The number of ways in which the name 'ESTABROK' can be made with no restrictions is 40, 320 ways.
<h3>How to determine the number of ways</h3>
Given the word:
ESTABROK
Then n = 8
p = 6
The formula for permutation without restrictions
P = n! ( n - p + 1)!
P = 8! ( 8 - 6 + 1) !
P = 8! (8 - 7)!
P = 8! (1)!
P = 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 × 1
P = 40, 320 ways
Thus, the number of ways in which the name 'ESTABROK' can be made with no restrictions is 40, 320 ways.
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Answer:
I think it may be 120 cm
Step-by-step explanation:
A F is 13 using pythagorean theorem on triangle CED and applying the hypotenuse length to A F.
ABC is similar to triangle CED and is dilated by a factor of 3 so the base is 36.
using pythagorean theorem on triangle ABC gets the hypotenuse length of 39 which can be applied to FE
add all the values together
39 + 13 + 15 + 36 +12 +5 = 120
Answer: 
Step-by-step explanation:
If you meant to solve for "z", then refer to the attachment below.
Answer:
42km
Step-by-step explanation:
Your answer to the length on JL is 22
