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Draw a line segment connecting the points \((0,0)\) and \((3,-1)\). This segment is perpendicular to the line \(y = 3x-10\).

The distance between the two points is \(\sqrt {10}\).

Note that the point \((3,-1)\) bisects chord AB. 

We know have a right triangle with legs \(\sqrt{10}\), 9, and a hypotenuse r. 

Using the Pythagorean Theorem, we find that \(r = \sqrt{91}\)

The formula for the area of a circle is \(\pi r^2\). 

You can take it from here, I suppose?

Rearrange as: \(x^2-x+y^2-y=4\)

Completing the square, we have: \((x-0.5)^2+(y-0.5)^2=4.5\)

The largest value of x is the center x co-ordinate + the radius

The center of the circle is 0.5 and the radius is \(\sqrt{4.5}\)

Can you take it from here?

Simplify to: \(-9x-7y+xy=118\)

Factor out the y: \(y(-7+x)-9x=118\)

Add 63 to both sides so we can apply Simon's favorite factoring trick: \(y(-7+x)-9x+63=118\)

Furthur factor: \((y-9)(-7+x)=181\)

181 is prime, so the only factors are 1 and 181

There are 2 cases for x and y for which the equation is satisfied: \((8,190)\) and \((10, 188)\)

Coincidentally, both sum to \(\color{brown}\boxed{198}\)

Hi, thanks for your response; however, this isn't supposed to be done using recursion. Would you care to show your method/explain it further?

The area of the triangle is:  A  =  ½ · (base) · (height)  =  ½ · ( 7 ) · ( 3 )   =  21/2

The equation of line(BA) is:  y  =  (3/10)(x + 6).

The line  x = k  is a vertical line that passes through the x-intercept  (k, 0).

It also passes through the point  ( k, (3/10)(k + 6) )  on the line(BA).

The region of the triangle to the left of the line  x = k  is a triangle with base length = k + 6

and whose height is  (3/10)(k  + 6).

The area of this triangle is:  ½ · (k + 6) · (3/10)(k + 6)

Since this area = one-half of the total area of the triangle:

     ½ · (k + 6) · (3/10)(k + 6)  =  ½ · (21/2)

                        (3/10)(k + 6)²  =  21/2

                                  (k + 6)²  =  35

                                    k + 6  =  sqrt(35)                       (you can ignore the other answer)

                                          k  =  sqrt(35) - 6                  (a point slightly to the left of the axis)

I answered the same question here: https://web2.0calc.com/questions/coordinates_83752#r2

Expand and convert as: \(3x^2-7x-17=0\)

Using the quadratic formula, we have: \( \large {7 \pm \sqrt{7^2-4 \times -17 \times 3} \over 6}\)

This simplifies to: \(\large {{7 \pm \sqrt{253} } \over 6 }\)

Thus, the answer is: \(253 + 7 +6 = \color{brown}\boxed{266}\)

The pink region is a square with a side length of \(\sqrt 5 \). 

This means that the area of the square is \(\color{brown}\boxed{5}\)

a=262,144

b=15,625

LCM[15,625, 262,144)=11,790,000

11790000 = 2^4 * 3^2 * 5^4 * 131

Number of ordered pairs=[ (4*2+1) + (2*2+1) + (4*2+1) + (2*1+1)]==9 * 5 * 9 *3==1,215

 4*10^0

Next time Elijah, click on the link icon so you can create a hyperlink so people can go directly to the website. 

Okay, so let's start off with an improper fraction like 35/34 ! We'd concert this fraction into a mixed number. How, you ask ? We'd first put a whole number :

1

and then we'd do 35-34, which equals 1 ! So, we'd do this :

1 1/34

well, it's blocked on my school network...

https://www.mathway.com/Calculus

yes you can recommend it to me

\(0.95\cdot9.57\cdot1.08=?\)

How would I figure out the 

Volume of the patio in cubic feet ?

Volume of the patio in cubic yards ?

set is (-infinity, 3)U(5, inifinity)

The angle is 30 degrees.

O_1 O_2 = 12*sqrt(3).

\(\large{36 \times 24 \times {3 \over 4} \times 55 = \text{____}}\)

You can rearrange this to form a rectangle with side lengths 13 and 14. The perimeter is thus \(13+13+14+14 = \color{brown}\boxed{57}\)

Note that \(x^2+y^2=(x+y)^2-2xy\)

Substituting what we know, we have: \(x^2+y^2=9^2-2 \times 10\)

Solving this, we find: \(x^2+y^2=\color{brown}\boxed{61}\)