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k cannot be an integer because this implies that a rational quantity - 45 -   would equal an irrational quantity - k√5

If this is what you meant, we can find "k" thusly :

45 = k √5   divide both sides by √5

45 / √5   = k

However......if you meant √45, then the guest's answer is correct.....

Find the area of the region enclosed by the graph of the equation x^2 + y^2 = 4x + 6y+13.

This will be a circle....we need to find the radius to find the area enclosed by the circle

Let's write this as    x^2 -4x + y^2 -6y - 13  = 0        complete the square on x and y 

x^2 - 4x + 4  + y^2 - 6y + 9  - 13  =  4 + 9

(x -2)^2 + (y -3)^2  - 13  = 13       add 13 to both sides

(x -2) ^2  + (y - 3)^2   = 26        26 = (radius)^2   so

The area of a circle =  pi * (radius) ^2

So....the area =  pi * 26 =   26* pi  ≈  81.68 units^2

The prime factorization of 45 is 5*3*3.

So, √45 can be re-written as √(5*3*3). Beacuse of order of operations, this yeilds exactly the same value as √(3^2) * √(5) .

The square root of 3 squared (√(3^2)) is 3.

Now our expression is 3√(5).

Therfore, k must be equal to 3

I've answered the first one elsewhere today........

Geometrically speaking, a parabola is defined as the set of points that are the same distance from a given point and a given line. The point is called the focus of the parabola and the line is called the directrix of the parabola. Suppose P is a parabola with focus (4,3) and directrix y=1 . The point (8,6) is on P because (8,6)  is 5 units away from both the focus and the directrix. If we write the equation whose graph is P in the form y=ax^2 + bx + c, then what is a*b*c ?

The vertex  will be found at (4,2)   and we can write that :

(y - 2) = (a)(x - 4)^2       and since (8,6)   is on the curve, we can solve for a

(6 -2)  = (a) (8 - 4)^2    simplify

4 = (a)(4)^2

4 = (a)16   →    a = 4/16  = 1/4

Since the x coordinate of the vertex is given by -b/ (2a)  we have that   -b/[2 (1/4)]  = 4   → -b / (1/2) = 4 → -b = 2 → b = -2

And using the fact that  (4,2)  is on the graph, we can find c, thusly :

y = ax^2 + bx + c  ......so......

2 = (1/4)(4)^2 -2(4) + c

2 = 4 - 8 + c

2  = -4 + c

c = 6

Then a*b*c =  (1/4) (-2) (6)    =  (1/4)(-12)    = -3

Here's the graph of the function :  https://www.desmos.com/calculator/1hi34pnvzw

4. The graph of (x-3)^2 + (y-5)^2=16 is reflected over the line y=2. The new graph is the graph of the equation x^2 + Bx + y^2 + Dy + F = 0 for some constants B, D, and F. Find B + D + F.

The first graph  is a circle with its center at (3, 5)   and a radius of 4

If it is reflected across y = 2, the new center lies at (3,-1)   with the same radius

So....the new equation becomes

(x - 3)^2 + (y + 1)^2    = 4        expand this   and simplify

x^2 -6x + 9  + y^2 + 2y + 1   = 4

x^2 -6x + y^2 + 2y + 6   = 0

And if

x^2 + Bx + y^2 + Dy  + F  = 0

Then, equating coefficients.........B = -6 ,  D = 2   and F = 6   and B + D + F =  2

Here's the graph of the original circle and its reflection across y = 2 :   https://www.desmos.com/calculator/d6l7hx2ye5

3. The grid lines in the graph are one unit apart. The red parabola shown is the graph of the equation y=ax^2 + bx + c. Find abc.

It appears that the following 3 points are contained on the graph : (-3, 5), (-1, 3)  and (-5, 3)

So....using the fact that   -b/2a = the x coordinate of the vertex, we have that -b / 2a =  -3   →   -b = -6a  →  b = 6a

So we can set up the following equations to find abc

a(-3)^2 + 6a(-3) + c  =  5    →    9a - 18a + c  =  5  →   -9a + c  = 5    (1)

a(-1)^2 + 6a(-1) +  c = 3   →    -5a +  c  =    3        (2)

Multiply  (1) by -1 and add to (2)  and we have

4a = -2    →   a = -2/4   →   a = -1/2

So b = 6(-1/2)   = -3

And using (1) to find c, we have     -9(-1/2) + c  = 5    →   9/2 + c  = 5   →   c = 10/2  - 9/2 = 1/2

So  abc =  (-1/2)(-3)(1/2)    =  3/4

Here's the graph to show that our function is correct :  https://www.desmos.com/calculator/gtspbs8qjg

a parabola curv have an equation as 3x + 1 = y^2 -y -1 which going through the y-axis in the positive side and also contain the smallest x value on the curv. Use quadratic equation to solve the point where the smallest x value is (coordinate).

The quadratic formula probably isn't necessary here....let's simplify the equation

3x + 1 = y^2 -y -1   subtract 1 from both sides

3x = y^2 - y - 2        divide both sides by 3

x = (1/3)y^2 - (1/3)y   - 2/3         complete the square on y   ....add 2/3 to both sides

x + 2/3 + 1/12  =  (1/3) (y^2  - y + 1/4)       factor the right side, simplify the left

(x + 3/4)  = (1/3) (y -1/2)^2         x takes on its minimum value at the  vertex which is    (-3/4, 1/2).....so x is minimum at -3/4

Here's the graph : https://www.desmos.com/calculator/j3plekvi77

2. The graph of the equation y =ax^2 + bx + c, where a, b, and c are constants, is a parabola with axis of symmetry x = -3. Find b/a.

If the axis of symmetry is x = -3, then the x coordinate of the vertex lies at (-3, m)  and we know that the x coordinate of the vertex can be found  by :   -b/2a  =  -3

Multiplying by 2 on each side, we have that     -b/a  = -6   →    b / a = 6

1 minute is 6 degrees

1. Find the vertex of the graph of the equation y = 3x^2 - 6x + 7.

The x coordinate of the vertex is  given by -b/(2a)  =  6 / (2*3)   =   6/6 =  1

The y coordinate of the vertex is given by substituting the x value of the vertex back into the function = 

3(1)^2 - 6(1) + 7  =    3 - 6 + 7 =    4

So.....the vertex lies at ( 1, 4)

Here's the graph :  https://www.desmos.com/calculator/q83yhluiyr

here is a example \(x^2-3+7-5^3=y-1, y-1=\begin{bmatrix} 12 && x2 \\ 1y && xy \end{bmatrix}\)

Let the cost of the skirt=S

The cost of the shirt=S - 18

S + S - 18 = 94

2S =94 + 18 =112 divide both sides by 2

S =112/2

S =56 rm - the cost of the skirt

56 - 18 = 38 rm - the cost of the shirt.

M-MENTAL

A-ATTACK

T-TO

H-HELATHY

S-STUDENTS

You use the sine function:

Sin(x)=Opposite / Hypotenuse

Melody.....this is a pretty good "primer" concerning polar plots  : 

http://tutorial.math.lamar.edu/Classes/CalcII/PolarCoordinates.aspx

(h • f) (x)    means to take the function "f"   and substitute it into the function "h"

Example :

f(x)   = 2x         h (x) =  3x + 5

So  (h • f) (x)   =     3(2x) + 5   =   6x +  5

Thanks Alan, I have played with Desmos and have managed to do it.

Desmos version:

I've just checked, and Desmos does allow polar plots. If you just use r and theta (use the Greek letter not the word "theta") Desmos automatically assumes you want a polar plot.

You can do polar graphs with wolfram alpha

but

I don't think you can do polar graphs or enter parametric equations in Desmos.   ://

Doesn't Desmos allow polar plots? (Just asking - I haven't checked).

If you calculate x = r*cos(theta) and y = r* sin(theta), you should get values of x and y that plot to give the curves I've shown.

Thanks Alan,

I don't know about polar plots.   

I suppose if I google it I will find a site where I can learn about them......

Got any great ideas?

I've graphed polar plots Melody.  r is the radius and theta is the angle.  This is usually what is wanted when an equation is specified using r and theta.

Log tables are as ancient and almost as difficult to interpret as hieroglyphs.

Just use a calculator :)

The area is given by :

6 * (1/2) 5^2 sin (60)   =

3 * 25 * √3 / 2   =

37.5 * √3   =  about 64.95 units ^2