3x^2 + 4x - 5 = 0
x = [-b ±√(b^2 - 4ac)]/2a
a = 3
b = 4
c = -5
x = [-4 ±√(16 + 60)]/6
x = [-4 ±√76]/6
x = [-4 ±√(2^2 * 19)]/6
x = [-4 ±2√19]/6
x = [-2 ±√19]/3
x = [-2 + √19]/3
x = [-2 + 4.35]/3
x = 2.35/3
x = 0.78 (rounded to the nearest hundredth)
x = [-2 - ±√19]/3
x = [-2 - 4.35]/3
x = -6.35/3
x = -2.12 (rounded to the nearest hundredth)
<span>∴ x = -2.12 , 0.78 (rounded to the nearest hundredth)</span>
The least number of rides to make sense to buy the monthly pass is 35
<em><u>Solution:</u></em>
Given that monthly transit pass costs $55
The regular price of each ride is $1.60
Cost of 1 ride = $ 1.60
To find: Inequality to determine the least number of rides, that it would make sense to buy the monthly pass
Let "x" be the number of rides
From given information, we can frame a inequality as:
![\text{ number of rides } \times \text{ Cost of 1 ride } \geq 55](https://tex.z-dn.net/?f=%5Ctext%7B%20number%20of%20rides%20%7D%20%5Ctimes%20%5Ctext%7B%20Cost%20of%201%20ride%20%7D%20%5Cgeq%2055)
Here we use greater than or equal to 55 . Since $ 55 is required to buy the monthly pass
![x \times 1.60 \geq 55\\\\1.60x \geq 55](https://tex.z-dn.net/?f=x%20%5Ctimes%201.60%20%5Cgeq%2055%5C%5C%5C%5C1.60x%20%5Cgeq%2055)
Thus the required inequality is found.
Let us solve the above inequality for "x"
![1.60x \geq 55\\\\x \geq \frac{55}{1.60}\\\\x\geq 34.375](https://tex.z-dn.net/?f=1.60x%20%5Cgeq%2055%5C%5C%5C%5Cx%20%5Cgeq%20%5Cfrac%7B55%7D%7B1.60%7D%5C%5C%5C%5Cx%5Cgeq%2034.375)
Therefore, least number of rides to make sense to buy the monthly pass is 35
Answer: me dont ride school bus
Step-by-step explanation:
Answer:
The yellow is 1 and a half
pink is 20
green is 65
gray is nine and 1/4
purple is 28
blue is 4
red is 35
orange is 35
????? I didn't really get some of it
Step-by-step explanation:
PEMDAS- Parenthesis, Exponent, Multiplication, Division, Addition, Subtraction.