Answer: A & C
<u>Step-by-step explanation:</u>
HL is Hypotenuse-Leg
A) the hypotenuse from ΔABC ≡ the hypotenuse from ΔFGH
a leg from ΔABC ≡ a leg from ΔFGH
Therefore HL Congruency Theorem can be used to prove ΔABC ≡ ΔFGH
B) a leg from ΔABC ≡ a leg from ΔFGH
the other leg from ΔABC ≡ the other leg from ΔFGH
Therefore LL (not HL) Congruency Theorem can be used.
C) the hypotenuse from ΔABC ≡ the hypotenuse from ΔFGH
at least one leg from ΔABC ≡ at least one leg from ΔFGH
Therefore HL Congruency Theorem can be used to prove ΔABC ≡ ΔFGH
D) an angle from ΔABC ≡ an angle from ΔFGH
the other angle from ΔABC ≡ the other angle from ΔFGH
AA cannot be used for congruence.
Okay, let's work this out...
What we know:
-32 stamps in all
- rows (horizontal)
- same # in each
What we "want to know" :
- # of combinations (different)
Problem Solving :
This is actually very easy its just the words than get ya!
1st : we need to figure out the factors of 32...
In other words, we need to figure out _x_=32 and how many different combinations and ways there are!
Note:(* means multiplication)
#1: What are the factors of 32?
32: 1*32 , 2*16 , 4*8
32: 32*1 , 16*2 , 8*4
The factors (not including 1 are 2,4,8,16)
Now, as you can see, there are 4 ways to get 32 as shown first.
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That ¦ is 1 way with 16 in 2 rows. Basic multiplication, 16*2=32 or 16+16=32.
But, this is also a way,
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Now there should be 2 in each row and 16 rows. Again 2*16=32 or 2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2=32
That's two ways so far.
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Another way which is 4 rows with 8 in each.
4*8=32 or 8+8+8+8=32
But, this is also a way,
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Now that is 8 rows with 4 in each. 8*4=32 or 4+4+4+4+4+4+4+4=32
That was our fourth way.
Again NOT including 1. If you include 1 then there will be 6 ways but aside from that there are 4 ways.
I hope that helped I worked hard typing this all for you. Any questions just ask!
Answer: the answers is a
Step-by-step explanation:
