Question

Kak, yolong bantu menjawab soal nomor 21 dong kk dgn penyelesaian

Answer 1

Jawaban berupa lampiran

Jawabannya

p = 7

2p = 14 (C)

mohon maaf apabila ada kesalahan

Answer 2

Diketahui:

y = 2x-√(4x-p)

ymaks = 3

Ditanyakan:

2p

Jawab:

y = 2x-√(4x-p)

y = 2x - (4x-p)^(1/2)

ymaks = y' = 0

2 - (1/2)(4)(4x-p)^(-1/2) = 0

2 - 2/√(4x-p) = 0

2 = 2/√(4x-p)

2√(4x-p) = 2

√(4x-p) = 1

kuadratkan kedua ruas

4x-p = 1

4x = 1+p

x = (1+p)/4

ymaks = 2x - √(4x-p)

3 = 2((1+p)/4) - √(4((1+p)/4)-p)

3 = ((1+p)/2)- √(1+p-p)

3 = ((1+p)/2)-√1

3+1 = (1+p)/2

4 = (1+p)/2

8 = 1+p

8-1 = p

7 = p

Maka,

2p = 2(7) = 14 (C)

Mapel: Matematika

Kelas: 11

Materi: Turunan