Answer:
A. EG = √3 × FG
D. EG = √3/2 × EF
E. EF = 2 × FG
Step-by-step explanation:
∵ tan 60 = √3
∵ tan60 = EG/GF
∴ EG/GF = √3
∴ EG = √3 × GF ⇒ A
∵ m∠F = 60°
∵ sin60 = √3/2
∵ sin 60 = EG/EF
∴ √3/2 = EG/EF
∴ EG = √3/2 × EF ⇒ D
∵ cos60 = 1/2
∵ cos60 = GF/EF
∴ GF/EF = 1/2
∴ EF = 2 × GF ⇒ E
Answer:
A
Step-by-step explanation:i took the test
Sorry but I don’t understand your language.
This is the easiest way to solve this problem:
Imagine this represents how many combinations you can have for each of the 4 wheels (each blank spot for one wheel): __ __ __ __
For the first situation it says how many combos can we make if no digits are repeated.
We have 10 digits to use for the first wheel so put a 10 in the first slot
10 __ __ __
Since no digit can be repeated we only have 9 options for the second slot
10 9_ __ __
Same for the third slot, so only 8 options
<u>10</u> <u> 9 </u> <u> 8 </u> __
4th can't be repeated so only 7 options left
<u>10</u> <u> 9 </u> <u> 8 </u> <u> 7
</u><u>
</u>Multiply the four numbers together: 10*9*8*7 = 5040 combinations
For the next two do the same process as the one above.
If digits can be repeated? You have ten options for every wheel so it would look like this: <u>10</u> <u>10</u> <u>10</u> <u>10
</u>
10*10*10*10 = 10,000 combinations
If successive digits bust be different?
We have 10 for the first wheel, but second wheel only has 9 options because 2nd number can't be same as first. The third and fourth wheels also has 9 options for the same reason.
<u>10</u> <u> 9</u><u> </u> <u> 9 </u> <u> 9 </u>
10*9*9*9 = 7290 combinations
