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secondary 3 | E Maths
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Someone help me please!! Thankssss
Date Posted: 3 years ago
Views: 181
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Alternatively, for part c, we can do guess and check, or a shorter idea as follows.
Since the gradient of AD is 5,
x-movement is k
y-movement must be 5k
By the Pythagoras' Theorem (actually, length formula itself comes from this theorem!)
k^2 + (5k)^2 = 104
k^2 + 25 k^2 = 104
26k^2 = 104
k^2 = 4
k = 2 or -2
If k = 2,
x-movement is +2
y-movement is +10
So D = (2 + 2, 5 + 10) = (4, 15)
If k = -2,
x-movement is -2
y-movement is -10
So D = (2 - 2, 5 - 10) = (0, -5)
Since the gradient of AD is 5,
x-movement is k
y-movement must be 5k
By the Pythagoras' Theorem (actually, length formula itself comes from this theorem!)
k^2 + (5k)^2 = 104
k^2 + 25 k^2 = 104
26k^2 = 104
k^2 = 4
k = 2 or -2
If k = 2,
x-movement is +2
y-movement is +10
So D = (2 + 2, 5 + 10) = (4, 15)
If k = -2,
x-movement is -2
y-movement is -10
So D = (2 - 2, 5 - 10) = (0, -5)
Date Posted:
3 years ago
Thank you! I still don’t really undedstand part C but issok i’ll ask mh teacher!
Part c is complicated.
We need to satisfy the two conditions.
A gradient of 5 means that for every 1 unit change in x, we get a 5 units change in y.
So if there is an increase in x by k units, there is an increase in y by 5k units.
Nic, I will need you to draw a right-angled triangle whose hypotenuse is AD. It is a right-angled triangle because horizontal lines are perpendicular to vertical lines.
If we let E be the third point at the right angle, then angle AED is 90 degrees.
Then, AE = k and DE = 5k.
We perform Pythagoras Theorem from there with AD being square root 104.
We need to satisfy the two conditions.
A gradient of 5 means that for every 1 unit change in x, we get a 5 units change in y.
So if there is an increase in x by k units, there is an increase in y by 5k units.
Nic, I will need you to draw a right-angled triangle whose hypotenuse is AD. It is a right-angled triangle because horizontal lines are perpendicular to vertical lines.
If we let E be the third point at the right angle, then angle AED is 90 degrees.
Then, AE = k and DE = 5k.
We perform Pythagoras Theorem from there with AD being square root 104.
I will redo part c with diagrams later
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A further explanation for part c
Date Posted:
3 years ago
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