Mixins must be called with @include mixin-name; inside selectors.
Step 2: Identify the error in usage
The code uses text-style; without @include, so it won't apply the mixin styles.
Final Answer:
Mixin is not called with @include keyword -> Option A
Quick Check:
Call mixins with @include [OK]
Hint: Always use @include to call mixins [OK]
Common Mistakes:
Forgetting @include before mixin name
Adding parameters when none needed
Placing mixin inside selector definition
5. You want to create a button style that can have different background colors but share the same padding and border radius. How can mixins help you avoid duplication?