Please do questions 1 through 5 from Exercise 7.4 on page 155 of your textbook. Remember, put down all that you think is important in trying to find the answer to questions, even if you are unsure about what you are doing.
Sir Kerai, I am kinda getting confused with the way to find out the equation of the line when u are given the gradient. If posibbloe can u please guide me oncce again? hira hameed grade 9th
guys q5 is very easy! :D look in the equation first u put x=0 and then when u get the points plot it and then in the same equation u put y=0 u will get the other set of points... and plot it! :)
Okay, so some very interesting questions have been posted. I like Hira's response to one of the questions about how to find the x and the y intercepts with each of the equations in number 5. I think the confusion for some of you must be because two of the equations in that question do not appear to be in the format you have seen equations of lines. If the format is confusing you then you can try and make y the subject of the equation (solve for y) and see that it is the same thing. Of course, Hira's suggestion is still correct since it is telling you how to find the points. Rearranging the equation will only help you see that they are also equation of lines.
Now addressing Hira's question in the beginning. Hira, since you are given the gradient then you can already start to form part of the equation of the line, right? Like if the gradient is 2, then you know that the equation will be y = 2x + (something) . That something is the y-intercept or the value of y when x = 0. So try using the point on the line given to you and find that (Something)!
hi:). i have a question about q 3 part d.in q 3 we r given points and have to find equation. i have done all the others but in part d there is also a unknown a. . the points are:-A(a,0), B(2a, 3a). what i did is 3a-0/2a-a and getting the gradient 3a. so to find intercept i chose point B and my eq was 3a=2a(3a)+x. my problem is that when 2a and 3a are multiplied wont the result be 6a2. so inshort it would be 3a=6a2+x. and i dont get how to remove the square so that the answer would equal -3a as answer would also get square rooted.
i as typing the above comment when i think i got the answer as x should now be something which makes 6a2 9a2 so when squarerooted the answer is 3a. so x will be +3a. so i got the answer. :)
rly sorry to keep posting when i rly should just sit down and think but just to make sure i tried it with points A as well but here the eq y=squareroot(x(3a) +3a) isnt working there. the answer is coming0=3a. by that eq the answer should equal 0. help?
Falah, your question is not making much sense but I think you're worried that the answer at the back is not matching your work. Maybe you should check your gradient calculation again. Last time I checked, 2a - a was just a and then 3a/a is equal to.... THINK!
Bushra, yes it is important to sketch the graphs in question 5. Again you're not making precise ones. As for question 3d, can you tell me what part is confusing you? You have to be more specific for me to help you. Start by telling me what you have already tried.
Sameel, did I not give you all a very clear indication that we will not be following the book and its order? Then why do you have to ask such an irrelevant question?
What I am teaching you is much better for your understanding and will help you do a lot more ,even from the book, much later on. Remember, the book assumes you have already done a lot of stuff with functions from before, so it would anyway never be a good idea to follow the book. Just trust me and focus on the assignment at hand. :)
And in question 3, if you are asked to draw a graph then draw one, but if you are not, then dont!
Please refrain from insulting comments to each other with "Anonymous" as your identification. Please do not try and spoil a good medium of communication.
Sir Kerai,
ReplyDeleteI am kinda getting confused with the way to find out the equation of the line when u are given the gradient. If posibbloe can u please guide me oncce again?
hira hameed grade 9th
sir kerai,
ReplyDeletei am getting confused with question number 5. could you please help me with it!
same here i dont get question 5 t00 :(
ReplyDeleteguys q5 is very easy! :D look in the equation first u put x=0 and then when u get the points plot it and then in the same equation u put y=0 u will get the other set of points... and plot it! :)
ReplyDeleteOkay, so some very interesting questions have been posted. I like Hira's response to one of the questions about how to find the x and the y intercepts with each of the equations in number 5. I think the confusion for some of you must be because two of the equations in that question do not appear to be in the format you have seen equations of lines. If the format is confusing you then you can try and make y the subject of the equation (solve for y) and see that it is the same thing. Of course, Hira's suggestion is still correct since it is telling you how to find the points. Rearranging the equation will only help you see that they are also equation of lines.
ReplyDeleteNow addressing Hira's question in the beginning. Hira, since you are given the gradient then you can already start to form part of the equation of the line, right? Like if the gradient is 2, then you know that the equation will be y = 2x + (something) . That something is the y-intercept or the value of y when x = 0. So try using the point on the line given to you and find that (Something)!
hi:). i have a question about q 3 part d.in q 3 we r given points and have to find equation. i have done all the others but in part d there is also a unknown a. . the points are:-A(a,0), B(2a, 3a). what i did is 3a-0/2a-a and getting the gradient 3a. so to find intercept i chose point B and my eq was 3a=2a(3a)+x. my problem is that when 2a and 3a are multiplied wont the result be 6a2. so inshort it would be 3a=6a2+x. and i dont get how to remove the square so that the answer would equal -3a as answer would also get square rooted.
ReplyDeletei as typing the above comment when i think i got the answer as x should now be something which makes 6a2 9a2 so when squarerooted the answer is 3a. so x will be +3a. so i got the answer. :)
ReplyDeleterly sorry to keep posting when i rly should just sit down and think but just to make sure i tried it with points A as well but here the eq y=squareroot(x(3a) +3a) isnt working there. the answer is coming0=3a. by that eq the answer should equal 0. help?
ReplyDeleteOh alright thanku sir....I think i've got it.
ReplyDeleteI dont get ques 2 do we have to find the y intercept or the gradient?
ReplyDeletenever mind i got it :)
ReplyDeletecan we do ques no 5 in our copy? or do we have to do it in the graph book?
ReplyDeleteis it important to make a graph in ques 5 and i do not get question 3d
ReplyDeletenai it says u have to sketch it! so u can do those rough sketches sir taught us ...!
ReplyDeleteyeah i sketched it in my copy only..
ReplyDeletesame ere.
ReplyDeletei did really small sketches beside the answers.
ReplyDeletestill waiting for answer of my posts above
Falah, your question is not making much sense but I think you're worried that the answer at the back is not matching your work. Maybe you should check your gradient calculation again. Last time I checked, 2a - a was just a and then 3a/a is equal to.... THINK!
ReplyDeleteTaha, listen to Hira, just do sketches in your copy, no graph paper needed.
ReplyDeleteBushra, yes it is important to sketch the graphs in question 5. Again you're not making precise ones.
ReplyDeleteAs for question 3d, can you tell me what part is confusing you? You have to be more specific for me to help you. Start by telling me what you have already tried.
i get question 3d
ReplyDeleteyes...listen to meh :p
ReplyDeleteisnt 3a/a equal to 3a?. or is it 2a?. . cause my gradeint is 3a
ReplyDeleteFalah, this is basic algebra....
ReplyDeleteWhat is 3a/a...
I would like someone other than myself to answer this question for her....
Okay try answer the question a/a! What's that equal to?
Hira, I don't have your e-mail address yet. Please send it to me via the blog.
ReplyDeletefalah 3a/a=3 because a and a gets cancelled!!!!
ReplyDeleteSir, Why did we skip exercise 7.1-7.3?
ReplyDeleteAnd in Q.3 do we draw the graph?
because a divided by a equals 1
ReplyDeleteSameel, did I not give you all a very clear indication that we will not be following the book and its order? Then why do you have to ask such an irrelevant question?
ReplyDeleteWhat I am teaching you is much better for your understanding and will help you do a lot more ,even from the book, much later on. Remember, the book assumes you have already done a lot of stuff with functions from before, so it would anyway never be a good idea to follow the book. Just trust me and focus on the assignment at hand. :)
And in question 3, if you are asked to draw a graph then draw one, but if you are not, then dont!
it doesnt say soo...sameel...-_-
ReplyDeletei am so stupid. thanks for helping bushra
ReplyDeleteyes falah yes u r
ReplyDeleteWhose anonymous?
ReplyDeletePlease refrain from insulting comments to each other with "Anonymous" as your identification. Please do not try and spoil a good medium of communication.
ReplyDelete