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.
Answer:
Diameter: the straight distance from one edge of a circle to the other
Radius: the straight distance from one edge of a circle to the center
I hope this helps :)
Answer:
3ab
Step-by-step explanation:
Answer:
Hence proved triangle ADE ≅ triangle BCE by Side Angle Side congruent property.
Step-by-step explanation:
Given:
AD ⊥ AB
CD
BC ⊥ AB
CD
AD = BC
∴ ∠ A = ∠ B = ∠ C = ∠ D =90°
∠ EDC = ∠ ECD
Solution
∠ C = ∠ BCE + ∠ ECD⇒ equation 1
∠ D = ∠ ADE + ∠ EDC⇒ equation 2
∠ C = ∠ D (given)
Substituting equation 1 and 2 in above equation we get
∠ BCE + ∠ ECD = ∠ ADE + ∠ EDC
But ∠ EDC = ∠ ECD (given)
∴ ∠ ADE = ∠ BCE
ED = EC (∵ base angles are same triangle is isosceles triangle)
Now, In Δ ADE and Δ BCE
AD =BC
∠ ADE = ∠ BDE
ED = EC
∴ By Side Angle Side congruent property
Δ ADE ≅ Δ BCE