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secondary 3 | E Maths
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I dont understand
Date Posted: 5 years ago
Views: 215
(a)
C: when x = 0, y = -6
C(0, 6)
A & B:
when y = 0,
x² − x - 6 = 0
(x-3)(x+2) = 0
x = 3 or -2
A(3, 0) & B(-2, 0)
(b)
ΔABC has height 6 and base 3+2 = 5
area of ΔABC = ½ x 6 x 5 = 15 sq units
(c)
length of BC = √(2² + 6²) = √40
let length of ⊥ from A to BC be AD,
area of ΔABC = 15 sq units
½ x BC x AD = 15
½ x √40 x AD = 15
AD = 30/√40
= 3/2(√10) units
C: when x = 0, y = -6
C(0, 6)
A & B:
when y = 0,
x² − x - 6 = 0
(x-3)(x+2) = 0
x = 3 or -2
A(3, 0) & B(-2, 0)
(b)
ΔABC has height 6 and base 3+2 = 5
area of ΔABC = ½ x 6 x 5 = 15 sq units
(c)
length of BC = √(2² + 6²) = √40
let length of ⊥ from A to BC be AD,
area of ΔABC = 15 sq units
½ x BC x AD = 15
½ x √40 x AD = 15
AD = 30/√40
= 3/2(√10) units
See 1 Answer
done
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I add a point D for easy reference.
Gradient 1/3 is solve from the Gradient -3 by pendicular relationship.
Solve 2 euqation by simultaneous method to get intersection point D.
Hope it help u out. All the best!!!
Gradient 1/3 is solve from the Gradient -3 by pendicular relationship.
Solve 2 euqation by simultaneous method to get intersection point D.
Hope it help u out. All the best!!!
Date Posted:
5 years ago
what if dont add 'd'?
Not necessarily must be D, u can just say is a perpendicular meet up point.
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