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= 2</b> </p> <p><img width="26" height="2" src="file:///C:/Users/jp%20computer/AppData/Local/Packages/oice_16_974fa576_32c1d314_f69/AC/Temp/msohtmlclip1/01/clip_image003.gif" v:XXXXXX="XXXXXXXXXXXX"> </p> <p> </p> <p><b>m=2</b> </p> <p> </p> <p><b>Equation in XXXXXXX is y=XX+b XXXXX b XX XXXXXXXXX</b> </p> <p> </p> <p><b>For calculating it XXX XX use XXXXX X(6,XX)</b> </p> <p> </p> <p><b>So we XXXX XXXX y=11 XXX x=X lets put in the XXXXXXXX</b> </p> <p> </p> <p><b>Y=mx+b</b> </p> <p> </p> <p><b>XX=2×6+b</b> </p> <p> </p> <p><b>b=-1</b> </p> <p> </p> <p> </p> <p><b>so XXXXXXXX is</b> <b>= 2 − 1</b> </p> <p> </p> <ul><li><b>Find XXX equation of XXX XXXXXXXXXXXXX XXXXXXXX XX line AB (Amks)</b> </li></XX> <p> </p> <p> </p> <p><b>For doing so first XXXXXXXXX slope of the XXXXXXXXXXXXX XXXXXXXX</b> </p> <p> </p> <table> <tbody><tr> <td> <p><b>XXXXXXXX of AB are (</b> </p></td><td> <p> </p></td><td> <p><b>,</b> </p></td><td> <p> </p></td><td> <p><b>)=(X,9)</b> </p></td><td> <p> </p></td><td> <p> </p></td></tr> <tr> <td> <p> </p></td><td> <p> </p></td><td> <p> </p></td><td> <p> </p></td></tr> <tr> <td> <p><b>XXXXX of XXX AB XXX calculate XXXX above points as</b> </p></td><td> <p> </p></td><td> <p><b>= = X</b> </p></td></tr> <tr> <td> <p> </p></td></tr> <tr> <td> <p><b>m=2</b> </p></td><td> <p> </p></td><td> <p> </p></td><td> <p> </p></td><td> <p> </p></td><td> <p> </p></td><td> <p> </p></td></tr> </tbody></table> <p><XXX width="X" XXXXXX="X" src="file:///C:/XXXXX/XX%20computer/AppData/Local/XXXXXXXX/oice_16_974fa576_32c1d314_f69/AC/Temp/msohtmlclip1/XX/XXXXXXXXXXXXX.jpg" v:XXXXXX="XXXXXXXXXXXX"> </p> <p> </p> <p><b>XXX XXXX XX XXX Perpendicular bisector XX−<XXX width="12" height="4" src="file:///X:/Users/XX%XXXXXXXXXX/XXXXXXX/XXXXX/XXXXXXXX/XXXXXXXXXXXXXXXXXXXXXXXXXXXXX/XX/XXXX/XXXXXXXXXXXX/XX/clip_image007.XXX" v:shapes="XXXXXXXXXXXX"> of XX so XXXX XX</b> <b>-<XXX width="X" height="4" XXX="XXXX:///X:/XXXXX/jp%20computer/XXXXXXX/Local/Packages/oice_16_974fa576_32c1d314_f69/AC/XXXX/msohtmlclip1/XX/XXXXXXXXXXXXX.jpg" v:shapes="_x0000_i1027"></b> </p> <p> </p> <p> </p> <p> </p> <p><b>XXX XX can calculate the intercept XXXXX (X, X) XXX m XX -1/2</b> </p> <p> </p> <p> </p> <p> </p> <p><b>XX get</b> </p> <p> </p> <p><b>X=− <img width="X" height="X" XXX="file:///C:/Users/XX%XXXXXXXXXX/XXXXXXX/XXXXX/Packages/XXXXXXXXXXXXXXXXXXXXXXXXXXXXX/AC/XXXX/msohtmlclip1/01/clip_image005.XXX" v:shapes="XXXXXXXXXXXX">×5+b</b> <b><br> </b> </p> <table> <tbody><tr> <td> <p><b>XXX y =-12hence XXXXX are the XXXXXXXXXX XX XXXXX X</b> </p></td></tr> </tbody></table> <p><b>b=11.5</b> </p> <p> </p> <p><b>now equation is</b> </p> <p> </p> <p> </p> <p> </p> <p><b>X= −<img width="X" height="X" XXX="file:///X:/Users/XX%20computer/AppData/Local/XXXXXXXX/XXXXXXXXXXXXXXXXXXXXXXXXXXXXX/XX/Temp/XXXXXXXXXXXX/01/XXXXXXXXXXXXX.jpg" v:XXXXXX="XXXXXXXXXXXX"> + 11.5 is</b> </p> <p> </p> <p> </p> <p> </p> <XX><li><b>Given that AC is XXXXXXXXXXXXX to AB and the equation XX XXXX XX XX y=-XX+45, find XXX co-XXXXXXXXX of</b> </li></ul> <p><b>(3ks)</b> </p> <p> </p> <p><b>XXX’s XXXX XXX equation XX XXXX AC XXXXX XX XXXXXXXXXXXXX XX XXXX XX XXXXXX equation XXXXX XX y=2x-1 perpendicular line XX XXX XXXX has a negative, XXXXXXXXXX XXXXX XX another.</b> </p> <p> </p> <p><b>So AC XXXX XXXX equation y=-X/2x-X</b> </p> <p> </p> <p> </p> <p> </p> <p><b>XX y will XX XXXX XXX XXX XXXXX c either we use XXXXXXXX XX XX or XXXXXXXX Ac so we XXX put XXXX equal</b> </p> <p> </p> <p><b>-1/XX-X=-XX+XX</b> </p> <p> </p> <p><b>2.XX=XX</b> </p> <p> </p> <p><b>X=92/X</b> </p>">
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