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| From: sahatapu | Reply 161 of 193 | Reply | |
| Subject: Contributed Answer/Explanation to Q. 17 |
55 degree
Posted at: Tue Oct 5 06:24:52 2010 (GMT)
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| From: liveusb | Reply 162 of 193 | Reply | |
| Subject: Contributed Answer/Explanation to Q. 17 |
55 degrees
Let the Angle CAD=x degreesAngle BCD=100 degrees Since, ABCD is a
quadrilateral 25+x+100=180 (Sum of opp. sides)So, x=55 degrees. Posted at: Sun Oct 31 06:47:17 2010 (GMT)
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| From: subhajitdas | Reply 163 of 193 | Reply | |
| Subject: Contributed Answer/Explanation to Q. 16 |
1
Posted at: Mon Nov 15 16:54:41 2010 (GMT)
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| From: akshitmantri | Reply 164 of 193 | Reply | |
| Subject: Contributed Answer/Explanation to Q. 17 |
55
80-25=55 Posted at: Fri Dec 3 17:21:10 2010 (GMT)
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| From: vishnuramesh | Reply 165 of 193 | Reply | |
| Subject: Contributed Answer/Explanation to Q. 17 |
55dgrees
Posted at: Sat Dec 4 01:37:22 2010 (GMT)
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| From: vishnuramesh | Reply 166 of 193 | Reply | |
| Subject: Contributed Answer/Explanation to Q. 25 |
15
Posted at: Sat Dec 4 01:47:40 2010 (GMT)
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| From: vishnuramesh | Reply 167 of 193 | Reply | |
| Subject: Contributed Answer/Explanation to Q. 29 |
68
Posted at: Sat Dec 4 01:53:40 2010 (GMT)
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| From: harshaktg | Reply 168 of 193 | Reply | |
| Subject: Contributed Answer/Explanation to Q. 7 |
19.35 square centimetres
radius of circle = 3cm
diameter of circle = 6cm (2*r)
no. of circles = 2
no. of semi-circles = 1
no. of rectangle = 1
Area of unshaded portion = Area of rectangle - ((2*(Area of circle)) + Area
of semicircle) = l*b - ( (2*(∏r2)) + ½∏r2 )
= 15 * 6 - ( (2 * (3.14 * 3 * 3)) + ½ * 3.14 * 3 * 3)
= 90 - ( (2 * 28.26) + 14.13)
= 90 - ( 56.52 + 14.13 )
= 90 - 70.65
= 19.35 cm2
Posted at: Mon Dec 13 02:34:28 2010 (GMT)
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| From: harshaktg | Reply 169 of 193 | Reply | |
| Subject: Contributed Answer/Explanation to Q. 11 |
M = [4 1 ; 1 -3]
[1-243] + 2M = 3[302-3]
2M = 3[302-3] - [1-243]
2M = [906-9] - [1-243]
2M = [822-6]
M = ½* [822-6]
M = [411-3]
Posted at: Mon Dec 13 02:42:13 2010 (GMT)
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| From: sahil_sks | Reply 170 of 193 | Reply | |
| Subject: Contributed Answer/Explanation to Q. 34 |
80 m
tan 30=20/basebase=20*3^tan 60=x/20*3^=60so height of pole = 60+20
= 80 m (ans) Posted at: Wed Dec 29 17:00:56 2010 (GMT)
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| From: shivam151 | Reply 171 of 193 | Reply | |
| Subject: Contributed Answer/Explanation to Q. 1 |
S.I = (P * R *T)/100 340 = (P * 4 *2) /100 340*100 = 8P P = 34000/8 P =
Rs. 4250
S.I = (P * R *T)/100340 = (P * 4 *2) /100340*100 = 8PP = 34000/8 P =
Rs. 4250 Posted at: Sat Jan 1 18:28:25 2011 (GMT)
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| From: shivam151 | Reply 172 of 193 | Reply | |
| Subject: Contributed Answer/Explanation to Q. 2 |
S.I = (P * R *T)/100 340 = (P * 4 *2) /100 340*100 = 8P P = 34000/8 P =
Rs. 4250
S.I = P*R*T/100340 = P*4*2/10034000 = 8PP = 34000/8P = Rs. 4250Again, S.I
= 4250 * 2 * 1\100S.I = 8500/100S.I =Rs.85Then Amount = Rs. 4250 + 85 =Rs.
4335Again,S.I. = 4335 *2*1/100S.I = Rs.( 8670/100)S.I = Rs 86.70amount =Rs.
(4335 +86.70) =Rs. 4421.70Compound Interest =Rs.(4421.70 - 4250)
= Rs. 171.70 Posted at: Sat Jan 1 18:46:11 2011 (GMT)
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| From: mukul2029 | Reply 173 of 193 | Reply | |
| Subject: Contributed Answer/Explanation to Q. 50 |
it wll be a rhombus..all sides r equal nd diagonals are per.bisectors
Posted at: Sun Jan 2 17:59:17 2011 (GMT)
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| From: simplygowda | Reply 174 of 193 | Reply | |
| Subject: Contributed Answer/Explanation to Q. 39 |
2y=x+2
Posted at: Thu Jan 6 14:57:36 2011 (GMT)
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| From: tfjames | Reply 175 of 193 | Reply | |
| Subject: Contributed Answer/Explanation to Q. 35 |
x+1/a
(x+1/a) * (x-1/a) = x2 - 1/a2 (x+1/a) * (x+1/a) = x2 + 2x/a +
1/a2 So H.C.F is x+1/a Posted at: Wed Jan 12 05:27:43 2011 (GMT)
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| From: tfjames | Reply 176 of 193 | Reply | |
| Subject: Contributed Answer/Explanation to Q. 47 |
12
In tri.ABD, AB=5 BD=4 ANG ABD=90o So AD=3 [Phythogoras
Theorem]AD * DE = CD * DB [Intersecting Chords] 3 * DE = 9 * 4 DE =
36 / 3 =12 Posted at: Wed Jan 12 05:39:27 2011 (GMT)
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| From: sooraj96 | Reply 177 of 193 | Reply | |
| Subject: Contributed Answer/Explanation to Q. 7 |
19.05
Area of 1st circle =πr2
=3.14*3*3 =28.26cm2Area of 2nd
circle=28.26cm2Area of third semi- circle=πr2/2
=3.14*3*3/2
=28.26/2
=14.13cm2Total Area=70.95cm2Area of rectangle=l*b
=15*6
=90cm2 ⇒Area of unshaded region = 90-70.95
=19.05cm2 Posted at: Wed Jan 12 08:54:50 2011 (GMT)
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| From: sooraj96 | Reply 178 of 193 | Reply | |
| Subject: Contributed Answer/Explanation to Q. 11 |
1
<p>[<font size="2" class="Apple-style-span">-4
-1</font></p><p> -1 6] </p> Posted at: Wed Jan 12 08:58:22 2011 (GMT)
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| From: sooraj96 | Reply 179 of 193 | Reply | |
| Subject: Contributed Answer/Explanation to Q. 17 |
55
Posted at: Wed Jan 12 08:59:28 2011 (GMT)
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| From: saloniadarsh2june | Reply 180 of 193 | Reply | |
| Subject: Contributed Answer/Explanation to Q. 1 |
Rs 4250
SI=Rs 340
R=4%pa
T=2 yrs
P=?
SI=P*R*T/100
340=P*4*2/100
=4250
Posted at: Sat Jan 15 07:09:03 2011 (GMT)
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