Area is 6000 yd2
LW = 6000
field must be 40 yards longer than its width
L = W + 40
Replace L with W+40 in 1st equation to solve for Width
(W+40)(W) = 6000
W2 + 40W - 6000 = 0
This is a quadratic but is factorable
Factors of -6000 that add to 40 are (100)(-60)
(W+100)(W-60) = 0
W = -100 or W = 60
Since the width will not be negative discard -100
The width is 60 yards
Length is W+40 = 100 yards
We can name an angle either by naming its vertex or by three letters , keeping the letter of vertex in between.
so here vertex is F
we can name it as ∠EFG, ∠F , ∠GFE
This strikes out the option B that is ∠G
so option B is the answer
V = lwh
2x³ + 17x² + 46x + 40 = l(x + 4)(x + 2)
2x³ + 12x² + 16x + 5x² + 30x + 40 = l(x + 4)(x + 2)
2x(x²) + 2x(6x) + 2x(8) + 5(x²) + 5(6x) + 5(8) = l(x + 4)(x + 2)
2x(x² + 6x + 8) + 5(x² + 6x + 8) = l(x + 4)(x + 2)
(2x + 5)(x² + 6x + 8) = l(x + 2)(x + 4)
(2x + 5)(x² + 2x + 4x + 8) = l(x + 4)(x + 2)
(2x + 5)(x(x) + x(2) + 4(x) + 4(2)) = l(x + 4)(x + 2)
(2x + 5)(x(x + 2) + 4(x + 2)) = l(x + 4)(x + 2)
(2x + 5)(x + 4)(x + 2) = l(x + 4)(x + 2)
(x + 4)(x + 2) (x + 4)(x + 2)
2x + 5 = l
Answer:
the desired range is 2 < t < 12.
Step-by-step explanation:
If the third side is 12 (the sum of 5 and 7), we don't really have a triangle, since the 5-unit and 7-unit sides add up to 12 and lie on the third side (of length 12).
Since triangle ABC is acute, all its angles are between 0 and 90 degrees. So the third side can be 12 but not greater. The smallest possible length of the third side is 2, since 5 + 2 = 7.
Thus, the desired range is 2 < t < 12.
