Here, triangle AEN and ALG are similar.
We know that the ratio of corresponding sides of two similar triangles are equal. So,
![\frac{21}{42} =\frac{2x-9}{(2x-9)+(x+7)}\\\\\frac{1}{2} =\frac{2x-9}{3x-2}\\\\ 3x-2=4x-18\\\\-2+18=4x-3x\\\\16=x\\\\x=16\\](https://tex.z-dn.net/?f=%5Cfrac%7B21%7D%7B42%7D%20%3D%5Cfrac%7B2x-9%7D%7B%282x-9%29%2B%28x%2B7%29%7D%5C%5C%5C%5C%5Cfrac%7B1%7D%7B2%7D%20%3D%5Cfrac%7B2x-9%7D%7B3x-2%7D%5C%5C%5C%5C%203x-2%3D4x-18%5C%5C%5C%5C-2%2B18%3D4x-3x%5C%5C%5C%5C16%3Dx%5C%5C%5C%5Cx%3D16%5C%5C)
Answer:
He will go 61 feet.
Step by Step Explanation:
You fill in variables for 6 them combine your like terms.
Answer:
3/8 (Decimal: 0.375)
Step-by-step explanation:
20
There are 10 feet of posts on each side. That means there are 40 feet of fencing in total. If you divide that by two (to represent posts being 2 feet apart) then you get 20. There are 20 posts needed.
Answer:
p=7x
Step-by-step explanation:
49x^[2] + 28x - 10 = p^[2] + 4p -10
This equation is in the form a^[2]x + bx + c.
<u><em>The 'c' is common for both equations, this means the 'a' and 'b' must also be common. </em></u>
There are two ways to find p: 'a' or 'b'
<u>a method</u>
49x^[2] = p^[2]
=> The square root of both sides = 7x = p
<u>b method</u>
28x = 4p
28x/4 = 4p/4
7x = p