const numbers = [1, 2, 3, 4, 5];
const result = numbers
.filter(n => n % 2 === 0)
.map(n => n * 2)
.reduce((acc, n) => acc + n, 0);
console.log(result);
Ответ:
12
JavaScript test | #JavaScript & Max
JA JavaScript test @js_test · 9.77K subscribers
const numbers = [1, 2, 3, 4, 5];
const result = numbers
.filter(n => n % 2 === 0)
.map(n => n * 2)
.reduce((acc, n) => acc + n, 0);
console.log(result);