Let the width be x.
Length = 2x -1
The area of rectangle = 45 cm²
x*(2x -1) = 45
2x² -x = 45
2x² -x - 45 = 0 This is a quadratic equation
comparing to ax² + bx + c = 0, a = 2, b = -1, c = -45
x = (-b + √(b² -4ac)) / 2a or (-b - √(b² -4ac)) / 2a
x = (- -1 + √((-1)² -4*2*-45)) / 2*2 or (- -1 - √((-1)² -4*2*-45)) / 2*2
x = (1 + √(1 +360)) / 2*2 or (1 - √(1 + 360) / 2*2
x = (1 + √361) / 4 or (1 - √361) / 4
x = (1 + 19) / 4 or (1 - 19) / 4
x = 20/4 or -18/4
x = 5 or -4.5
x can't be negative since we are solving for side.
x = 5 as the only valid solution.
Recall, the width = x = 5.
Length = 2x - 1 = 2*5 - 1 = 10 - 1 = 9
Hence the length = 9cm , and width = 5cm
Answer:
are the green boxes the ones you can't drag a box to?
The answer of the problem is 3.6
Answer:
(4, -4)
Step-by-step explanation:
Using the midpoint formula;
M(X,Y) = {(ax1+bx2)/a+b, (ay1+by2)/a+b}
X = (ax1+bx2)/a+b
Y = (ay1+by2)/a+b
Substitute
X = 4(2)+2(8)/4+2
X = 8+16/6
X = 24/6
X = 4
Also
Y = 4(-3)+2(-6)/6
Y = -12-12/6
Y = -24/6
Y = -4
Hence the required partition is (4, -4)
