i will mark brainliest, thank, and 5star Find the smallest positive value of the nonnegative number \lambdaλ such that the inequality
\frac{a+b}{2}\ge \lambda \sqrt{ab}+(1-\lambda) \sqrt{\frac{a^2+b^2}{2}}
2
a+b
​
≥λ
ab
​
+(1−λ)
2
a
2
+b
2

​

​

holds for all positive real numbers a, ba, b.

unpaid balance is : $149.99-$50 = $99.99 new balance = 149.99+2.62-50 = $102.61 finance charge = 99.99- 0.21/12= $1.75 new balance = 99.99+1.75 = $101.74 average daily balance method average daily balance= 3399.70/30 = $113.32 finance charge = $ 0.21/12ă—113.32 = $1.98 new balance = $149.99-50+1.98 = $101.97

x=half

step-by-step explanation:

the correct answer is $7.38

step-by-step explanation: