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| 71251. |
34 In fig. tangent segments PS and PT are drawn to a circle with centre O such thatSpT = 12o".Prove that OP-2 PS. |
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| 71252. |
(2) From a solid cylinder of height 14 cm and base diameter 7 cm, two equal conicalholes each of radius 2.1 cm and height 4 cm are hollowed out. Find the volume of theremaining solid. |
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| 71253. |
triangle ABC, LA = <B¡62| | then the smallest sidB) BCC) ABD) |
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Answer» In such a triangle, the shortest side is always opposite the smallest angle. (These are shown in bold colorabove) Similarly, the longest side is opposite the largest angle. In the figureabove,dragany vertex of the triangle and see that whichever side is the shortest, the opposite angle is also the smallest.so according to this nomenclature AB will be smallestas angle c is smallest |
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| 71254. |
Prt-B andared a cute kay㎞2A-BC. |
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| 71255. |
Fromasolid cylinder of height 14 cm and base diameter 7 cm, two equal24.Conicavolume of the remaining solid.Lholes each of radius 2.1 cm and height 4 cm are cut off. Find theCBSE 2011 |
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| 71256. |
Ptively. Î1ld the valul ard ican.Ifthe difference of mode and median of a data is 24, then find thedifference of median and mean.3. |
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| 71257. |
In an election between two candidates 10% of thevotes polled are invalid. A candidate got 60% ofthe valid votes and won the election by a majority5,400 valid votes. Find the total number of validvotes.(a) 20,000(c) 24,000(e) of these(b) 22,000(d) 27,000 |
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Answer» 10% are invalid so 90% are validamong valid A got 60% so another got 30%difference is 30% which is 5400so valid votes are=(5400×90)/30=16200 |
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| 71258. |
if mean=24 ,median=26, then mode=? |
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Answer» Mode = 3Median - 2Meanmode = 3(26) - 2(24) = 78 - 48 = 30 |
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| 71259. |
3IfA- cos'0 + sin'o, than prove that for all values of0, 4 SAS |
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Answer» A= sin⁴∅+cos²∅ = sin⁴∅ +1-sin²∅ for maximum and minimum value f'(∅) =0 => 4sin³∅cos∅-2sin∅cos∅=0=> 2sin∅cos∅(2sin²∅-1) = 0=> sin2∅(-cos2∅) = 0=> 1/2*(sin4∅) = 0 => 4∅ = 0 ∅ = 0, π/4, π/2 ... now value of A at ∅ = 0 is sin0+cos0 = 1and value at ∅ = π/4 is (1/√2)⁴+(1/√2)² = 1/4+1/2= 3/4value at ∅ = π/2 = 1+0 = 1 so, minimum Value is = 3/4 and maximum Value is 1 |
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| 71260. |
Math 7 Q2 R1.Write 10-Sas a fraction and decimal.ute value of 242 |
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Answer» 10 power - 5 In fraction 1/100,000 In decimal 0.00001 |
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| 71261. |
givesEXERCISE 6.2n Fig. 6.28, find the values of x and y and thenhow that AB CD.50°or130°, Fig. 6.28 |
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Answer» x + 50° = 180° ( Linear pair) x = 180° - 50° x = 130° y = 130° x = y , alternate angle are equal so AB||CD |
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| 71262. |
Prove that LPOSPRTR T |
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Answer» as angle PQR angle PQS and Angle PRT and PRQ are linear pairs of angleshence there sum is 180now is angle PQR and angle PRQ are equalthenangle PQS and Angle PRT will also be equal |
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| 71263. |
n fig., PS = PR ,TPS = LQPR. Prove that PT = PQ |
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| 71264. |
n Fig. 6.14, lines XY and MN intersoot at O. fPOY90° and a: b 2:3, find c.9y10lX.Fig. 6.14 |
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| 71265. |
inder wrien divided by Ir¡2)lind the valuby' (sby irhe polynoindal >3 +2x5x-7 when aided byT2), lind the value of k.remaindet a. ithe polynomini r3 + a 12x + 6 hen2p +4 6, then find the valye ni, 12/1~vesdivided by ( 2) leaves thedgd by (x 2) and remainder |
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Answer» R1 = x3+2x2-5kx-7R2 = x3+kx2-12x+6 for R1 x = -1 and for R2 x = 2 put the values of x R1 = (-1)^3+2(-1)^2-5k(-1) -7R1 = -1 +2 + 5k -7p = 5k - 6 R2 = (2)^3+k(2)^2-12(2)+6R2 = 8 + 4k -24 +6q = 4k - 10 2p+q = 6 2(5k-6)+(4k-10) = 6 10k-12+4k-10 = 6 14k = 6+12+10 k = 28/14 k = 2 Like my answer if you find it useful! |
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| 71266. |
OULUInder having radius and18. The volume of a cylinder having aheight 4 cm and 8 cm respectively, will128(A) 128 ar cm3 (B) cm(C) 144 a cm (D) 64 cm. |
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Answer» ans is 'A'volume of cylinder=π×r×r×h=4×4×8π=128cm3 option A is the best answer me🙋🙋 sorry 64 is the best answer me Given thatradius of a cylinder-= 4height of a cylinder= 8volume of a cylinder= ?but we know thatvolume of a cylinder= πr.rh = 4.4.8π = 64π cm³ 128 is the correct answer options D correct answer 128pi cubic centimeters c is right answer sorry a is right answer A option correct answer 128 is the right answer A) 128 cm*3 is a right answer volume of cylinder=πr²h =π4×4×8 =128π =ans is =A 128π =volume of cylinder=π×r×r×h=4×4×8π=128π volume of cylinder = π×r×r×h =π×4×4×8 =128π (A) 128 π cm³ is the best answer 128cm is the right answer (A) 128 π cm³ is the correct answer I think option 'A' is the right answer option A is best answer 😉 option a is the correct answer option A is the correct answer. A is the right answers 64π cm¶ answer your question 128π correct answer your question the answer is a 123cm3 Option (A) 128πcm3 is correct answer for me answer of this question is A ans option a right answer hoga 128 π cubic centimeter (A) 128 π cm³ is best answer volume of cylinder=r²*h*pie = 4²*8*pie = 128 pie cm³hence (A) is correct option (A) 128 π cm³ is the right answer (A) option is right in my opinino π*4*4*8= 128π A) 128π is right answer correct answer is option (A) Given that radius of cylinder = 4 cmheight of cylinder = 8 cm so volume of cylinder =128 pai (22/7 or 3.14) cubic of centimeter 0ption (A) is hundreds present correct option A is correct answer here, r=4h=8 volume of clyinder=π×r×r×h π×4×4×8 π×16×8 π128so, 128πcm3. A option is the correct answer. The correct answer of this question is A Volume of Cylinder = (A) 128 π cm³ Radius of cylinder=4cmHight of the cylinder=8cmVol of cylinder=Π×4×4×8cmC = 64cm cube 128 is correct answer your question option A is right answer. |
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| 71267. |
6.In Fig. 6.44, the side QRofAQR is produced toYouYou('TrexasidIna point S. If the bisectors of Z PQR andZ PRS meet at point T, then prove thatR s |
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| 71268. |
In Fig. 6.15, < PQR = < PRQ, then prove that<PQS = < PRT. |
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| 71269. |
In Fig. 6.15,<PQR = < PRQ, then prove that <PQS = < PRT. |
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Answer» <PQR=<PRQ<PQS=180-<PQR.......{angles on St line<PRT=180-<PRQ.......{angles on St linesince <PQR=<PRQso <PQS=<PRT |
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| 71270. |
In Fig. 6.15, PQR = < PRQ, then prove thatPQS = PRT. |
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| 71271. |
4. BEAR is a parallelogram in which ZRBO - 50°and ZEBO = 25° and ZAOR = 60°. Find ZBEO,ZORA, ZBAE and ZAER. |
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Answer» angle BEO=95' angle ORA=95'ANGLE BAE=50'ANGLE AER=10' |
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| 71272. |
Write one difference between the SAS andthe RHS rule in a right-angled triangle. |
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| 71273. |
lWnby9600votes. Find etotal number of votes polled.If Ravi's income is 60% more than that of Sushil's. What per cent is Sushil's incomeincome?The population of a town increases 15% every year. If the percent population is 20700.Findthepopulatuyear ago.A man saves 10 % of his monthly income. After a year, he finds that he has? 16200 ashissavings.Fmmonthly income..11., |
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Answer» let monthly income be y12 x y x 10/100 = 16200y = 13500 rs |
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| 71274. |
5. If the difference of mode and median of a data is 24, then find the difference of medianE 24.4and mean. |
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| 71275. |
EXERCISE 9.31. In Fig.9.23, E is any point on median AD of aA ABC. Show that ar (ABE)-ar (ACE)DC Fis the mid-point of median |
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| 71276. |
The method used for solving an assignmentproblem is Sas sectie marisd |
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Answer» The Hungarian method is a combinatorial optimization algorithm that solves the assignment problem in polynomial time and which anticipated later primal-dual methods. |
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| 71277. |
Fig. 6.14ig. 6.15, PQR PRQ, then prove thatPQS PRT |
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| 71278. |
If angle pqr =angle prq then prove that angle pqs=angleprt |
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Answer» Bhai I don't know please like 👍 plz write the question again I didn't understood |
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n Fig. 6.15, 2 POR = < PRQ, then prove thatPQS-4 PRT |
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Answer» you left your figure away. where is figure 6.15 ??? |
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| 71280. |
3. In Fig. 6.15, Z PQRZ PQS- PRTPRQ, then prove that |
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| 71281. |
Q.5. If the poly normal ft)-X4-6x + 16x2-25x t 10 įs divided by another polynomial6x 16x25x 10 is divided by another polynomialr. 2x + K the remainder comes out to be x ㅜ a. Find K and a |
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| 71282. |
Two acute angles of aligiiFind the marked angles as shown in the figure50°40°50DB |
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Answer» x+20+40=180 (In triangle ABC) x=180-60 x=120y=20+40 (Exterior opposite angle is sum of interior angles)y=60 y+z+50=180 ( In the smaller triangle) 60+50+z=180 z=180-110z=70 i. x=180°-(40°+20°){angle sum property of a triangle} =180°-60° =120°. y=180°-x{linear pair} =180°-120° =60°. z=180°-(50°+y) =180°-(50°+60°) =180°-110° =70°. ii.click the picture correctly |
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| 71283. |
polynomial 616r2 -25x+ 10 is divided by another polynomial 2 -2inder comes out to be x + a, find k and a. |
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| 71284. |
3.The difference between two acute angles of a right angle triangle israd. Find the angles in degree. |
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Answer» Let, two acute angles be x and y so, ATQ, X - y= 20°......(1) x+ y= 90°......(2) solving 1 & 2, we get; =>2x= 110° => x= 55° putting the value of x in 2, we get; => 55°+ y= 90° => y= 35° so, Two acute angles are 55° & 35° |
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| 71285. |
11. Can a triangle have(i) two right angles?(ii) two obtuse angles?(iii) two acute angles?(iv) each angle more than 60°?(v) each angle less than 60°?(vi) each angle equal to 60°? |
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| 71286. |
The difference between two acute angles of right angled triangle is2 \pi^{\mathrm{C}} / \mathrm{s}Find the angles in degees. |
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| 71287. |
The difference between two acute angles of a rightangled triangle is 3- Find the angles in degrees.(9)10 |
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Answer» thanks |
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| 71288. |
g || How mory 00» dew o M?,uflm}\ polggen किक|E san रथ 1. |
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8. In the given figure, ADAand AD2-BD Ă EC.Prove that triangles ABD and CAE aresimilar. |
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| 71290. |
The monthly income (in rupees) of 7 houscholds in a villinare 1200, 1500, 1400, 1000,000, 1600, 10000. ) Find thee median income of the house holds. (ii) If one moreincomeotousehold with monthly income of 1500 is added, what will the median income be? |
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Answer» How to find median :1. Put all the numbers in numerical order.2. If there is an odd number of results, the median is the middle number.3. If there is an even number of results, the median will be themeanof the two central numbers. given, 1200, 1500, 1400, 1000, 1000,1600, 10000.now one more household with monthly income of Rs. 1500.then there are 8 results and in numerical order : 1000, 1000, 1200, 1400, 1500,1500, 1600, 10000.here number of results is 8(even) so, median = (1400+ 1500)/2 = 1450 Hence, median = 1450 |
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6cmFig. 13.22Look at the following figures. By the applicatiof SAS congruence condition state whichcongruent:4.65°8 cm8 cm |
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Answer» BC=EFAngle A=Angle DBA=FD Therefore, by SAS congruency,Both triangles are congruent. |
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| 71292. |
How is the SAS rule different from the ASA rule? |
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Answer» SASstands for “side, angle, side” and means that we have two triangles where we know two sides and the included angle are equal. If two sides and the included angle of one triangle are equal to the corresponding sides and angle of another triangle, the triangles are congruent. ASA (angle, side, angle) ASAstands for “angle, side, angle” and means that we have two triangles where we know two angles and the included side are equal. If two angles and the included side of one triangle are equal to the corresponding angles and side of another triangle, the triangles are congruent. |
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| 71293. |
Write in symbolic form howtriangles PQS and PRS arecongruent by the SAS rule. G |
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| 71294. |
Use the given information to mark the diagram appropriately. Name thetriangle congruence and then identify the theorem or postulate (SSS, SAS,ASA, AAS, HL) that would be used to prove the triangles congruent.c,Given: IN M NAa MKReason:Reason: |
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| 71295. |
Q1. If a and B are the zeroes of the polynomial x2 - 5x + k and a - B = 1, then find k. |
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| 71296. |
5. If the polynomial x-6+16x2-25x + 10 is divided by another polynomial x2-2x+k,the remainder comes out to be x + a, find k and a. |
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Answer» LET x^4–6x³+16x²-25x+10 = q(x)(x²-2x+k)+(x+a)..【q(x≡ quotient 】 x^4–6x³+16x²-26x+(10-a) =(x²-2x+k)(x²-4x+m)【-6x³=-2x³-4x³】 Now,(i) comparing x-terms,2m+4k=26orm+2k=13...(A) (ii) comparing x²-terms ,m+k+8=16m+k=8m+2k=13 k=5m=3 BUT comparing constant terms, mk=10-a=15a=10–15=-5 |
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| 71297. |
r 150,-53 are 2 ÂąV3, Find other zeroes.If the polynomial x4-6r3 + 16x2remainder is x+ a, find k and a.25x + 10 is divided by another polynomial x2-2x+k, then |
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| 71298. |
e hypotenuse of a right triangle with its legs of lengh 3x and 4xof a right triangle with its legs of length 3x and 4x5x(b) 7x(c) 16x |
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Answer» sir I am asking the hypotenus of the triangle Using PGT √(9+16) x = √25 x = 5 x. thank you so much |
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| 71299. |
3Example 37. If α and β are the zeroes of the polynomial x2-5x + k, suchthat a -B- 1 then find the value of k. |
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| 71300. |
22 + b2 = 12Is the triangle a right triangle? |
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Answer» 8^2 + 15^2 = 17^264 + 225 = 289289 = 289 Yes, triangle is right triangle |
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