Answer:
+5
2x+5
3x-5
2(n+5)
3(+5)
5x(x^3)
Step-by-step explanation:
How do i solve this radical expression? It has to do with factoring i know that.
1 answer:
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This appears to be about rules of exponents as much as anything. The applicable "definitions, identities, and properties" are
i^0 = 1 . . . . . as is true for any non-zero value to the zero power
i^1 = i . . . . . . as is true for any value to the first power
i^2 = -1 . . . . . from the definition of i
i^3 = -i . . . . . = (i^2)·(i^1) = -1·i = -i
i^n = i^(n mod 4) . . . . . where "n mod 4" is the remainder after division by 4
1. = -3^4·i^(3·2+0+2·4) = -81·i^14 = 81
2. = i^((3-5)·2+0 = i^-4 = 1
3. = -2^2·i^(4+2+2+(-1+1+5)·3+0) = -4·i^23 = 4i
4. = i^(3+(2+3+4+0+2+5)·2) = i^35 = -i
i^0 = 1 . . . . . as is true for any non-zero value to the zero power
i^1 = i . . . . . . as is true for any value to the first power
i^2 = -1 . . . . . from the definition of i
i^3 = -i . . . . . = (i^2)·(i^1) = -1·i = -i
i^n = i^(n mod 4) . . . . . where "n mod 4" is the remainder after division by 4
1. = -3^4·i^(3·2+0+2·4) = -81·i^14 = 81
2. = i^((3-5)·2+0 = i^-4 = 1
3. = -2^2·i^(4+2+2+(-1+1+5)·3+0) = -4·i^23 = 4i
4. = i^(3+(2+3+4+0+2+5)·2) = i^35 = -i
Answer:
Hello,
Step-by-step explanation: