Answer:
Area of rectangle is
Perimeter of Rectangle is
.
Step-by-step explanation:
Given:
Let the width of the rectangle be 'w'.
Also Given:
A rectangle has a length that is 5 meters greater than the width.
Length of rectangle = ![w+5\ meters](https://tex.z-dn.net/?f=w%2B5%5C%20meters)
We need to write expression for Area of rectangle and Perimeter of rectangle.
Solution:
Now we know that;
Perimeter of rectangle is equal to twice the sum of the length and width.
framing in equation form we get;
Perimeter of rectangle = ![2(w+5+w)=2(2w+5) =4w+20](https://tex.z-dn.net/?f=2%28w%2B5%2Bw%29%3D2%282w%2B5%29%20%3D4w%2B20)
Also We know that;
Area of rectangle is length times width.
framing in equation form we get;
Area of rectangle= ![w(w+5) = w^2+5w](https://tex.z-dn.net/?f=w%28w%2B5%29%20%3D%20w%5E2%2B5w)
Hence Area of rectangle is
and Perimeter of Rectangle is
.
Answer:
D.
Step-by-step explanation:
![{x}^{2} = {18}^{2} + ( {12}^{2} + {8}^{2} ) \\ \\ {x}^{2} = 324 + (144 + 64) \\ \\ {x}^{2} = 324 + 204 \\ \\ {x}^{2} = 528 \\ \\ x = \sqrt{532} \\ \\ x = \sqrt{4\times 133} \\ \\ x = 2\sqrt {133}](https://tex.z-dn.net/?f=%20%7Bx%7D%5E%7B2%7D%20%20%3D%20%20%7B18%7D%5E%7B2%7D%20%20%2B%20%28%20%7B12%7D%5E%7B2%7D%20%20%2B%20%20%7B8%7D%5E%7B2%7D%20%29%20%5C%5C%20%20%5C%5C%20%20%7Bx%7D%5E%7B2%7D%20%20%3D%20324%20%2B%20%28144%20%2B%2064%29%20%5C%5C%20%20%5C%5C%20%20%7Bx%7D%5E%7B2%7D%20%20%3D%20324%20%2B%20204%20%5C%5C%20%20%5C%5C%20%20%7Bx%7D%5E%7B2%7D%20%20%3D%20528%20%5C%5C%20%20%5C%5C%20x%20%3D%20%20%5Csqrt%7B532%7D%20%20%5C%5C%20%20%5C%5C%20x%20%3D%20%20%5Csqrt%7B4%5Ctimes%20133%7D%20%20%5C%5C%20%20%5C%5C%20x%20%3D%202%5Csqrt%20%7B133%7D)
Answer:
D divide both the numerator and denominator by cos(x)cos(y)
Step-by-step explanation:
just took the test
Answer:
wheres the image or text
Step-by-step explanation: