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The minimum area is 15 5/6 square feet.

The minimum vaue occurs when a = b = c = d = e = f = 7/6.  Then 1/a + 4/b + 9/c + 16/d + 25/e + 36/f = 78.

AP = 4*sqrt(3).

\(a^2+b^2=c^2\)

\(12^2+3^2=c^2\)

\(144+9=c^2\)

\(153=c^2\)

\(\sqrt{153}=\sqrt{c^2}\)

\(c=12.37\)

the ladder reaches12.37 feet up the side of the house

The height of a parallelogram is 4.5 cm. The base is twice the height. What is the area?

\(A=2cm\times 2\times 2cm=8cm^2\)

  !

Let a and b be the roots of \(y^2+5y-11=0\)

Find \((a+3)(b+3)\)

First

\(a+b=-5\)

\(ab=-11\) 

\((a+3)(b+3)\) Expand this.

\(ab+3a+3b+9\)  Factor 3 from middle terms

\(ab+3(a+b)+9\)

Now substitute the values 

\(-11+3(-5)+9\)

\(-11-15+9\)

\(=-17\)

General note:

Vieta's formula states that: 

for any quadratic equation (Even for any polynomial ) 

\(ax^2+bx+c=0\)

With roots: \((x_1,x_2)\)

\(x_1+x_2=-\frac{b}{a}\)

\(x_1x_2=\frac{c}{a}\)

Does it work for cubic equation?

\(ax^3+bx^2+cx+d=0\)

with roots: \((y_1,y_2,y_3)\)

\(y_1+y_2+y_3=-\frac{b}{a}\)

\(y_1y_2y_3=-\frac{d}{a}\)

and

\(y_1y_2*y_2y_3*y_1y_3=\frac{c}{a}\)

I think if you are using a mac, then there is an option key. :)

What is  'option' ?  Is that on a phone?

In Latex I just use       ^\circ

\(18^\circ\)

Yeah... I was just searched up how to do it in LaTeX. Does anyone know?

The answer is 11.62

Whatever works for you is good.

I am sure some people will take note of what you have suggested and learn from you.  

I actually have only heard of the 1st one and I have never used any of them.

If I could remember them they would be very useful.

if 40% is 80 pi, then 100% (or the whole lap) is 80pi*100/40=200pi; because the circumference is pi * diameter, the diameter is 200pi/pi=200 meters

Any tips on #1

I think that would be your opinion! :)

A garden has dimensions 2x+3 ft by x+6 ft. If the garden is 95 ft^2, then what are the dimensions. Solve by factoring.

Hello Guest!

\((2x+3)\cdot (x+6)=95\\ 2x^2+12x+3x+18-95=0\\ 2x^2+15x-77=0\)

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

\(x = {-15 \pm \sqrt{15^2+4\cdot 2\cdot 77} \over 2\cdot 2}\\ x=\frac{-15\pm 29}{4}\)

\(x_1=3.5\\ x_2\ not\ applicable\)

The dimensiones are \(10ft\times 9.5ft\)

  !

Is this two different questions?

1)  What times what  =  2  but when added  =  -4:

      What times what  =  2   --->   x · y  =  2

      When added  =  -4        --->   x + y  =  -4

      Solving  x · y  =  2  for  y:  y  =  2/x

      Sustituting this into the other equation:  x + y  =  -4   --->   x + 2/x  =  -4

                 multiplying both sides of this equation by x:             x² + 2  =  -4x

                 adding 4x to both sides:                                    x² + 4x + 2  =  0    

     Using the quadratic formula when  a = 1, b = 4, and c = 2:   x  =  [ -4 +/- sqrt( 4² - 4·1·2 ) ] / ( 2·1)

                                                                                                    x  =  [ -4 +/- sqrt(8) ] / 2

                                                                                                    x  =  [ -4 +/- 2sqrt(2) ] / 2

     One answer:  x = -2 + sqrt(2)

     Other answer:  x = -2 - sqrt(2)

2)  When the linear-quadratic equation is -1/2x² - 2x + 7  =  -2x + 5

                                                                         -1/2x² + 2  =  0

      multiply both sides by -2:                                  x² - 4  =  0

                                                                     (x + 2)(x - 2)  =  0

     One answer:  x = -2

     Other answer:  x = 2

Thank you so much, but I entered the answer after checking if it was correct, and it still says it was wrong... 

Sorry its cosθ= (−√3)/4))*****

Thank you!

Problem #2:     If you are allowed to use coordinate geometry:

Since triangle(ACB) is a right triangle, the center of the circumcenter is the midpoint of the hypotenuse.

Place the diagram on a coordinate plane with the midpoint of the hypotenuse the origin.

Then point C will be the point (-4, -3).

The equation of the bisector of angle(C) has a slope of 1.

This line has an equation of y + 3  =  1(x + 4)   --->   y  =  x + 1.

The circumcenter has its center at (0, 0) and a radius of 5   --->   x² + y²  =  25.

Combining these two equations will give us the coordinates of point M.

x² + y²  =  25     and     y  =  x + 1     --->     x² + (x + 1)²  =  25     

                                                          --->     x² + x² + 2x + 1  =  25

                                                          --->     2x² + 2x - 24  =  0

                                                          --->     x² + x - 12  =  0

                                                          --->     (x - 4)(x + 3)  =  0

Either  x = 4  or  x = -3  (impossible)

Since  x = 4, by substitution,   x = 3   --->   M  =  (3, 4)

Using the distance formula to find the distance from C(-4, -3) to M(3, 4),

we get the distance of CM to be:  7·sqrt(2).

This is a joke not a prank.... I am sorry

Joke:

Why was the piece of popcorn so buttery?

Answer:

The peice  of popcorn was so salty that it needed to be lathered up!

you have the x and y coordinates (5,-6) and slope -5 you just need to find b by plugging it into 

y=mx+b

-6=-5(5)+b

-6=-5(5)+b

-6=-25+b

b=19

so the equation of the line is:

y=-5x+19