I need help with Question 34,24,59,60,25,30,31,27,28,46,47,48,52,53. Please explain how you arrived at the answer to. Thank You! http://www.eduweb.vic.gov.au/edulibrary/public/commrel/aboutschool/selectentry/mathspractice.pdf LINK
are those the questions numbers?
Yes
for 34) you have two points: (0,4) and (6,-1) can u compute the slope?
What do you mean. You can do anything to arrive at the answer I suppose.
Can you skip to Question 25, I am most puzzled about that on. Apparently, Pythagorean theorem is involved.
YES! the idea is work on the triangle AEG. apply pythagoras to compute EG first
What happens after I worked out EG which is 10?
now apply pythagoras again you have EG=10, AE=2 from the triangle AEG
Its a bit confusing. Can you please explain?
OK. you have a reactangle box wich means that all angles between the sides are 90º right?
Yes
now imagine you draw two lines: AG and EG the question is now what kind of triangle is AEG?
Ok, what else? Thank you, I really appreciate this, this question I just am really stuck on! What's next? The next step?
what kind of triangle is AEG?
I don't know?
what do u think is the measure of the angle between AE and the bottom of the box? the bottom of the box is the square EFGH.
90^o?
yes. if a triangle has one angle =90º, what is its name?
for #24 \(x\) is the hypotenuse of a triangle where one side is \(5-3=2\) and the other side is 4 pythagoras give the hypotenuse as \[\sqrt{2^2+4^2}=\sqrt{20}=2\sqrt{5}\]
The answer is A = 2/26 the answers say
59 is pythagoras again solve \[(x+1)^2+(x+3)^2=(x+5)^2\] \[x^2+2x+1+x^2+6x+9=x^2+10x+25\] \[2x^2+8x+10=x^2+10x+25\] \[x^2-2x-15=0\]
T
yes u r right!
Thanks everyone!
you get \[(x-5)(x+3)=0\] so \(x=5\)
Thanks!
yw
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