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<blockquote data-quote="akolla" data-source="post: 5191365" data-attributes="member: 33077"><p>TASK 1 </p><p></p><p>(a) i can't recall anything for this, sory for that. </p><p></p><p>(b) 1. -3x2 +3x -15</p><p> 2. x4 -4x3 +2x2 -5</p><p> 3. -30x2 -31x -5</p><p> 4. 31x -29</p><p> 5. 8x2 -15x +13</p><p></p><p>(c) 1. x(m+n)</p><p> 2. x(m-n)</p><p> 3. x(mn)</p><p></p><p>(d) 1. 2(4).3(2) = 144</p><p> 2. 4x(-12)/ 2x(-6)</p><p> 3. x(2.25).Y(-3.5).z(-1.25)</p><p> 4. x(6).y(6).z(-3) what you've got know is that when the power is minus you could interchange the minus to plus. for example say you got x^(-1), then that is as same as the 1/[x^(+1)]. i.e, x^(-1) = 1/[x^(+1)].and also carefuly see and use the addition, substraction and multiplication as done in the part (c) where they applicable. </p><p></p><p></p><p>TASK 2</p><p></p><p></p><p>(a) 1. 3y(x-2y)</p><p> 2. (x-4)(x+4)</p><p> 3. (x+1)(x+2)</p><p> 4. (2x-1)(x+3)</p><p> 5. x[(x-1)2 +1] -1 this may have exact cubic factors, but am lazy to derive them and test.</p><p></p><p>(b) 5. x = -0.5 in solving this take a common factor. i.e 3(x+2). am sure 1-4 you can solve it youself, give it ago first.</p><p> </p><p>(c) 1. -7</p><p> 2. 19 </p><p></p><p>(d) from the first equation, x= (k1 - b1y)/ a1. now substitute this in the second equation and get the answer for "y". do vise-versa for "x".</p><p></p><p>(e) name the three equations 1,2 and 3 in there original order. from equation 1 and 3 derive the values of "y" and "x" interms of "z" and the real numbers. now use that values in the equation number 2 to get the value of "z". now freely substitute the value of "z" in the equation number 1 and 2 to get the values of "x" and "y".</p><p></p><p></p><p>TASK 3 </p><p></p><p>for this hopefully someone will give you the answer. if not i'll ckeck this thread later and give you the answer.</p><p></p><p></p><p>TASK 4</p><p></p><p>can't remember any equations for these shit. but i'll check for them for you if incase no one gives the answers. given the equations these are very straight forward ones, nothing to workout.</p><p></p><p></p><p>TASK 5</p><p></p><p>for most of it i've gotta refresh ma memory of them, am outa touch of these as i haven't come across them for years. probably someone will give you the answers, if not i'll refer to them later for you. but that will take more than a week atleast. coz am outa ma home these days. </p><p></p><p></p><p></p><p>there is another guy given the answers in steps for some of it. i haven't checked them, but most probably they might be right. all the sums of this first two tasks are very fundemental stuff. so you should be able to do those in your own seen done one of them by someone else. so check the answers of your own mate.even ma answers may have mistakes in them, coz i just did it ma head and type it in without serious checking or calculations. however i'll do the task 3 for you later if no one turns out to give you the answer. am not comming to EK these days at all and saw this very accidently logging after awhile, coz am out on some other work for last few weeks and the comming week. and the task 4 and 5 i'll refer to those and do when i get back to ma ususal place if you still coudn't solve them by then.</p></blockquote><p></p>
[QUOTE="akolla, post: 5191365, member: 33077"] TASK 1 (a) i can't recall anything for this, sory for that. (b) 1. -3x2 +3x -15 2. x4 -4x3 +2x2 -5 3. -30x2 -31x -5 4. 31x -29 5. 8x2 -15x +13 (c) 1. x(m+n) 2. x(m-n) 3. x(mn) (d) 1. 2(4).3(2) = 144 2. 4x(-12)/ 2x(-6) 3. x(2.25).Y(-3.5).z(-1.25) 4. x(6).y(6).z(-3) what you've got know is that when the power is minus you could interchange the minus to plus. for example say you got x^(-1), then that is as same as the 1/[x^(+1)]. i.e, x^(-1) = 1/[x^(+1)].and also carefuly see and use the addition, substraction and multiplication as done in the part (c) where they applicable. TASK 2 (a) 1. 3y(x-2y) 2. (x-4)(x+4) 3. (x+1)(x+2) 4. (2x-1)(x+3) 5. x[(x-1)2 +1] -1 this may have exact cubic factors, but am lazy to derive them and test. (b) 5. x = -0.5 in solving this take a common factor. i.e 3(x+2). am sure 1-4 you can solve it youself, give it ago first. (c) 1. -7 2. 19 (d) from the first equation, x= (k1 - b1y)/ a1. now substitute this in the second equation and get the answer for "y". do vise-versa for "x". (e) name the three equations 1,2 and 3 in there original order. from equation 1 and 3 derive the values of "y" and "x" interms of "z" and the real numbers. now use that values in the equation number 2 to get the value of "z". now freely substitute the value of "z" in the equation number 1 and 2 to get the values of "x" and "y". TASK 3 for this hopefully someone will give you the answer. if not i'll ckeck this thread later and give you the answer. TASK 4 can't remember any equations for these shit. but i'll check for them for you if incase no one gives the answers. given the equations these are very straight forward ones, nothing to workout. TASK 5 for most of it i've gotta refresh ma memory of them, am outa touch of these as i haven't come across them for years. probably someone will give you the answers, if not i'll refer to them later for you. but that will take more than a week atleast. coz am outa ma home these days. there is another guy given the answers in steps for some of it. i haven't checked them, but most probably they might be right. all the sums of this first two tasks are very fundemental stuff. so you should be able to do those in your own seen done one of them by someone else. so check the answers of your own mate.even ma answers may have mistakes in them, coz i just did it ma head and type it in without serious checking or calculations. however i'll do the task 3 for you later if no one turns out to give you the answer. am not comming to EK these days at all and saw this very accidently logging after awhile, coz am out on some other work for last few weeks and the comming week. and the task 4 and 5 i'll refer to those and do when i get back to ma ususal place if you still coudn't solve them by then. [/QUOTE]
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