Answer:
<em>See Reasoning Below</em>
Step-by-step explanation:
To prove that AB = BL, or in other words AB ≅ BL, let us consider the triangles CED and BEL. If we were to prove they were congruent, then by CPCTC ( corresponding parts of congruent triangle are congruent ) DC ≅ BL. As AB ≅ DC by " Properties of Parallelogram " it would be that through transitivity, AB ≅ BL / AB = BL;
Now for " part 2 " we can consider that AB = DC, from part 1. If AB = BL, then AL = 2 ( AB ) by the Partition Postulate. AB = DC, so we can also say that AL = 2 ( DC ) - Proved
See attachment in statement reasoning form for part 1;
Answer:
Step-by-step explanation:
3x + 4(x - 8) - x = 3/5 (10x + 15)
3x + 4x - 32 - x = 3(2x + 3)
6x - 32 = 6x + 9
This is what I did. I'm not sure that this problem has a solution
Are<span> at (2,0) & (0,1) & the eqn of one side is x=2, then the orthocentre of the triangle is a) (3/2, 3/2) ...</span>
? it’s already rounded, is there supposed to be a decimal?
In this case, h(x) = sqrt(x) + 3
A. f(x)=x+3; g(x)=√x
B. f(x)=x; g(x)=x+3
C. f(x)=√x; g(x)=x+3
D. f(x)=3x; g(x)=√x
Again, you need to find a function f(x) that once evaluated in g(x) gives us h(x)
h(x) = g(f(x))
Looking at the options, the answer is C.
g(f(x)) = f(x) + 3 = sqrt (x) + 3 = h(x)
