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To account for the conditional at the beginning, let's just give the two youngest kids four candies off the bat. That leaves you with 5 candies to divide between 5 kids, which is a simple stars and bars problem (if you don't know what it is, look it up it's really cool :) )

This means the answer is 9 choose 4 which is (9!)/(4!)*(5!) = (9 * 8 * 7 * 6) / 24 = 9 * 7 * 2 = 112 ways

\(-2(3x + 6) ≥ 4(x + 7)\)

\(-6x-12\ge4x+28\)

\(10x\le-40\)

\(x\le-4\)

\(8(6x − 7) < 5(9x − 4)\)

\(48x-56<45x-20\)

\(3x<36\)

\(x<12\)

s = (a+b+c)/2       Multiply both sides by '2' 

2s = a* b + c        Subtract  b and c from both sides 

a = 2s - b - c 

15000 drops *  1 minute / 41 drops = ~ 366 minutes

Interior angle + exterior angle = 180 degrees 

  23 e     + e = 180 

    e = 7.5 degrees     so interior angle = 172.5 

Sum of interior angles of n-gon is  ( n-2)*180 

   each angle is then   (n-2)*180 / n    and this equals  172.5  degrees   solve for n 

         (n-2)*180 / n = 172.5 

           180n -360 = 172.5 n 

                n = 48 sides 

Maybe a bit simpler to realize exterior angles sum to 360 

                        n (7.5) = 360 

                             n = 48 sides 

a = v^2 / r        multiply both sides by 'r '

ar = v^2           divide both side by 'a'

r = v^2 /a 

87 kg *  1 pound / .45 kg =  ~ 193 pounds 

\(f(x) = \frac{2x + 5}{x - 4} + \frac{x^2 + 1}{4x - 3}\)

\(y = \frac{2x + 5}{x - 4} + \frac{x^2 + 1}{4x - 3}\)

\(x= \frac{2y + 5}{y - 4} + \frac{y^2 + 1}{4y - 3}\)

\(x(y-4)(4y-3)= (2y + 5)(4y-3) + (y^2 + 1)(y-4)\)

\(4xy^2-19xy+12x= y^3+4y^2+15y-19\)

\(y^2(4x-4)-y(19x+15)+12x= y^3-19\)

Honestly, Idk what to do from here, but you have to solve for y in this equation. My idea would be to pile them into a cubic and try solving for x that way but.... I have doubts.

The length of $\overline{AB}$ is $4,$...Find the length $AB.$

The answer is 4. :)

Analyzing the Problem

Understanding the Geometry:

We have two concentric circles (circles with the same center).

A chord of the larger circle is tangent to the smaller circle. This means the chord is perpendicular to the radius of the smaller circle at the point of tangency.

We need to find the area of the ring-shaped region, which is the difference in areas between the larger and smaller circles.

Key Points:

The chord of the larger circle is the diameter of the smaller circle.

The radius of the smaller circle is half the length of the chord.

Solution

Find the radius of the smaller circle:

Since the chord is the diameter, the radius of the smaller circle is 10/2 = 5 units.

Find the area of the smaller circle:

Area of a circle = πr², where r is the radius.

Area of the smaller circle = π(5)² = 25π square units.

Find the radius of the larger circle:

The radius of the larger circle is the sum of the radius of the smaller circle and the distance from the center to the chord (which is the same as the radius of the smaller circle).

Radius of the larger circle = 5 + 5 = 10 units.

Find the area of the larger circle:

Area of the larger circle = π(10)² = 100π square units.

Find the area of the ring-shaped region:

Area of the ring-shaped region = Area of the larger circle - Area of the smaller circle.

Area of the ring-shaped region = 100π - 25π = 75π square units.

Therefore, the area of the ring-shaped region is 75π square units.

The radius of a circle can be found with the area of the circle.  The area of a sector is equal to n/360 * a where n is the angle of the sector and a is the area of the circle.

Using this formula,

60pi = 83/360 * a

a = 360/83 * 60pi

a ~ 260.24pi

The formula for the area of a circle is A = r^2 pi.

260.24pi = r^2 pi

260.24 = r^2

r ~ 16.13

So the radius of the circle is about 16.13.

A rectangular room measures 12-feet by 6-feet. How many square   

yards of carpet are needed to cover the floor of the room?  

    12 ft       1 yd   

    ––––  x  –––––  =  4 yd    

       1          3 ft    

     6 ft        1 yd   

    ––––  x  –––––  =  2 yd    

       1          3 ft    

       4 yd  x  2 yd  =  8 yd²

_.   

\(a^3 + 3ab^2 = 679 ,3a^3 - ab^2 = 615\)

\(10a^3 = 3\cdot615+679\)

\(10a^3=2524\)

\(a = \sqrt[3]{252.4}\)

Uhh this doesn't look right. Maybe a moderator check? If it is, @Kittykat you can use this to find a-b, just plug in

\(x+y=10 \)

\(x^3 + y^3 = 300 + x^2 + y^2\)

\(x^3+y^3=(x + y)(x^2 − xy + y^2)\)

\(10(x^2-xy+y^2)=300+x^2+y^2\)

\(9x^2-10xy+9y^2=300\)

\(9(x+y)^2-28xy=300\)

\(28xy=600\)

\(xy=\frac{150}{7}\)

\(x+y=10,xy=\frac{150}{7}\)

I think you can solve on your own now. If you are having trouble, leave a note and ill do the rest.

A line and a circle intersect at A and B as shown below. Find the distance between A and B      .
The line is x = 4, and the equation of the circle is x^2 + y^2 = 25.   

To find the point(s) common to both functions, substitute the equal value that you know.     

                                                  x² + y² = 25    

you know that x=4                      4² + y² = 25    

                                                         y² = 25 – 16 = 9     

                                                         y   = +3      

and it's given that x=4    

so the points of intersection are       +4,+3 &  +4,–3   

so, of course, the distance between the points is 6 units    

_.   

Sorry, I had a typo. its A circular table is pushed into a corner of the room, where two walls meet at a right angle. A point P on the edge of the table (as shown below) has a distance of 8 from one wall, and a distance of 9 from the other wall. Find the radius of the table.