3x^2 + 16x + 4
3 ( x^2 + (16/3)x + 4/3) complete the square inside the parentheses
3 ( x ^2 + (16/3)x + 64/9 + 4/3 - 64/9)
3 [ ( x + 8/3)^2 + 12/9 - 64/9) ]
3 [ ( x + 8/3)^2 - 52/9 ]
3 ( x + 8/3)^2 - 52/3
3 ( x - -8/3)^2 - 52/3
(a , h , k) = (3 ,- 8/3 , -52/3 )
The y intercept is - 4 and the slope is 1/2
So
(0 , -4) + (2 , 1) = (2 , - 3) = J
(2, -3) + ( 2 ,1) = ( 4, -2)
(4 , -2) + (2, 1) = (6 , -1) = D
And that's all.....!!!
x ( x + 6) > 16 rearrange as
x^2 + 6x - 16 > 0 factor as
(x + 8) ( x - 2) > 0
This will true when x < -8 and x > 2
The value of what ????
V = pi (d/2)^2 * h
V = pi (d^2 / 4) * h
h = 4V / ( pi * d^2) (meters)
2 3sqrt 5 + sqrt 11
____________________ * ________________ =
3sqrt 5 - sqrt 11 3sqrt 5 + sqrt 11
2 ( 3sqrt 5 + sqrt 11)
____________________ =
(3sqrt 5)^2 - (sqrt 11) ^2
6 sqrt 5 + 2sqrt 11
___________________ =
45 - 11
6sqrt 5 + 2 sqrt 11
________________ =
34
3 sqrt 5 + 1 sqrt 11
_____________________
17
A + B + C + D + E =
3 + 5 + 1 + 11 + 34 = 54
Very nice work by everyone, here......!!!!!
QP must lie on the line with a slope of 1
Call Q = (x1 , 0)
(6 - 0 ) / ( 1 - x1) = 1
6 = 1 - x1
x1 = 1 - 6
x1 = -5
So Q =( -5,0)
Likewiae RP must lie on the line with a slope of 3
Call R = (x2 , 0)
(6-0) / (1 - x2) = 3
6 = 3 ( 1 - x2)
2 = 1 - x^2
x2 = 1 - 2
x2 = -1
So R =( -1, 0)
So QR = 4= base of PQR
And the height of the triangle = 6
Area = (6 * 4) / 2 = 12
A= 49
B D C
Agree with UTS.....if D is on BC, then BC cannot = CD
The slope of the gven line is -2
The slope of a perp line = 1/2
Equation of this line
y = (1/2) ( x - 5) + 5
y = (1/2)x - 5/2 + 5
y = (1/2)x + 5/2
Setting this = to the given line to find the x intersection coordimate
-2x + 5 = (1/2)x + 5/2
(- 2 - 1/2)x = 5/2 - 5
( -5/2)x = -5/2
x = 1
And y = -2(1) + 5 = 3
Intersection pont = (1, 3)
