If you know what a factorial is, it would be (4!-.4)/.4+4!
4!=3*2*1
5!=4*3*2*1
99!=98*97*96*........*3*2*1
_ (4/0.4)^sqrt4-4 =77 9 81 =77
(4+4)*(4+4)=8*8=64
(44 - 4)/4
77 + 7/7 = 78
The only solution that I have ever seen uses some notation that is not so commonly used, namely a bar over 4 to mean the infinitely repeating decimal .44444... which is 4/9 . So, if x = 4/9 is written with one four, then sqrt(4/x) 4 - 4 uses four fours and equals 77 since 4/x = 4 * 9/4 = 9 and sqrt(9)4 - 4 = 34 - 4 = 81 - 4 = 77.
_ (4/0.4)^sqrt4-4 =77 9 81 =77
The answer is 4! - [(4+4)/4]
(4x4)-(4/4) = 15
4+4+4x4 by the power of 0.
4! - sqrt 4 + 4/4
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+4)*(4+4)=8*8=64
4/.4 + 4/4 [= 10+1 = 11]
(44 - 4)/4
44/4 + 4! = 11 + (4x3x2x1) = 11 + 24 = 35
4! - sqrt(4)*sqrt(4)/4 = 23
how do i make 17 using only 2 4 6 8