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Arc XY is 142 degrees.

"Parametrize the circle of radius 9 centered at (0,4) in the clockwise direction starting at (0,−5)."

The dot, square and diamond illustrate that the circle starts at (0, -5) and proceeds clockwise.

She spent 1-1/4-1/2 = 1/4 of her time reading a book. She also spent an hour reading a book. So that means if we let x equal the total time, 1/4x=1, so x=4 hours.

"At westwood Middleschool, 648 students responded to a survey asking about their favorite type of exercise. Twice as many students chose jogging as swimming. The same number of students chose yoga as jogging. Twice as many students chose biking as jogging. How many students chose each type of exercise?"

Let j = number of joggers, s = number of swimmers, y = number os yogis, b = number of bikers

We are told:

j = 2s

y = j

b = 2j

j + s + y + b = 648

Use the first three equations in the fourth:

j + j/2 + j + 2j = 648  or  (9/2)j = 648  hence j = 2*648/9  = 144

Can you take it from here?

If \(x\) and \( y\) are positive integers for which \(3x + 2y + xy = 115\), then what is \(x + y\)?

\(\begin{array}{|lrcll|} \hline & 3x + 2y + xy &=& 115 \quad | \quad + 3*2 \\ & 3x + 2y + xy + 3*2 &=& 115 + 3*2 \\ \text{First solution}& \mathbf{(x+2)(y+3)} &=& \mathbf{121} \\ \text{Second solution}& \mathbf{[-(x+2)][-(y+3)]} &=& \mathbf{121} \\ \hline \end{array}\)

\(\text{The divisors of $\mathbf{121}: 1 |\ 11 |\ 121$ }\\ \begin{array}{|rclcl|} \hline 1&*&121 &=& 121 \\ 11&*&11 &=& 121 \\ 121&*&1 &=& 121 \\ \hline \end{array} \)

\(\text{First solution}\quad \mathbf{(x+2)(y+3)} = \mathbf{121} \)

\(\begin{array}{|r|r|r|r|r|} \hline x+2 & y+3 & x & y & x+y \\ \hline 1 & 121 & -1 & 118 & 117 \\ 11 & 11 & \color{red}9 & \color{red}8 & \color{red}17 & \checkmark \text{ $x$ and $y$ are positive integers} \\ 121 & 1 & 119 & -2 & 117 \\ \hline \end{array} \)

\(\text{Second solution}\quad \mathbf{[-(x+2)][-(y+3)]} = \mathbf{121}\)

\(\begin{array}{|r|r|r|r|r|} \hline -(x+2) & -(y+3) & x & y & x+y \\ \hline 1 & 121 & -3 & -124 & -127 \\ 11 & 11 & -13 & -14 & -27 \\ 121 & 1 & -123 & -4 & -127 \\ \hline \end{array}\)

I am taking this straight from EPs answer.  I have not looked if EP's answer is correct

\(y = \frac{2x^2}{ 3} - 4x + 4 \\ y -4= \frac{2x^2}{ 3} - 4x \\ \frac{3}{2}(y -4)= \frac{3}{2}( \frac{2x^2}{ 3} - 4x) \\ \frac{3}{2}(y -4)= x^2 - 6x \\ \frac{3}{2}(y -4)+3^2= x^2 - 6x +3^2 \\ \frac{3}{2}(y -4)+9= x^2 - 6x +9 \\ \frac{3}{2}(y -4)+9= (x-3)^2 \\ \frac{2}{3}(\frac{3}{2}(y -4)+9)= \frac{2}{3} (x-3)^2 \\ y -4+6= \frac{2}{3} (x-3)^2 \\ y+2= \frac{2}{3} (x-3)^2 \\ \frac{3}{2}(y+2)= (x-3)^2 \\ (x-3)^2=4*\frac{3}{8}(y+2)\)

This is a concave up parabola with vertex = (3,-2)   and focal length a=3/8

I have not checked my answer.

The directrix of a parabola is x = 4. Its focus is (2,6)
What is the standard form?
Answer in the form x-h=a(y-k)^2

\(\text{Let a general point on the parabola $(x,y)$ }\)

1. The distance between \((x,y)\) and focus \((2,6) \) is \(\sqrt{\left(x-2 \right)^2+ \left(y-6 \right)^2}\)

2. The distance between \((x,y)\) and the directrix \((x=4,y)\) is \(\sqrt{\left(x-4 \right)^2+ \left(y-y \right)^2} =x-4\)

On the parabola, these distances are equal:

\(\begin{array}{rcll} x-4 &=& \sqrt{\left(x-2 \right)^2+ \left(y-6 \right)^2} \\ \left(x-4 \right)^2 &=& \left(x-2 \right)^2+ \left(y-6 \right)^2 \\ x^2-8x+16 &=& x^2-4x+4 + \left(y-6 \right)^2 \\ -8x+16 &=& -4x+4 + \left(y-6 \right)^2 \\ -4x+12 &=& \left(y-6 \right)^2 \quad | \quad :(-4) \\ x+ \dfrac{12}{-4} &=& \dfrac{ \left(y-6 \right)^2}{-4} \\ \mathbf{ x -3 } &=& \mathbf{ \left(-\dfrac{1}{4}\right) \left(y-6 \right)^2 } \\ \end{array}\)

How far apart were the two artifacts?

Hi, how can that be changed to \((y-k)^2=x-h\)

The general form of an ellipse is 3x^2+4y^2+18x−8y−5=0 .

What is the standard form?

Find constants A and B such that
\(\dfrac{x + 7}{x^2 - x - 2} = \dfrac{A}{x - 2} + \dfrac{B}{x + 1}\)

for all \(x\) such that \(x\neq -1\) and \(x\neq 2\).
Give your answer as the ordered pair \((A,B)\).

\(\text{Let $x^2 - x - 2=(x-2)(x+1)$ }\)

\(\begin{array}{|lrcll|} \hline &\mathbf{ \dfrac{x + 7}{x^2 - x - 2} } &=& \mathbf{ \dfrac{A}{x - 2} + \dfrac{B}{x + 1} } \quad | \quad x^2 - x - 2=(x-2)(x+1) \\\\ &\dfrac{x + 7}{(x-2)(x+1)} &=& \dfrac{A}{x - 2} + \dfrac{B}{x + 1} \quad | \quad \times (x-2)(x+1) \\\\ &\dfrac{(x + 7)(x-2)(x+1)}{(x-2)(x+1)} &=& \dfrac{A(x-2)(x+1)}{(x-2)} + \dfrac{B(x-2)(x+1)}{(x+1)} \\\\ & x + 7 &=& A(x+1) + B(x-2) \\\\ \mathbf{x=-1:}& -1 + 7 &=& A(-1+1) + B(-1-2) \\ & 6 &=& A\times 0 -3B \\ & 3B &=& -6 \\ & B &=& -\dfrac{6}{3} \\ & \mathbf{B} &=& -2 \\\\ \mathbf{x=2:}& 2 + 7 &=& A(2+1) + B(2-2) \\ & 9 &=& 3A +B\times 0\\ & 3A &=& 9 \\ & A &=& \dfrac{9}{3} \\ & \mathbf{A} &=& 3 \\ \hline \end{array}\)

\(\mathbf{\boxed{(A,\ B) = (3,\ -2)}}\)

check:

\(\begin{array}{|rcll|} \hline \dfrac{x + 7}{(x-2)(x+1)} &=& \dfrac{A}{x - 2} + \dfrac{B}{x + 1} \quad | \quad A=3,\ B=-2 \\\\ \dfrac{x + 7}{(x-2)(x+1)} &=& \dfrac{3}{x - 2} - \dfrac{2}{x + 1} \\\\ \dfrac{x + 7}{(x-2)(x+1)} &=& \dfrac{3(x+1)-2(x-2)}{(x-2)(x+1)} \\\\ \dfrac{x + 7}{(x-2)(x+1)} &=& \dfrac{3x+3-2x+4}{(x-2)(x+1)} \\\\ \dfrac{x + 7}{(x-2)(x+1)} &=& \dfrac{x+7}{(x-2)(x+1)} \ \checkmark \\ \hline \end{array}\)

x = 3 and y = 4.

1 in/3ft  = x /12ft

1/3  x 12 = x =  4 inches

Help me

That did not work I think it is wrong

y = -x^2 + 5x      max point will be   - b/2a

                                -5/(-2) = 2.5    this is the value of 'x' at the high point (y)

y high point =  - (2.5)^2 +5(2.5) = - 6.25 + 12.5 = 6.25 ft   = 6 ft 3 inches         DUCK ! 

Standard form    y = ax^2 + bx + c

 2x^2−12x−3y+12=0    re-arrange so y term is on one side

3y = 2x^2-12x+12        divide through by 3

y =   2/3 x^2 - 4x + 4          done.

Three angles of a triangle add to 180 degrees

middle angle = m

smallest  = m-10

largest = 2m

add 'em     m-10   +  m  + 2m = 180

                4m -10 = 180

                  4m = 190

                    m = 47.5 degrees

                    smallest = m-10 =   37.5 degrees

                    largest = 2m = 95 degrees

Use something like this

Tan 37 = opp/adj  =   1250/x    where x = horiz distance

.7535 = 1250/x      

x=1658.8 ft

A = x votes

B = 2x votes

C = x-15000 votes     added together = 109000

x   + 2x   +  x - 15000 = 109000

4x - 15000 = 109 000

4x = 124000

x = 31 000        B = 2x = 62000 votes

57 : 100% = X : 95%        X = 54.15

Discounted price will be  $ 54.15  +tax  

Let  t = amount of time in the park.

So,  1/4 of her time  =  ¼t

       1/2 of her time  =  ½t

Equation:  ¼t  +  ½t  +  1  =  t

                       (3/4)t  +  1  =  t

Subtract  (3/4)t  from both sides:

                                      1  =  ¼t 

Multiply both sides by 4:

                                      4  =  t

The total amount of time was 4 hours.

  Interesting..... I was going to graph them on desmos....just had not done so yet!  

The probability that Alice wins the first time is  1/6.

If Alice doesn't win and Bob doesn't win and then Alice wins is  [(5/6)(5/6)](1/6)  =  (25/36)(1/6)

If Alice wins on the next round:  [(5/6)(5/6)][(5/6)(5/6)](1/6)  or (25/36)²(1/6)

If Alice wins on the next round:  [(5/6)(5/6)][(5/6)(5/6)][(5/6)(5/6)](1/6)  =  (25/36)³(1/6)

etc.

We have:                                                                  (1/6) + (25/36)¹(1/6)+ (25/36)²(1/6)  + (25/36)³(1/6) + ...

If we replace (25/36) with x, we have:                      (1/6) + (1/6)x + (1/6)x² + (1/6)x³ + ...

Factoring:                                                                  (1/6)[ 1 + x + x² + x³ + ... ]

But, the sum of 1 + x + x² + x³ + ...  is  1 / (1 - x),

So we have                                                                (1/6)[ 1 / (1 - x) ]

Since  x = 25/36                                                         (1/6)( 1 / (1 - 25/36)

                                                                                   (1/6)( 1 / (11/36) )

                                                                                   (1/6)(36/11)

                                                    =  6/11

From 3 pounds to 9 pounds    is a gain of 6 pounds

6 pounds/ 1 pound/month = 6 months