Area of a rectangle of length L and width W is A = LW.
Here, A = 60 ft^2 = (12 ft)(2x-3 ft). Let's find the values of x for which the area =60 ft^2, and then the values of x for which the area is greater than 60 ft^2.
60 = 12(2x-3) becomes 5 = 2x - 3 after dividing both sides by 12.
Adding 3 to both sides of this last equation results in 8 = 2x, so x = 4.
Does (12 ft)(2[4 ft] - 3) = 60? Does (12 ft)(5 ft) = 60 sq ft? YES.
The area of this rectangle will be greater than 60 sq ft if x>4 ft. (answer)
Complementary angles are angle pairs whose measures sum to one right angle (14 turn,90°, or π2 radians)
A) 11/15 = .733....
B) 3/5 = .6
C) 2/3 = .666....67
D) 19/30 = .63
B, D, C, A
We know that
the equation of the parabola is of the form
y=ax²+bx+c
in this problem
y=1/4x²−x+3
where
a=1/4
b=-1
c=3
the coordinates of the focus are
(-b/2a,(1-D)/4a)
where D is the discriminant b²-4ac
D=(-1)²-4*(1/4)*3-----> D=1-3---> D=-2
therefore
x coordinate of the focus
-b/2a----> 1/[2*(-1/4)]----> 2
y coordinate of the focus
(1-D)/4a------> (1+2)/(4/4)---> 3
the coordinates of the focus are (2,3)
Answer:
-4(8)=x^2-9x+12
-32=x^2-9x+12
Step-by-step explanation: is this what u want us to do?
