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Subject: Home work section
Replies: 14 Views: 1818
fanking 4.08.08 - 08:20am
Drop ya questions here nd ill see wat we can do. *
filifulu 15.12.09 - 05:02am
The base of a triangle is 3cm more than the height. The area is 35sqcm. Find its height.(show ur work) *
robinnoh 16.12.09 - 12:01pm
A=bh/2, 35=(3*3h)/2, 35*2=6h, 70=6h, 70/6=6h/6, there4 h=70/6 =11.67(4th sf) *
filifulu 17.12.09 - 09:45am
Step 1. A=b/2(h); h= (b-3) b/2 + (b/2 -3/2)=35cmsq *apply quadratics* 2(b/2)sq - 3/2=35 *multiply bth sides by 2* b2 - 3 - 70cmsq = (70 - 70)cmsq; product=-70 cm2; sum=-3 step 2. *look fr two no. Whn multiplied gives -70b and whn added gives -3b;which are -10b and +7b* step 3. *rearange the equation* b2 + 7b - 10b - 70= 0; which is the same as b(b+7)-10(b + 7)=0 (b-10) (b+7) = 0 so the base is either -7 or +10.bt rem length cnt be negative. So the base is 10 cm. The height is 10 -3cm=7cm *tht wz jst my version, i need urz* *
robinnoh 17.12.09 - 11:48am
Sori abt d 1st 1, i mss intrprt it. Newy; A=bh/2,b=h+3, there4: 35=(h+3)h/2,70=h2+3h(*2on both side), h2+3h-70=0(chng2 quadratic n dcompos it), h2+10h-7h-70=0,h(h+10)-7(h+10)=0,(h+10)(h-7)=0,there4 h z either h=-10 o h=7 bt like u sade length cnnt b -ive so h=7. *
filifulu 18.12.09 - 08:37pm
i like ur version too. More math on the way! *
filifulu 21.12.09 - 09:25am
Find the value of x in the equation (8x+1) + (23x+1) = 160 *
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