Answer:
No real solutions.
Step-by-step explanation:
729s^-5 = 729/s^5
729/s^5 = 3^2(1-s) = 9(1-s)
729=9(1-s)(s^5)
(1-s)(s^5)=81
s^5-s^6=81
Because 81 > 0 this means s^6 must be less than s^5. This is never possible as 6 is even so is always positive and is always greater in magnitude than s^5. The only time this isnt the case is when s is a decimal less than 1, but then the difference would be also less than 1 and not 81.
If the discriminant is negative no real solution exists.
If the discriminant is equal to 0 only one real solution exists.
If the discriminant is positive 2 real solutions exist.
The discriminant:
D = b² - 4 a c
1 ) - 7 x² + 6 x + 3 = 0
D = 6² - 4 · ( - 7 ) · 3 = 36 + 84 = 120 > 0
Answer: b) two solutions
2 ) - 8 x² - 8 x - 2 = 0
D = ( - 8 )² - 4 · ( - 8 ) · ( - 2 ) = 64 - 64 = 0
Answer: a ) one solution
Answer:
Length of the boxes are 17 in, 35 in, 17 in.
Step-by-step explanation:
The given question is incomplete: please find the complete question in the attachment.
A). From Box - (1)
Area of the box = 289 in.²
Expression that denotes the area of the box = 36x² - 12x + 1
So, 36x²- 12x + 1 = 289
(6x)² - 2(6)x + (1)² = (17)²
(6x - 1)² = (17)² [ Since (a² - 2ab + b²) = (a - b)²]
Given box is a square having sides = (6x - 1) in.
∵ (6x - 1)² = 17
⇒ 6x = 18
⇒ x =
= 3
Now length of the sides = (6x - 1) = 17 in.
B) Box - (2)
Similarly, 25x²- 50x + 25 = 1225
(25x² - 50x + 25) = (35)²
(5x - 5)² = (35)²
5(x - 1) = 35 ⇒ x = 8
Here length of one side of the box is 5(x - 1) = 5(8 - 1) = 35in.
C). Box - (3)
49x² - 56x + 16 = 289
(7x)² - 2×(7x)(4) + 4²= (17)²
(7x - 4)² = (17)²
Length of each side = (7x - 4) in.
From, (7x - 4)² = (17)²
7x - 4 = 17
⇒ x = 
⇒ x = 3
Therefore, length of each side of the box = (7x - 4) = 17 in.
14/6 fourteen sixths
2 2/6 two and two sixths
simplified 2 1/3 two and a third
all u do is multiply L*W*H
hope this helps