Answer:
cocomelon . 3/5 - 4
Step-by-step explanation:
1. Given that the width of the rectangle is x, and the area of the rectangle may be represented by the equation x^2 + 5x = 300, we can solve this equation for the width (x) as such:
x^2 + 5x = 300
x^2 + 5x - 300 = 0 (Subtract 300 from both sides)
(x - 15)(x + 20) = 0 (Factorise x^2 + 5x - 300)
From this, we get: x = 15 or x = -20
Since the width must be a positive length (ie. more than 0), -20 would be an invalid answer in the given context and thus the width is given by x = 15.
2. If we know that the length is 5 inches more than the width, we simply need to add 5 to the width we found above to obtain the length:
Length = x + 5
Length = 15 + 5 = 20
Thus, the width of the rectangle is 15 inches and the length of the rectangle is 20 inches.
Answer:
The maximum number of toys you can afford to buy is 40
Step-by-step explanation:
Let
x -----> the number of toys
we have the compound inequality
![150\leq 11x+10\leq 450](https://tex.z-dn.net/?f=150%5Cleq%2011x%2B10%5Cleq%20450)
Divide into two inequalities
-----> inequality A
----> inequality B
<em>Solve the inequality A</em>
![150-10\leq 11x](https://tex.z-dn.net/?f=150-10%5Cleq%2011x)
![140\leq 11x](https://tex.z-dn.net/?f=140%5Cleq%2011x)
Divide by 11 both sides
![12.7\leq x](https://tex.z-dn.net/?f=12.7%5Cleq%20x)
Rewrite
![x\geq 12.7](https://tex.z-dn.net/?f=x%5Cgeq%2012.7)
<em>Solve the inequality B</em>
![11x\leq 450-10](https://tex.z-dn.net/?f=11x%5Cleq%20450-10)
![11x\leq 440](https://tex.z-dn.net/?f=11x%5Cleq%20440)
Divide by 11 both sides
![x\leq 40](https://tex.z-dn.net/?f=x%5Cleq%2040)
The solution for x is the interval ----> [13,40}
Remember that the number of toys must be a whole number
The domain is all whole numbers greater than or equal to 13 toys and less than or equal to 40 toys
therefore
The maximum number of toys you can afford to buy is 40