Question

find the value of k in equation: kx^2-√2^(x+1)=0 (when x=1)​

Answer 1

Answer:

k = 2

Step-by-step explanation:

Given, x=1

k×1² - root 2 ^(1+1) = 0

k-root 2² = 0

k - 2 = 0

k = 2

Answer 2

HII

K=2

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