(4/.4)^sqrt(4)-4=77
with a dot over the 0.4 to show 0.4444 recurring
_ (4/0.4)^sqrt4-4 =77 9 81 =77
One possible way: Make a 10 using five 4s as follows: 4 + 4 + (4 + 4)/4 = 10 Make another two 10s using five 4s in each. Multiply these three 10s.
The answer is 4! - [(4+4)/4]
4! - sqrt 4 + 4/4
(4x4)-(4/4) = 15
It could be: 4+4+4-4/4 = 11
4+4+4x4 by the power of 0.
4/.4 + 4/4 [= 10+1 = 11]
4! - 44/4 = 24 - 11 = 13
To make 77 using four 4s, you can use the following expression: ( 4 \times 4 \times 4 + 4 ). This simplifies as follows: ( 4 \times 4 = 16 ), then ( 16 \times 4 = 64 ), and finally ( 64 + 4 = 68 ) (This is incorrect). A correct expression is ( 4! \times 4 - 4/4 ). Here, ( 4! = 24 ), so ( 24 \times 4 = 96 ), and ( 96 - 4/4 = 96 - 1 = 95 ) (This is also incorrect). The correct method is: ( (4 \times 4 \times 4) + 4 = 64 + 4 ) (This is incorrect too). However, a simpler way is to use exponentiation or other mathematical operations creatively to reach 77. Unfortunately, it's challenging to reach exactly 77 with just four 4s and standard operations without more context or flexibility in mathematical operations.
4! - sqrt(4)*sqrt(4)/4 = 23
(44 - 4)/4