - 201 ] / 100 = 240/100 = 2.4$

However, in the quadratic, we also have that $v/5$

Thus, we have 

$v/5 = 2.4\\ v = 12$

SO 12 is the final answer. 

Thanks! :)

We can essentially complete this problem with a nice and simply graph. 

Here, I borrowed Melody's. 

In the yellow area, it is where the point would satisfy the conditions given in the problem. 

As shown, we have split the graph into triangles. 

There are 18 congruent triangles here, with 12 of them are in the yellow zone.

Thus, we have a $12/18 = 2/3$ chance. 

Thanks! :)

When Dirk finishes, Edith has run 14km  and Foley has run 9 km

This means that Foley runs 9/14 as fast as Edith

When Edith has run 24km, Foley has run $24 (9/14) ≈ 15.43 km$

So Foley is $24 - 15.43 ≈ 8.57$ km behind!

Thanks! :)

We can define some variables from the information given from the problem. 

First, since we know that AD bisects BAC, let's name variables

Let  BD  = 3 - x

Let  CD =  x

Thus, we can plug the two numbers in to get the equation

$BD / AB = CD / AC \\ (3 - x) / 4 = x / 5 \\ 5 (3 - x) = 4x\\ 15 - 5x = 4x \\ 15 = 9x \\ x = 15/9 = 5/3 = CD$

Finally, plugging in these values for the area, we find that

$[ ADC ] = (1/2) (CD) ( AB) = (1/2) ( 5/3) ( 4) = 10 / 3$

Thus, 10/3 is our answer. 

Thanks! :)