Since EF=FG, you can set 6x - 10 = to 3x + 11

Then add like terms

so,

Now you can divide by 3 to get

:)
then put it back into


so,
Answers:
If an angle is labeled with a single letter, that letter represents the <u> vertex </u> of the angle.
If more than one angle has the same vertex, you must use <u> 3 </u> points to name the angle. The <u> </u><u>middle </u> point named must be the vertex.
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Explanation:
If we have a single triangle, and no other extra lines, then we can use single letters to name the three angles. Each vertex of the triangle corresponds to the vertex of that angle.
If you were to draw many triangles, in which some may or may not overlap, you'll mostly likely need to name the angle using 3 letters. This is so you are very specific about which angle you're talking about. The middle letter is always the vertex. The left and right letters are points on the arms of the angle. The order of the left and right letters doesn't matter as long as the middle letter stays the same. So something like angle ABC is the same as angle CBA.
Answer: 
Step-by-step explanation:
Given : The height of the rectangle = 
The width of the rectangle = 
Formula : Area = height x width
Therefore , the area of triangle in terms of polynomial will be :
![6k^3\times( 2k^2+4k+5)\\\\= 6k^3(2k^2)+6k^3(4k)+6k^3(5)\ \ [\text{Using Distributive property}]\\\\=12k^{3+2}+24k^{3+1}+30k^3\ \ [\text{Using exponents rule}:\ a^n\times a^m=a^{n+m}]\\\\=12k^5+24k^4+30k^3](https://tex.z-dn.net/?f=6k%5E3%5Ctimes%28%202k%5E2%2B4k%2B5%29%5C%5C%5C%5C%3D%206k%5E3%282k%5E2%29%2B6k%5E3%284k%29%2B6k%5E3%285%29%5C%20%5C%20%5B%5Ctext%7BUsing%20Distributive%20property%7D%5D%5C%5C%5C%5C%3D12k%5E%7B3%2B2%7D%2B24k%5E%7B3%2B1%7D%2B30k%5E3%5C%20%5C%20%5B%5Ctext%7BUsing%20exponents%20rule%7D%3A%5C%20a%5En%5Ctimes%20a%5Em%3Da%5E%7Bn%2Bm%7D%5D%5C%5C%5C%5C%3D12k%5E5%2B24k%5E4%2B30k%5E3)
Hence, the area of the entire rectangle =
Do 11/24 + 2/5. then subtract your answer from 240
Answer:
See below ~
Step-by-step explanation:
Given :
⇒ m∠1 = m∠2
⇒ HD = GF
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To Prove :
<u>Δ EHD ≅ Δ EGF</u>
<u />
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Solving :
⇒ m∠1 = m∠2 (Given)
⇒ HD = GF (Given)
⇒ ∠E = ∠E (Common angle)
⇒ ΔEHD ≅ ΔEGF (AAS congruence)