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Two rectangles

56 x 14   +    28 x 28 =                   yd²

The ratio of the area of triangle BXY to the area of triangle ABC is 4/3.

You are right guest. Here is a more detailed ex

1.  We complete the square twice. \begin{align*} x^2 + 6x + y^2 + 6y &= 18 \\ x^2 + 6x + 9+ y^2 + 6y +9 &= 18  +9+9\\ (x +3)^2 + (y + 3)^2 &= 36 \end{align*} This is in the form $(x - h)^2 + (y - k)^2 = r^2$, so the center is $\boxed{(-3, -3)}$.

2.  We are given $x^2 - 10x + y^2 - 10y + 25 = 0$. Factoring, this is $x^2 - 10x + (y - 5)^2 = 0$. We use completing the square and add $25$ to both sides of the equation, giving us $x^2 - 10x +25  + (y - 5)^2 = 25$. Factoring again, this is $(x - 5)^2 + (y-5)^2 = 25$. This equation represents a circle of radius $5$, centered at $(5, 5)$. Thus, this circle will touch the $y$-axis at $\boxed{(0, 5)}$.

2. The blue graph is shifted two units left, so y = f(x - 2).

x^2 + 6x          + y^2 + 6y   = 18       complete the square for x and y

(x+3)^2           + (y+3)^2    = 18 + 9 + 9

center h,k = -3,-3

x^2 -10x        + y^2 -10y     +25 = 0

(x-5)^2         + ( y-5)^2          =   - 25 + 25 + 25       y intercept occurs when x = 0   put in x = 0

      25         + y^2 -10y +25    = 25

y^2 -10y +25 = 0 

(y-5)^2 = 0                          y = 5         when x = 0        0,5

1.  First, subtract $4x$ and $6y$ from both sides to set up completing the square: $$x^2 -4x + y^2 -6y \leq 13.$$We add $((-4)/2)^2 = 4$ to complete the square in $x$ and $((-6)/2)^2 = 9$ to complete the square in $y$: $$x^2 -4x + 4 + y^2 -6y + 9 \leq 13+4+9,$$so $$(x-2)^2 + (y-3)^2 \leq 26.$$The graph of this inequality is a solid circle with center $(2,3)$ and radius $\sqrt{26}$, so the desired area is $\pi\left(\sqrt{26}\right)^2=\boxed{26\pi}$.

2.  First, we factor out the constants of the squared terms to get $9(x^2-2x)+9(y^2+4y)=-44.$ To complete the square, we need to add $\left(\dfrac{2}{2}\right)^2=1$ after the $-2x$ and $\left(\dfrac{4}{2}\right)^2=4$ after the $4y,$ giving$$9(x-1)^2+9(y+2)^2=-44+9+36=1.$$Dividing the equation by $9$ gives$$(x-1)^2+(y+2)^2=\frac{1}{9},$$so the radius is $\sqrt{\frac{1}{9}}=\boxed{\frac{1}{3}}.$

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The output is the y value so we find the point on the graph with a y value of 4.

That point is (3,4).

25,000*(1-0.13)^3=  16,462.58

haha LOL😎😎😎

Should there be any brackets here?

Please make it clear what the numerator and denominators are.

If there are 5 more in chinese than spanish, you'd say  49 are taking chinese only

    but 44 + 49 = 93    there are 4 students missing ...they must be taking both

Thank you can you tell me how you got 50 because thats were i dont understand?

Oh Arthur, Arthur, Arthur.....    You shouldn't tip the taxed amount !

Anyway ...

13.72   x   1.058 = bill amount after tax

(13.72)*(1.058)   x   1.15 = total amount after tax and tip= $             

Thanks, Melody! 

The sum of the interior angles of a triangle is 180^o   

I'll set up the equation for you:   (16x + 4) + (14x + 6) + 50  =  180  

combine like terms                       30x + 60  =  180  

subtract 60 from both sides                 30x  =  120  

divide both sides by 30                            x  =  4  

Check the answer     16x + 4  =  64 + 4  =  68  

                                 14x + 6  =  56 + 6  =  62  

                                                                  50  

 Total ...................................................... 180  

Where did the 50 come from? 

That unlabeled interior angle is the supplementary angle of 130^o

Supplementary angles total 180^o   

_.

Nice answer Textot    

Given parallelogram ABCD, find x, y, and z.

Hello Guest!

\(z-15°+ 2z=180°\\ 3z=165°\)

\(z=55°\)

\(x=90°+(90°-(z-15°))\\ x=90°+(90°-(55°-15°))\)

\(x=140°\)

\(y=90°-(2z-90°)\\ y=90°-(2\cdot 55°-90°) \)

\(y=70°\)

  !

Make this into a number line like this:

The sum of the distances from P to A, B, C, D, and E, respectively, are \(|x| + |x-1| + |x-2| + |x-5| + |x-7|\). 

P will be equal to the x value that will result in the minimum possible distance. 

The global minimum will only occur in critical points, which in this case are 0, 1, 2, 5, and 7.

Testing them out, 2 will result in the smallest possible sum of the distances of 11. \(\)

That means P will be on point C, which means AP will be equal to \(\boxed{2}\)

(by the way, I'm not too good at calculating global minima, so maybe there is a faster way to do it than to just test out critical points)