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Mrs Tibble’s / Mr Atkins’ Year 9728 comments to Mrs Tibble’s / Mr Atkins’ Year 9 |
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thought you said something like that at the end of class
thankyou
amelia
i am confused again, the answer i am getting is 358, is that at all right?
Hi Amelia,
It’s not 358 – that’s almost 360 and the home is not as north as that.
What measurements do you have for that last triangle with D-home as the hypotenuse?
JT
teh length to dicks home is 45km and from C to D is 19.8. I think the angle at dicks home in the triangle is 34 degrees and the D angle is 90 rounded
Amelia,
CD=19.8 is correct. D to home is further than that. Can you work out the horizontal distance across the whole diagram from D? There are 3 bits to it: one is 40km right of A, one is 14km left of D and the other is the top of the triangle with BC as hypotenuse. Add those to get the base of the triangle at the top.
so is the total distance 88.6km and if it is, do i then use trig. to work out teh angle at D then add it to 270 to get the bearing?
Yes
Well done.How much work have the rest have your group done?!! That’s the kind of perseverance we like to see in our A level mathematicians.
eerrrrrrrrr, would i get away with ssaying no comment?
one final qu. is the bearing 374?
Isn’t 374 bigger than 360?
lol, yeh, help me!! i am losing the will to look at numbers! i did 19.8/88.6 then put the answer into that sin-1( I got 77.1 from that.
The 19.8 is wrong. You need to work out the vertical side below home down as far as where the line across from D meets it (above A). Then do tan of angle.
If you can’t see it straight away, bring it to me in school: don’t spend any more time on it tonight.
JT
hehe, i drew my triangle the wrong way round i thnink, i will try it and see if i get a ‘sensible’ answer, if not, see ya tommorow
lol! Okay – have fun
For the small questions on card, I am stuck on trying to work out the last triangle.
How do I work out the answer using Pythagorus’s theorm?
If a2 + b2= c2 and I fond the square route of 14, do I have to add the answers together to get the sum of both sides? The c2 is 14 so 14x 14 is 196. How do those figures help me work out the answer?
Hi Ella,
Not sure why you are trying to square-root 14. That isn’t one of the sides in the final road from D to Home.
JT
Oh, ok… I am really confused! I don’t understand why I don’t need to use 14. When you checked my work 14 was one of the sides of the triangle. However, there are 2 sides that I need to work out the sides for- how do I do that?
Can I ask you in the lesson?
Mr. Atkins what was the homework? I’ve asked people but they can’t remember!
Hi Ella,
Can’t help you there; have sent Mr A a text message so if he doesn’t reply I guess you wont have to do anything
JT
hi Mrs Tibble,
I am really stuck on the maths homework. I don’t know what to go up in for the area on the graph.
Harriet
Hi Harriet,
What’s your biggest area value? Would every big square being 5000 work?
JT
Hii
for the graph, on the bottom do we do length of sides and do we go upto 500??
Yes and yes, Emily.
Hiiii
I was wondering about the Mock exams you can get off this website, I clicked on the hyperlink and it didn’t let me get through to seeing them. Do you no why???
Zoe
Hi Zoe,
I found that too and have let Mr B know so he can sort it out.
JT
Hi Mrs Tibble,
I was a bit stuck on some of the things we have to do for revision. It was compound interest, Speed/Distance/Time, Parallel and perpendicular lines on a graph and travel graphs.
Sorry it’s quite a lot but i was wondering if you could tell me what to revise in those topics and how to revise them!!
Zoe
Hi Zoe,
Travel graphs – be able to ‘read’ the graph ie understand what the data is telling you, take values off it, calculate average speeds using data from the graph, understand the time axis (60 sec in a min not 100, sometimes it’s journey time whereas sometimes it’s time on the clock), be able to plot journey data onto the graph. If you look at the questions you did for homework (on the A3 yellow sheets) you’ll get the idea.
Compound interest – these are percentages questions where, for example, you invest some money at 4% interest p.a. for 5 years and have to work out what your account is worth after the 5 years.
Parallel and perpendicular graphs – understanding y=mx+c. m is the gradient, c is the y-intercept. Parallel lines have the same slope so the value of m will be the same in their equations. If 2 lines are perpendicular, the product of their gradients is -1. So, for example, if one line has a slope of 3, the other one will have a slope of -1/3 because -1/3 x 3 = -1.
Hope that helps.
JT
Okay thank you very much!! And also for the scatter graphs do just have to identify weather it’s a strong/weak Correlation??
ZW
Zoe, no – whether it’s positive, negative or no correlation.
I was wondering on transformations, i understand translation, reflection and rotation…. I’m just a bit confused by enlargement. I no that you have to have the scale factor but I’m not sure about the centre of enlargement and how you plot it and how you plot the enlarged/smaller shape once you have done everything else.
Zoe
Hi Zoe,
If you have to draw the enlarged shape then you will be given the coordinates of the centre of enlargement. Mark that on the diagram. Then look at the scale factor. It tells you how many times bigger and how many times further FROM THE CENTRE the enlargement should be drawn. They are on the same side of the centre if the scale factor is positive, and on opposite sides if it’s negative.
Example: scale factor 3 means 3 times as big and 3 times as far from C on the same side.
Scale factor -1/2 means half as big and half as far from C but in the opposite direction (so C will be between them)
If you have both diagrams and have to find C, join the matching corners of both shapes and extend them across the page till you find where they all cross.
Does that help?
Have a look on Bitesize for illustrations
http://www.bbc.co.uk/schools/gcsebitesize/maths/shapes/transformationshirev1.shtml
JT
Hi miss tibble, i was wondering how u work out question 10 on that mock paper you gave us ?/? … its about co ordinates … ” a straight line L, passes through the point with cordinates (4,7) and is perpendicular to the line with equation y = 2x+3
Ellie
Hi Ellie,
This is what I wrote for Zoe further up the page
Parallel and perpendicular graphs – understanding y=mx+c. m is the gradient, c is the y-intercept. Parallel lines have the same slope so the value of m will be the same in their equations. If 2 lines are perpendicular, the product of their gradients is -1. So, for example, if one line has a slope of 3, the other one will have a slope of -1/3 because -1/3 x 3 = -1.
That’s the general theory. In relation to your question, you know the line has to have an equation of the form y=mx+c. The point (4,7) lies on the line so these x and y values can be put into the equation. You work out the m value using the rules above so that can be substituted in as well. That means you can solve the equation to find out what c is.
Once you know m and c the general equation for the whole line is y=mx+c with your m and c values replacing their letters.
Have a go and then tell me what you got.
JT
hi Mrs Tibble,
I am really confused on what compound intrest is.
Harriet
Hi Harriet,
Go up to what I posted for Zoe on June 1st.
JT
thank you miss i understand it now! Harriet
Hi Mrs Tibble,
sorry i have just read the perpendicular line thing and i don’t understand it.
Harriet
Hi Harriet,
When you multiply the gradients of perpendicular lines, the answer is -1.
JT
so what type of question would come up? Harriet
Harriet, try Q10 on the practice paper (like Ellie said). It would really help if you read through everyone else’s questions and answers first before asking the same thing again!
Have fun,
JT
Sorry! ok i think i understand it! is the answer y=2X+11! Harriet
Hi Harriet,
Gradient would be 2 if the line was parallel. However, it’s perpendicular to the one with slope 2 so its slope is -1/2 (since 2 x -1/2 = -1)
Plug that into y=mx+c and you get c=9
so equation of line is y = -0.5x + 9
JT
Hello,
Sorry to interupt, but how did you get the 9?
Amelia
Hi Amelia,
The line y=2x+3 has slope 2 (the m value is the gradient) so a line perpendicular to it will have slope -1/2. This is the m value for line L that you are trying to find.
We therefore know now that for L
y=-0.5x + c
but we don’t yet know what the c for this line is.
However, we know that the point (4,7) lies on the line so its coordinates must obey the equation of the line.
Therefore, substituting in these x and y values gives
7 = -0.5 x 4 + c
7 = -2 + c
therefore c = 9
Equation of L is thus y = -0.5x + 9
JT
Ohhhhhh, thankyou
Hi Amelia,
the other way you could get the 9 would be to sketch the graph. If the slope is -1/2 then it is a downhill graph, and the y values are dropping in halves. If you start at (4,7) and work back up the slope to the y axis, when you reach x=0, y will have gone up to 9.
Hope that helps.
JT
HELLO MRS TIBBLE
How were your holidays! my computer is not letting me see the mock exam you have got for us, would it be ok if you sent it to me if I gave you an email??
also, do you choose your profile picture or do you get any automatically? because mine is really weird!!!
Hi Claire (er do I teach a Claire in year 9?)
Which paper is it you are trying to do? If it’s the one on here from last year, it won’t open for some reason unknown to me. I have passed it up to higher expertise (ie Mr B) but I don’t think even he has been able to sort it. You should have been given some practice questions before we broke up, but if not, then try the past GCSE papers online – just pick out the topics from the revision list I did.
Here’s one: https://eiewebvip.edexcel.org.uk/Reports/Confidential%20Documents/0506/5523_03_que_20050607.pdf
JT