Answer:
square root 45 lies between 6 and 7
Step-by-step explanation:
I hope it's helpful!
Answer:
Yes you are right
Step-by-step explanation:
1st of all you said ''tell me if im right please'' =3
Answer:
![4ln [\frac{x^2 (x^3-1)}{x-5}]](https://tex.z-dn.net/?f=%204ln%20%5B%5Cfrac%7Bx%5E2%20%28x%5E3-1%29%7D%7Bx-5%7D%5D)
Step-by-step explanation:
For this case we have the following expression:
![4[ln(x^3-1) +2ln(x) -ln(x-5)]](https://tex.z-dn.net/?f=%204%5Bln%28x%5E3-1%29%20%2B2ln%28x%29%20-ln%28x-5%29%5D)
For this case we can apply the following property:
And we can rewrite the following expression like this:
And we can rewrite like this our expression:
![4[ln(x^3-1) +ln(x^2) -ln(x-5)]](https://tex.z-dn.net/?f=%204%5Bln%28x%5E3-1%29%20%2Bln%28x%5E2%29%20-ln%28x-5%29%5D)
Now we can use the following property:
And we got this:
![4[ln(x^3-1)(x^2) -ln(x-5)]](https://tex.z-dn.net/?f=%204%5Bln%28x%5E3-1%29%28x%5E2%29%20-ln%28x-5%29%5D)
And now we can apply the following property:
And we got this:
And that would be our final answer on this case.
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.
