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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! :)

Identify Right Triangles: Since ∠ABC=45∘, triangle ABC is a 45-45-90 triangle. This means ∠BAC=45∘ and ∠ACB=90∘. Additionally, since ∠CAD=30∘ and ∠ACB=90∘, triangle ACD is a 30-60-90 triangle.

Find AC in Triangle ACD: Because triangle ACD is a 30-60-90 triangle, we know that the ratio of side lengths is constant (i.e., AC/AD = 3​). We are given that AD = 2, so:

AC = 3​⋅AD=3​⋅2=23​

Find AB in Triangle ABC: Since triangle ABC is a 45-45-90 triangle, we know that the ratio of side lengths is also constant (i.e., AB/AC = 1). We found AC in the previous step, so:

AB = AC = 2√3

Find BD: Now that we know AB, we can find BD by subtracting AD from AB:

BD = AB - AD = 2√3 - 2

Therefore, the length of BD is 2*sqrt(3) - 2.

Right Triangle Identification: We are given that D is the foot of the altitude from A to BC. This implies that ∠ADC=90∘. Therefore, triangle ADC is a right triangle.

Soh Cah Toa: We are asked to find sin B, which is the opposite side (AC) divided by the hypotenuse (in this case, we don't know the length of BC, but we can find it).

Pythagorean Theorem in Right Triangle ADC: We know AC = 10 and CD = 6. Since this is a right triangle, we can use the Pythagorean Theorem to find the length of the missing side (AD):

AD^2 + CD^2 = AC^2

Substitute the known values: AD^2 + 6^2 = 10^2

Solve for AD: AD^2 = 100 - 36 = 64 AD = 64​=8

Pythagorean Theorem to Find Hypotenuse (BC): Now that we know the length of a leg (AD) in right triangle ABC, we can find the length of the hypotenuse (BC) using the Pythagorean Theorem again:

BC^2 = AB^2 + AC^2 (We don't know the length of AB, but we can find BC)

Substitute the known values: BC^2 = 8^2 + 10^2

Solve for BC: BC^2 = 64 + 100 = 164 BC = 164​ (We can leave it in this form for now)

Finding sin B: Now that we know both AC and BC, we can find sin B:

sin B = AC / BC

Substitute the known values: sin B = 10 / sqrt(164​)

Since we cannot simplify 164​ further, the answer is sin B = 10/sqrt(164).

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! :)

I'm not great at direct addition, so let's convert everything to base 10. We have

\((221)_3 = (2 × 3^2) + (2 × 3^1) + (1 × 3^0) = (25)_{10} \)

\((122)_3 = (1 × 3^2) + (2 × 3^1) + (2 × 3^0) = (17)_{10} \)

Adding these two numbers up, we get \(25+17=42\)

So now we want to convert 42 back to base 3. 

First, we do 42/3, which has a quotient of 14 and remainder of 0.

Next, we take 14 and have 14/3, which has a quotient of 4 and remainder of 2. 

Next, we do 4/3, which has a quotient of 1 and remainder 1. 

Lastly, we do 1/3, which has quotient 0 and remainder 1. 

So our number is \(42_{10}=1120_3\)

Our final answer is \(1120_3\)

Thanks! :)

We know that \(f(3)=4\), this will be very important. 

Now, in the second equation, notice that we have \(f(2x/3)\). The only value of f(x) we know is when the x is 3. 

So, we want \(2x/3=3\). 

Solving for this, we get \(x=9/2\)

So, now we have 

\(y=(5 + f(3))/11\\ y=(5 + 4)/11\\ y=9/11\)

This means that \((9/2, 9/11)\) must be on the graph.

So our answer is \((9/2, 9/11)\)

Thanks! :)

I think the problem is meant to be \(0.1101_2\)

To solve this problem, let's note something really important. We have

\(\\0.1_2=\frac{1}{2}\\\\ 0.01_2=\frac{1}{2^2}=\frac{1}{4}\\\\ 0.0001_2=\frac{1}{2^4}=\frac{1}{16}\\\\\)

This is important because notice that \(0.1101 = 0.1+0.01+0.0001\)

So, now we just have \(\frac{1}{2}+\frac{1}{4}+\frac{1}{16}=\frac{8}{16}+\frac{4}{16}+\frac{1}{16} = 13/16\)

So 13/16 is our answer. 

Thanks! :)

First, let's note something really important for the problem. 

We have \(P_1 P_2 P_3 \dotsb P_{10}\), which actually just forms a regular decagon!

\(P_1 P_2 + P_2 P_3 + P_3 P_4 + \dots + P_9 P_{10} + P_{10} P_1\)is just the perimeter of that decagon. 

The length of a decagon can be found with formula \(\frac{\text{radius}}{2( -1 + \sqrt {5})}\)

Thus, the perimeter is \(10 (1/2) ( -1 + \sqrt{ 5}) = 5 ( -1 + \sqrt{ 5}) ≈ 6.18\)

So our answer is approximately 6.18

Thanks! :)

We can use Simon's Favourite Factoring Trick to solve this problem.

\(3x + 2y + xy = 18 + 7x + 5y\)

\(xy-4x-3y-18=0\)

\(xy-4x-3y+12=30\)

\(x(y-4)-3(y-4)=30\)

\((x-3)(y-4)=30\)

The pairs of factors of 30 is (1, 30), (2, 15), (3, 10), (5, 6), (6, 5), (10, 3), (15, 2), and (30, 1). We choose x and y such that the values in the parentheses is equal to one of the two factors of 30. This makes all values of (x, y) equal to (4, 34), (5, 19), (6, 14), (8, 10), (9, 9) (13, 7), (18, 6), or (33, 5). The sum of x and y for each pair respectively is 38, 24, 20, 18, 18, 20, 24, and 38. Therefore, all possible sums of x and y are 18, 20, 24, and 38.

You could've just used desmos for that since there are only 2 Points where those to graphs meet.

(-4.193/-38.541), (1.193/4.541)

https://web2.0calc.com/questions/algebra_56153

Simplify as

x + y + 3z = 10

7x + 9y + z  = 18

kx + 14y +7z = 8

If the system has no solutions, the determinant =  0

So

1  1   3   1   1

7  9   1   7   9

k  14  7  k   14

[ 1*9*7 + 1*1 *k + 3*7 *14 ] - [ 9*3*k + 14 * 1 * 1 + 7*7*1 ]  = 0

[ 63 + k + 294] - [ 27k + 14 + 49 ] =  0

294 - 26k   = 0

k = 294/26   =147 / 13

It is impossible to eliminate both the \(x^5\) term and the \(x^6\) term at once, so we set \(b = 0 \rightarrow f(x) + b \cdot g(x) = f(x) = 2x^5 - 6x^4 - 4x^3 + 12x^2 + 7x + 5\). \(\text{deg}(f(x) + b \cdot g(x) = \boxed{5})\)

Here are the answers:
\[\mathbf{A}^{-1} \mathbf{w} = \begin{pmatrix} 2 \\ 3 \end{pmatrix}\]

\[\mathbf{A}^{-1} \mathbf{x} = \begin{pmatrix} 4 \\ 0 \end{pmatrix}\]

\[\mathbf{A}^{-1} \mathbf{y} = \begin{pmatrix} -1 \\ 2 \end{pmatrix}\]

\[\mathbf{A}^{-1} \mathbf{z} = \begin{pmatrix} 5 \\ 1 \end{pmatrix}\]

AB + AC + BC = 28

Draw AE and BF  perpemdicular to BC

Let DE  = x

Let CF = [ 104 - ( 78 + x ) ] = 26 - x 

We have rwo right triangles ADE and BCF

Let the height of the  trapezoid  = h

So

h = sqrt [  10^2 - x^2]

h = sqrt [ 24^2 - (26  -x)^2 ]

So

sqrt [ 10^2 -x^2]  = sqrt [ 24^2 - (26 -x)^2 ]      square both sides

10^2-x^2  = 24^2 - (26 -x)^2

100 - x^2  = 576 - x^2 + 52x - 676  simplify

52x - 200 = 0

13x - 50 = 0

x = 50/13

h = sqrt [ 10^2 - (50/13)^2]  = 120/13

Area of  trapezoid   =  (1/2) (120/13) (78 + 104)   =  840

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https://web2.0calc.com/questions/polynomial_8831

                      C

         E                      D

                     H

A  20          144                       B

ABH forms a triangle

Angle ABH = 180° - 144° - 20°  = 16° 

There are 2 ways to do the problem. 

First, we could take a look at all the possible conbinations we could have to give us x^2. 

First off, we have 8x*x = 8x^2. 

Next, we have 7x*x=x^2

Lastly, we have x^2*1 = x^2

Adding all the coefficients up, we get \(8+7+1=16\)

We could also just factor it, which isn't reccomended, but we get

\(x^{7}+4x^{6}+4x^{5}+19x^{4}+19x^{3}+16x^{2}+16x+1\)

Which shows the coefficient of x^2 is 16. 

So our final answer is 16

Thanks! :)

We could probably do this really hard way. 

By distributing every number, we get

\({x^{3}\left(x^{4}+3x^{3}+4x^{2}+8x+7\right)+x^{2}\left(x^{4}+3x^{3}+4x^{2}+8x+7\right)+x\left(x^{4}+3x^{3}+4x^{2}+8x+7\right)+1\left(x^{4}+3x^{3}+4x^{2}+8x+7\right)}\)

Simplfying, we have

\(x^{7}+4x^{6}+8x^{5}+16x^{4}+22x^{3}+19x^{2}+15x+7\)

As it clear shows\(15\) is the coefficient of x. 

So our final answer is 15. 

Thanks! :)

Notice that \(x^2 + y^2 = 4x + 5y\) is very similar to the equation of a circle. 

We have \(x^2-4x+y^2-5y=0\). 

Now, completing the square for both x and y, we get

\((x-2)^2+(y-5/2)^2=25/4+16\)

This is the official equation of a circle. 

It has the center at \((2, 5/2)\) and a radius of \(\sqrt{89/4}\)

The largest value of x is basically the x value of the center added to the radius. 

Thus, we have \(2+\frac{\sqrt{89}}{2} = \frac{4+\sqrt{89}}{2}\)

So our final answer is \(\frac{4+\sqrt{89}}{2}\)

Thanks! :)