To solve this, create an equation that follows 45% of a value is 18. This would look like x*.45=18. Now, solve for x, x*.45=18, divide by .45, x=18/.45=40, making your value 40. To check, plug in your value to the equation 40*.45=18 and see if it stays true.
23 lines × 5 trees= 115 tree. or. 5 lines × 23 trees = 115 trees
If the area is 10 square meters than double that is area of a rectangle with height = 5m and base of 4m because height • base should produce 20m^2.
Hope this helps.
R = ((3V)/((pi)(h^2))+h)/3
We have a rectangle with length L that is 3 inches more than the width W. Then we can write this as:

The area of the rectangle is 180 square inches.
We have to find the width W.
As the area is equal to the product of the length and the width, we can write this equation and solve for W as:

We have a quadratic equation. The roots of this equation will be the mathematical solutions.
We can find the roots using the quadratic formula:
![\begin{gathered} W=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ W=\frac{-3\pm\sqrt[]{3^2-4\cdot1\cdot(-180)}}{2\cdot1} \\ W=\frac{-3\pm\sqrt[]{9+720}}{2} \\ W=\frac{-3\pm\sqrt[]{729}}{2} \\ W=\frac{-3\pm27}{2} \\ W_1=\frac{-3-27}{2}=-\frac{30}{2}=-15 \\ W_2=\frac{-3+27}{2}=\frac{24}{2}=12 \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20W%3D%5Cfrac%7B-b%5Cpm%5Csqrt%5B%5D%7Bb%5E2-4ac%7D%7D%7B2a%7D%20%5C%5C%20W%3D%5Cfrac%7B-3%5Cpm%5Csqrt%5B%5D%7B3%5E2-4%5Ccdot1%5Ccdot%28-180%29%7D%7D%7B2%5Ccdot1%7D%20%5C%5C%20W%3D%5Cfrac%7B-3%5Cpm%5Csqrt%5B%5D%7B9%2B720%7D%7D%7B2%7D%20%5C%5C%20W%3D%5Cfrac%7B-3%5Cpm%5Csqrt%5B%5D%7B729%7D%7D%7B2%7D%20%5C%5C%20W%3D%5Cfrac%7B-3%5Cpm27%7D%7B2%7D%20%5C%5C%20W_1%3D%5Cfrac%7B-3-27%7D%7B2%7D%3D-%5Cfrac%7B30%7D%7B2%7D%3D-15%20%5C%5C%20W_2%3D%5Cfrac%7B-3%2B27%7D%7B2%7D%3D%5Cfrac%7B24%7D%7B2%7D%3D12%20%5Cend%7Bgathered%7D)
The solutions are W = -15 and W = 12.
The first one is not valid, as W has to be greater than 0.
Then, the solution to our problem is W = 12 in.
Answer: the width is W = 12 inches.