Algebra problems help pleaseee(: Even if you dont know all of them?(:

5.A manufacturer determines that the number of drills it can sell is given by the formula D= -3p^2 + 180p-285, where p is the price of the drills in dollars.
a. At what price will the manufacturer sell the maximum number of drills?
b.what is the maximum number of drills that can sold?

6.factor -15x^2-21X
7. factor x^2-2x-63
8. factor 9x^2-16

5.
get the maxima of the curve but if you don't know calculus, just get the vertex of this parabola, since the parabola opens downward, the vertex should be the highest point in the graph.
to make it easier
-D/3 = p^2 - 60p +95
-D/3 = p^2 - 60p +900 + 95 - 900
-D/3 +805 = (p-30)^2
D - 2415 = -3(p-30)^2
p = 30, D = 2415
or you can do the shortcut
x = -b/2a = -180/2(-3) = 30
D = -3(30)^2 +180(30) -285 = 2415
6. -15x^2 -21x
-3x(5x-7)
7.x^2 - 2x - 63
( x + 7 )( x - 9 )
8. 9x^2 - 16 (difference of two squares)
= ( 3x - 4 )( 3x +4 )

D= -3p^2 + 180p-285

D'= -6p+180

set it equal to zero: -6p+180=0
-6p=-180
p=30

the dude sells the maximum amount of drills at 30 dollars.

plug it into the equation: -3(30)+180(30)+ 285= -90+5400+285= 5595 drills sold. :D.



to factor -15x^2-21x: -3x(5x-9)

to factor x^2-2x-63: (x-9)(x+7)

to factor 9x^2-16: (3x-4)(3x+4)

5)

a)

diff(-3*p^2+180*p-285, p) = - 6*p + 180

solve( - 6*p + 180 = 0)

p = 30

b)

f(p):=(-3 p^2 + 180 p-285)

f(30) = 2,415

6)

- 3*x*(5*x + 7)

7)

(x - 9)*(x + 7) = 0

8)

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