It will move it from one section to another and make the x axis 4 units more to the right
Answer:
The Law of Cosine : cos C = 
Step-by-step explanation:
See the figure to understand the proof :
Let A Triangle ABC with sides a,b,c,
Draw a perpendicular on base AC of height H meet at point D
Divide base length b as AD = x -b and CD = x
By Pythagoras Theorem
In Triangle BDC And In Triangle BDA
a² = h² + x² ( 1 ) c² = h² + (x-b)²
c² = h² + x² + b² - 2xb ...(. 2)
From above eq 1 and 2
c² = (a² - x²) + x² + b² - 2xb
or, c² = a² + b² - 2xb .....(3)
Again in ΔBDC
cos C = 
Or, cos C = 
∴ x= a cos C
Now put ht value of x in eq 3
I.e, c² = a² + b² - 2ab cos C
Hence , cos C =
Proved Answer
What you're looking at in this image are two alternate exterior angles.
Alternate exterior angles, are, essentially, exterior angles of a transversal that runs through two parallel lines. These angles are on alternating sides. These angles, assuming the lines are parallel, must be equal.
There's a proof behind why these angles are equal, but I won't bore you with the specifics as the question doesn't require it.
So - we know that these two angles must be equal to prove that the lines A and B are parallel. Knowing this, we can write an equation:
5x + 20 = 3x + 60
<u>Subtract 20 from both sides:</u>
5x + 20 - 20 = 3x + 60 - 20
5x = 3x + 40
<u>Subtract 3x from both sides:</u>
5x - 3x = 3x - 3x + 40
2x = 40
x = 20
If you have any questions on how I got to the answer, just ask!
- breezyツ
Answer/Step-by-step explanation:
AC = 1.2
AB = 4
BC = 2.6
DF = 3
DE = 10
EF = 6.5
Thus:



The ratio of their corresponding sides are all equal to ⁵/2. Therefore, both triangles are similar.
I give a good guess of 42