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2x+3y = 15 (x3) --> 6x+9y = 45 _
-5y = 41
y = -41/5
3x+2(-41/5) = 43
3x-82/5 = 43
3x = 215/5+82/5
3x = 297/5
x = 99/5
(x+y) = 99/5 + (-41/5) = 58/5
(x-y) = 99/5 - (-41/5) = 140/5 = 28
(x-y)(x+y) - (x+y)² = 28. 58/5 - (58/5)²
= 1624/5 - 3364/25
= 8120/25 - 3364/25
= 4756/25
2) misal bil pertama: x dan bil kedua: y, maka
x+y ≥ 100
y = 3x
substitusi
x+3x ≥ 100
4x ≥ 100
x ≥ 25
jadi batas bil pertama tidak kurang dari 25
3) eleminasi
3x+5y = 21 (x2) --> 6x+10y = 42
2x+3y = -5 (x3) --> 6x+9y = -15 _
y = 57
substitusi
3x+5(57) = 21
3x+285 = 21
3x = 21-285
3x = -264
x = -88
Hp = {(-88,57)}
4) keliling = 2(p+l)
140 = 2(p+l) --> kedua ruas dibagi 2
70 = p+l
l = 70-p
luas = p.l ≥ 600
p.(70-p) ≥ 600
70p-p² ≥ 600
-p²+70p-600 ≥ 0 kedua ruas dikali -1 --> berubah tanda
p²-70p+600 ≤ 0
(p-60)(p-10) ≤ 0
p = 60 atau p = 10
++++ - - - - - ++++ uji p = 0 diperoleh hsil +
10 60
tanda ≤ maka yg mrpkn penyelesain adlh yg bertanda --
interval panjang = { p | 10 ≤ p ≤ 60}
semoga membantu ya.. :)