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Maths, Addmaths and Statistics: Post your doubts here!

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please help with these questions
PQ = 7 cm , MN = 14 cm
Let A and B be midpoints of MN and PQ respectively.
Therefore AB = 3 cm (given), PB=BQ=3.5 cm, MA=AN=7cm
Now,
Considering Triangle OAN,
Let OA=x, ON=r,
Using Pythagoras Theorem,
(ON)^2=(OA)^2+(AN)^2
(r)^2=(x)^2+(7)^2
(r)^2=(x)^2+49...... (i)

Considering Triangle OBQ,
OB=x+3, OQ=r (same radius as ON)
Using Pythagoras Theorem,
(OQ)^2=(OB)^2+(BQ)^2
(r)^2=(x+3)^2+(3,5)^2....... (ii)

Solving Simultaneously,
(x)^2+49=(x)^2+6x+9+12.25
6x=27.75
x=4.625 cm

From (i),
(r)^2=(4.625)^2+49
(r)^2=70.39
r=8.39 cm (3 sig. fig.)

See diagram for better understanding.

Ques.1.png
 
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please help with these questions
XY= 16 cm, NP = 2 cm, OP=OX=OY (radii)
Let ON=x, OY=r,
XN=NY=8 cm (Midpoint),
OP=ON+NP
OP=x+2
Therefore,
OY=x+2
r=x+2.... (i)

Considering Triangle ONY,
Using Pythagoras Theorem,
(r)^2=(x)^2+(8)^2(r)^2=(x)^2+64.... (ii)

Solving Simultaneously,
(x+2)^2=(x)^2+64
x^2+4x+4=x^2+64
4x=60
x=15 cm

Therefore,
r=15+2
r=17 cm

Hope that helps. :)

Ques.2.png
 
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Look at these two transformations:
1. Stretch with y-axis as invariant line and stretch factor of -1.
2. Reflection in the y-axis.
Aren't 1 and 2 the same?
 
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PQ = 7 cm , MN = 14 cm
Let A and B be midpoints of MN and PQ respectively.
Therefore AB = 3 cm (given), PB=BQ=3.5 cm, MA=AN=7cm
Now,
Considering Triangle OAN,
Let OA=x, ON=r,
Using Pythagoras Theorem,
(ON)^2=(OA)^2+(AN)^2
(r)^2=(x)^2+(7)^2
(r)^2=(x)^2+49...... (i)

Considering Triangle OBQ,
OB=x+3, OQ=r (same radius as ON)
Using Pythagoras Theorem,
(OQ)^2=(OB)^2+(BQ)^2
(r)^2=(x+3)^2+(3,5)^2....... (ii)

Solving Simultaneously,
(x)^2+49=(x)^2+6x+9+12.25
6x=27.75
x=4.625 cm

From (i),
(r)^2=(4.625)^2+49
(r)^2=70.39
r=8.39 cm (3 sig. fig.)

See diagram for better understanding.

View attachment 23460
yay i had tried this yesterday and its correct :p
 
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You can write both but you MUST describe them correctly, i.e. Reflection with line of reflection y=x OR A stretch with y-axis invariant line and stretch factor -1.
 
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PQ = 7 cm , MN = 14 cm
Let A and B be midpoints of MN and PQ respectively.
Therefore AB = 3 cm (given), PB=BQ=3.5 cm, MA=AN=7cm
Now,
Considering Triangle OAN,
Let OA=x, ON=r,
Using Pythagoras Theorem,
(ON)^2=(OA)^2+(AN)^2
(r)^2=(x)^2+(7)^2
(r)^2=(x)^2+49...... (i)

Considering Triangle OBQ,
OB=x+3, OQ=r (same radius as ON)
Using Pythagoras Theorem,
(OQ)^2=(OB)^2+(BQ)^2
(r)^2=(x+3)^2+(3,5)^2....... (ii)

Solving Simultaneously,
(x)^2+49=(x)^2+6x+9+12.25
6x=27.75
x=4.625 cm

From (i),
(r)^2=(4.625)^2+4
(r)^2=70.39
r=8.39 cm (3 sig. fig.)

See diagram for better understanding.

Thanks :)
 
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i always thought that integration by parts, u-substitution and trig substitution are not in syllabus as they are not in book but i recently saw a question which has to be done by u-substitution. so does it mean that others are also included(by parts and trig substitution)
 
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i always thought that integration by parts, u-substitution and trig substitution are not in syllabus as they are not in book but i recently saw a question which has to be done by u-substitution. so does it mean that others are also included(by parts and trig substitution)
Yeah, our book doesn't have them either. Our teacher solved questions and helped us with these topics.
 
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Trigonometric Substitution is hard. I've only done I guess a couple of questions but they won't be a part of the exams surely!
 
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