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Find the area of the rectangle.

\(2(a+b)=40\\ a=\frac{40}{2}-b\\ 10^2\cdot \sqrt{2}^2=a^2+b^2\\ 200=400-40b+2b^2\\ b^2-20b+100=0\\ b=10\pm \sqrt{100-100}\\ b=10\\ a=10\\ A=ab=10\cdot 10\\ \color{blue} A=100\)

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AX² + CX²  =  BX² + DX²   

3² + 3²  =  3² + DX²   

DX²  =  3² + 3² – 3²    

DX²  =  9     ====>>          DX = +3    

discard negative value      DX = 3    

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Circumradius of equilateral triangle with side length 'a' is given by 

2 x  h/3   = a * sqrt (3) / 6 

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! :)

You cannot find a specific number due to variables being the sidelengths, but, it would be this if you can calculate the area of said triangle.

R = (a*b*c)/(4*(area of triangle))

The system of equations is:

p - 2q = 3 q - 2r = -2 p + r = 9

We can solve this system using Gaussian Elimination.

First, we can add the first and second equations to get:

p - 2q + q - 2r = 1

This simplifies to:

-q - 2r = 1

Now, we can subtract the third equation from this equation to get:

-q - 2r - (p + r) = 1 - 9

This simplifies to:

-2q = -8

Therefore, q = 4.

Now, we can substitute this value into the second equation to get:

4 - 2r = -2

This simplifies to:

2r = 6

Therefore, r = 3.

Finally, we can substitute these values into the first equation to get:

p - 2(4) = 3

This simplifies to:

p = 5

Therefore, the ordered triple that satisfies the system of equations is (5, 4, 3).

Will has covered 45 minutes worth of lake distance  

       45 min  * 50 m/min = 2250 m meters before Grace starts ....so there is only   2800 - 2250 = 550 meters to cover before they meet 

               their closing rate is (50 + 30 )  = 80 m/min 

                              distance / rate = time 

                                         550 m / 80 m/min = 6 7/8 minutes they will meet after Gracie starts 

                  which would be   2:45 + 6 7/8  = 2:51  and 7/8 = 2:51 and 52.5 seconds  (almost  2:52 pm) 

The answers are 4, 11, 12, and 15.

The answer is A: 38x - 1450 > 76.

The statement that is true is D: x + 12 <= y - 5.

The inequality is 4x + 85 >= 40.

Let's analyze the situation:

George usually arrives at the train station at 5:00 pm to pick up Ava.

Ava arrived early, so she started walking home an hour earlier, which means she started walking at 4:00 pm.

George and Ava met along the route between the train station and their house.

They arrived home at 4:48 pm, which means they drove for 12 minutes after picking up Ava.

Since they drove for 12 minutes after picking up Ava, Ava must have been walking for 4:48 pm - 12 minutes = 4:36 pm. Therefore, Ava had been walking for 4:36 pm - 4:00 pm = 36 minutes before George picked her up.

Let's analyze the situation:

George usually arrives at the train station at 5:00 pm to pick up Ava.

Ava arrived early, so she started walking home an hour earlier, which means she started walking at 4:00 pm.

George and Ava met along the route between the train station and their house.

They arrived home at 4:48 pm, which means they drove for 12 minutes after picking up Ava.

Since they drove for 12 minutes after picking up Ava, Ava must have been walking for 4:48 pm - 12 minutes = 4:36 pm. Therefore, Ava had been walking for 4:36 pm - 4:00 pm = 36 minutes before George picked her up.

Let's analyze the situation:

George usually arrives at the train station at 5:00 pm to pick up Ava.

Ava arrived early, so she started walking home an hour earlier, which means she started walking at 4:00 pm.

George and Ava met along the route between the train station and their house.

They arrived home at 4:48 pm, which means they drove for 12 minutes after picking up Ava.

Since they drove for 12 minutes after picking up Ava, Ava must have been walking for 4:48 pm - 12 minutes = 4:36 pm. Therefore, Ava had been walking for 4:36 pm - 4:00 pm = 36 minutes before George picked her up.