2058+245+28+2 = 2333
The number 2058 is divisible by several integers, including 1, 2, 3, 6, 343, 1029, and 2058 itself. It is even, so it is divisible by 2, and the sum of its digits (2 + 0 + 5 + 8 = 15) is divisible by 3, indicating that 2058 is also divisible by 3. Thus, 2058 has various divisors.
The prime factors of 2058 are 2*3*7*7*7, or 2 × 3 × 73.
The prime factorization of 2058 is 2 x 3 x 7 x 7 x 7 or 2 x 3 x 73.
-2 + 28 = 26
-6z 2 = 3z+ 4z+28 -6z+2 +3z = 3z+ 4z +28 +4z -6z+4z+2 = 3z+ 28 -10z+2 -3z= 28 - 2 -10z + -3z = 26 -13z/13 = 26/-13 z = -2
The prime factors of 2058 are 2*3*7*7*7, or 2 × 3 × 73.
The prime factorization of 2058 is 2 x 3 x 7 x 7 x 7 or 2 x 3 x 73.
It is 2+28 = 30
-2 + 28 = 26
2 x 3 x 7 x 7 x 7 = 2058
1/7 + 1/2 + 5/28 = 4/28 + 14/28 + 5/28 = 23/28
-6z 2 = 3z+ 4z+28 -6z+2 +3z = 3z+ 4z +28 +4z -6z+4z+2 = 3z+ 28 -10z+2 -3z= 28 - 2 -10z + -3z = 26 -13z/13 = 26/-13 z = -2
28*6+2= 170 Thank you
120 + 13 + 3 + 8 + 2 + 2 + 353 + 42 + 20 + 28 + 87 + 8 + 10 = 696
16 + 24/2 = 28/1 or 28
Do you mean, 2 + 3^x = 245 ?? 2 + 3^x = 245 subtract 2 from each side 3^x = 243 use natural logs ( I always do, though you can use logs ) ln(3^x) = ln(243) you have the right in this operation to bring the exponent before the natural log x(ln3) = ln(243) divide x = ln(243)/ln(3) ( not ln(243/3) !!!! ) x = 5.493061443/1.098612289 x = 5 ----------check 2 + 3^5 = 245 2 + 243 = 245 245 = 245 ===========checks
122.5