If # (a,b,c,d)means you raise b,c and d to the power of a and then take the sum of the resulting values, and # (2,3,4,d)=89,then d=?
2^a + 3^a + 4^a = 89
1^a + 1^a + 1^a = 89
3^a = 89
a = 86
an I do it right
If # (a,b,c,d)means you raise b,c and d to the power of d and then take the sum of the resulting values, and # (2,3,4,d)=89,then d=?
2^d + 3^d + 4^d = 89
1^d + 1^d + 1^d = 89
3^d = 89
d = 86
If # (a,b,c,d)means you raise b,c and d to the power of a and then take the sum of the resulting values, and # (2,3,4,d)=89,then d=?
a. 8
b. 9
c. 64
d. 81
e. 90
please answer and explain
2 + 3^a + 4^a + d^a = 89
2 + 1^a + 1^a + d^a = 89
2^a + d^a = 87
I don't know how to do
Your reasoning is incorrect. Let's solve the equation step by step to find the value of 'd'.
Given: #(a, b, c, d) = b^a + c^a + d^a = 89
1. Substitute the given values (#(2, 3, 4, d) = 89) into the equation:
3^2 + 4^2 + d^2 = 89
Simplify: 9 + 16 + d^2 = 89
Combine like terms: 25 + d^2 = 89
2. Subtract 25 from both sides of the equation to isolate d^2:
d^2 = 89 - 25
d^2 = 64
3. Take the square root of both sides to solve for 'd':
d = √64
4. Evaluate the square root:
d = ±8
Therefore, 'd' can be either 8 or -8.