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73051.

W’ e1o ] o TO=0l+ XS~ % | ु 1w

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x² - 3√5x + 10 = 0

x² -√5x -2√5x + 10 = 0

x(x - √5) -2√5 ( x - √5) = 0

(x - 2√5) ( x -√5) = 0

So, x = 2√5 and x = √5 are roots

73052.

Short Answer Que4. Find the zeroes of the following quadraticpolynomials and verify the relationshipbetween the zeroes and the coefficients(i) f(x) x -3x-28(ii) f(x) = 2x?-x-6[CBSE-12-MA2-032]RGREEN 100% SUCCESS!NMATHEMATICS-10

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i)x²-3x-28=f(x)f(x)=x²-7x+4x-28f(x)=x(x-7)+4(x-7)f(x)=(x+4)(x-7)zeroes are -4 , 7

ii)2x²-x-6=f(x)f(x)=2x²-4x+3x-6f(x)2x(x-2)+3(x-2)f(x)=(2x+3)(x-2)zeroes are -3/2 , 2

73053.

VERY SHORT ANSWER TYPE QUESTIONSng a circle. Find the length of BC.In fig. AABC is circumscribi3 cm9 cmin4 cmo cm21 .25c

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73054.

in the triangle ABC the bisector of Angle B and angle C intersect each other at point O prove that angle BOC barabar 90 degree 1upon 2 < A

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Ray BO is the bisector of angle CBE Therefore,angle CBO=1/2of angle CBE=1/2(180-y)=90-y/2 (1)Similarly,ray OC is the bisector of angle BCD Therefore,angle BCO=1/2 of angle BCD=1/2(180-z)=90-z/2 (2)In triangle BOC,angle BOC+BCO+CBO=180 (3)Substituting (1,2,3) you getAngle BOC+90-z/2+90-y/2=180Angle BOC=z/2+y/2=1/2(y+z)But,x+y+z=180 (angle sum property)y+z=180-xAngle BOC=1/2(180-x)=90-x/2=90-1/2angle BAC

73055.

A well d domeler 3m is dua n deep. The courts lakenout olt Was been serea evens al aand tlle shape

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73056.

5. The sum of two numbersis 18. Tlle SuiTO6.If α , β are the roots of the quadratic equation 6x2-6x + 1-0, then

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73057.

2000.7% less than the cost of the trouser, then what is the COST UNU STIIL20

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question is not come clear..

73058.

is adrgye lind he value of y301M5 is a multiple of 9 and two values are obtaincd by M. Why? Where Misagit.

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when a number is divisible by 9 sum of digits is divisible by 931M5 simplified to 3+1+5+M= 9+M= so possible M are 9, 0

73059.

git u a metres from the ground4322a33. Atoweris503 mhigh. The angle of elevation of its top from a point 50maway from its foot has measure

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As height of tower is 50√3 m

Distance of observer 50 m

In a right angle triangle perpendicular and base is given.

and we know that from trigonometric ratio

tan thetha = perpendicular/ base

tan thetha = 50√3/50tan thetha = √3

angle of elevation = tan-1 (√3)angle of elevation = 60°

73060.

Very Short Answer Type Que4. Determine which of the folii) +32 +3x +15. Use factor theorem to det(i) , p(x) = x4-

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Kindly post one question per post to experience the instant solution feature of scholr at its best.

73061.

99 = xs-x7

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2x-5x = 66-3x = 66x = 66/(-3)= -66/3= -22

2x-5x=66-3x=66x= 66/(-3)x= -66/3x= -22the answer is correct.like my answer

2 x minus 5 is equals to be minus b equals to b b x equals to 66 upon 3 is equals to minus 22 thank you

2x-5x=66-3x=66x=-66/3x=-22 is the correct answer.

-22

x=-11is the correct answer

- 22 answer please give me a like

73062.

NTEGER:8. The reading of a spring balance when a block issuspended from it in air is 60 N. This reading ischanged to 40 N when the block is submerged in water. The specific gravity of block is

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73063.

In the figure given ; angle A-75 degree angle B- 45 degree find the measure of angle ACD

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73064.

In Fig, 6.17,(0) and (i), DE| BC. Find EC in (i) and AD in (ii).1.8 cm1.5 cm 1 cm7.2 cm3 cm5.4 enFig. 6.17

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DE is parallel BcAD/DB=AE/ECso, 1.5/3=1/ECEC1.5=3Ec=3/1.5Ec=2

73065.

2. Identify the following sets as finite or infinite.(1) X = The set of all districts in Tamilnadu.(ii) Y = The set of all straight lines passing through a point.(iii) A = { x:xe Z and x <5}(iv) B = {x:x2-5x+6 = 0, xe N}

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1st finite 2nd infinite 3rd infinite 4th finite

1) finite set2) infinite set3) finite Ste4) finite set

73066.

TCISe.) .Find the equation of a straight line passing through the point (4, 5) and equallyinclined to the lines 3x 4y+7 and 5y 12x+6.ancale between the fines

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73067.

.Theratiobetweenthecurvedsurfaceand total surface of a cylinder is 1 :2. Find thevolume of the cylindet, given that its total surface area is 616 cm2.

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lovely

73068.

respectively.9 If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum offirst n terms

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73069.

\frac{2^{3} \times 3^{4} \times 4}{3 \times 32}

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Ans=27

73070.

Show that any positive odd integer is of the form 6+1, or 6q+3, or 6q+5, where q issume nteger2.-3. or

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73071.

\frac { 2 ^ { 3 } \times 3 ^ { 4 } \times 4 } { 3 \times 32 }

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73072.

11. In the given figure, AD 13 cm, BC 12 cm,AB 3 cm and angle ACD angle ABC 90Find the length of DC.

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73073.

m me given figure,. AD1 3 cm, BC st 12 on,3ent and angle ACD = angle ABC = 900Fihd the length of Do

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73074.

(1) 2 tan®45° + cos? 30° — sin? 60°sin 30° + tan 45° — cosec 60°sec 30° + cos 60° + cot 45°(४)

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Tan45=1cos30=√3/2sin60=√3/2cosec60=2/√3sec30=2/√3

hence 1)=2(1)+3/4-3/4=22)1/2+1-2/√3/2/√3+1/2+1=3/2-2/√3/2/√3+3/2=3√3-4/3√3+4

73075.

38.1&.DFind the equation of the straight lines passing through the point (1.2) and making aangle of 60 with the line x+y+2

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73076.

A park is 40m long and 30m broad. Find its area and the cost of grassing it 0Rs. 100 per square metre.

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73077.

)12.6 ellnd the radius of a circle whose circumference is given below110 cm(ii) 132 cnm(ii) 4.4 m(iv) 11 m

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73078.

\frac { \operatorname { tan } 45 ^ { \circ } } { \operatorname { cosec } 30 ^ { \circ } } + \frac { \operatorname { sec } 60 ^ { \circ } } { \operatorname { cot } 45 ^ { \circ } } - \frac { 5 \operatorname { sin } 90 ^ { \circ } } { 2 \operatorname { cos } 0 ^ { \circ } } = 0

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73079.

amount spent by RasRadhuika's mother gave her t 10.50 and her father gave her t 15.80, find the totalnount given to Radhika by the parens

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73080.

Thus, the number of. Find the value of x:. Are the following in proportion. . Are the following in continued. Find the third proportion to.. Find the fourth proportion to(a) 21:28:: x 52(a) 30, 35, 40, 45(a) 4, 6, 9(a) 9 and 4(a) 8, 12, 16The ratio of

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21:28::X:523/4=X/52X=3*52/4=3*13=39

find the value of x if x-8=21

73081.

Vill 12796 (viii) 55555 () 509025. Observe the following pattern1+3 = 221+3+5 = 321+3+5+ 7 = 42and write the value of 1+3+5 + 7 +9+... upton terms.

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73082.

( \sqrt { 32 } - \sqrt { 5 } ) ^ { 1 / 3 } ( \sqrt { 32 } + \sqrt { 5 } ) ^ { 1 / 3 }

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73083.

1. State the property that is used in each of thefollowing statements?G) If all b, then 21 45.(ii) If 24|= 1.6, then a ll b.(ifi) If 44+ 45 180°, then a ll b.4

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ans

73084.

tion should a line be drawn through the point (1, 2) so that its pointshou15, In whatofintersection with the line x + y = 4 is at a distance-from316. A straight line moves so that the sum nf t

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ho gya kr li bakwas ye tumhari bakwas ke liye nhi h agr nhi aata toh reply krna zaroori nhi h

73085.

2. Find the equation of the straight line at a distance of 3 units from the origin suchthat the perpendicular from the origin to the line makes an angle tan1with12

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73086.

5. The distance of the line 3x -y 0 from thepoint (4, 1) measured along the straight linemaking an angle of 1350 with the x-axis isg (a) 11/211/2411/22(d) 9/2

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73087.

EXERCISE 15.11. A parkis 40 m long and 30 m broad. Find its area and the cost of grassingit 1 Rs 00 per square metre.

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Thanks

73088.

5. A swimming pool is 50 m in length, 30 m in breadth and 2.5 m in depth. Find the cuof cementing its floor and walls at the rate of27 per square metre.

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73089.

ofwatercahasquarecohtaiferislue20ella0. How many litres1. The length, breadth and height of a room are in the ratio 3:2: 1.Ifits volume is 1296 m3, find its breadth.

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Volume = lbh Let l, b, h be in the ratio 3:2:1. Then, l = 3h, b = 2h. Volume = (3h)(2h)(h)=> 6h³ = 1296=> h³ = 216=> h = 6cm. Breadth = 2h = 12cm

Please hit the like button if this helped you.

73090.

tan 45Q. 21. Evaluate cosec30+sec 60cot 452 sin 90cos 60

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tan45/cosec30 + sec30/ cot45 - 2sin90/cos60= (1)/(2) + (2/V3)- (2)(1)/(1/2)=1/2+2/V3-(2/2)=1/2+2/v3-1=V3-4-V3/V3=v3(1-4-+1)/v3=2

tan45/ cosec30=1/2; sec30/cot45=2/V3/1=V3/2; 2sin90/ cos60; 2(1)/1/2=4, ; 1/2 x V3/2x4=1/2 × V3/2×4=V3/4×4=V3

73091.

I prove that Ras ARITtem' Te = I205 (II)

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LHS = tan -1(root x) = 1/2 x 2tan-1(rootx) = 1/2 x cos-1 ((1- (root x)2)/ (1 + (root x)2 ) ) = 1/2 cos-1 ( 1- x / 1+ x) = RHS

Hence proved

73092.

A fraction bears the same ratio toas u does27 7to The fraction is:9745453521

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73093.

7.Showthat2- 1 is divisible by 8, if n is an odd positive integer.18. If the zeroes of the nolvnomial 3 321 are a-h a'nt h find a and

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Ans :- Given⇒ n is an odd positive integer.Proof⇒We know that an Odd Positive Integer nis always written in the form of (4k + 1) or (4q + 3) or (4q + 5) and so on.

If n = 4k + 1Then, n² - 1 = (4k+ 1)² - 1= 16 k² + 1 + 8k - 1[∵(a + b)² = a² + b²+ 2ab]= 16k² + 8k= 8k(2k + 1)

Hence, it is divisible by 8.

If n = 4k + 3Then, n² - 1 = (4k + 3)² - 1= 16k² + 9 + 24k - 1= 16k² + 24k + 8= 8(2k² + 3k + 1)

Hence, it is also divisible by 8.

If n = 4k + 5Then, n² - 1 = (4k + 5) - 1= 16k² + 25 + 40k - 1= 16k² + 40k - 24= 8(2k² + 5k - 3)

Hence, it is also divisible by 8.

Now, For any value of n, n² - 1 is always be divisible by 8.

73094.

(viii)37x+ 29y = 45;29x + 37y=21

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73095.

EXAMPLE 6 After covering a distance of 30 km with a uniform speed there is some defect in a trainengine and therefore, its speed is reduced to 4/5 of its original speed. Consequently, the train reachesits destination late by 45 minutes. Had it happened after covering 18 kilometres more, the train wouldhave reached 9 minutes earlier. Find the speed of the train and the distance of journey

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73096.

In the figure, ABC-69 and ACB = 31°. Find BDC.

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73097.

4 The costof fencing a rectangular field atfenc30 per metre is? 2400. If the length of the field is24 m, then its breadth islaj 8 m(b) 16 m(c) 18 m(d) 24 m

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73098.

encing it at ?30 per metre. The area of a rectangular field is 3584 m2 and its length is 64 m. A boy runs atfeld at the rate of 6 km/h. How long will he take to go 5 times around 112roundurnach measuri

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73099.

A wooden sheet is 15 m long and 6 m broad. It has to be painted on bothsides. Find the cost of painting at 30 per square metre.

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Given:-the wooden sheet is 15 m long and 6 m broad.

=> Area of the wooden sheet=(15×6) m²=90 m²

∴Cost of painting both the sides at Rs 30 per square metre=Rs(2×90)×30=Rs 5400

73100.

By what number should(-3/2)^-3 be multiplied so that product is 9/8^-2

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hi