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68451.

er o/Limits, Continuity, and Series8.FtSnBy using the identityk(k +1) k k+1prove that sn =n/(n + 1), and then find the sum of the infinite senes

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prove???

i have already showed that. please specify ur question.

ohh.. that last one

thanks buddy!! you are a genius

hehehe..love to help in math..

68452.

Find the value of k, for which one root of the quadratic equatio n k x2-14x + 12is six times the other.1.0

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68453.

हे... (ाडडिड o 2% “toensBl llow fddle . cqest fl*u‘fl 3BD < ond Ak Sum|| ही पड ९00 SO नह

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Let the first term and the common difference of the A.P areaanddrespectively.

Since the A.P contains 37 terms. So, the middle most term is (37+1)/2 th term = 19thterm.

Thus, three middle most terms of this A.P.are 18th, 19thand 20thterms.

Givena18+a19+a20= 225

⇒ (a+ 17d) + (a+ 18d) + (a+ 19d) = 225

⇒ 3(a+ 18d) = 225

⇒a+ 18d= 75

⇒a= 75 – 18d… (1)According to given information

a35+a36+a37= 429

⇒ (a+ 34d) + (a+ 35d) + (a+ 36d) = 429

⇒ 3(a+ 35d) = 429

⇒ (75 – 18d) + 35d= 143

⇒ 17d= 143 – 75 = 68

⇒d= 4

Substituting the value ofdin equation (1), it is obtained

a= 75 – 18 × 4 = 3

Thus, the A.P. is 3, 7, 11, 15 …

68454.

lts area.gular field is 60 m wide and its perimeter is 320 m. If laying of grassleaving a bare path 8 m wich would it cost for grass to be laid on the field 1

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B=60mp=320mnow,2(l+b)=320l+60=160l=160-60l = 100mnow,new length =100-(2×8)=84mnew bredth=60 -16=44marea=84×44=3,696 m squrecost =3,696×7.5=27,720

68455.

A machine can fill 48 containers in 6 hours. How many containers can be filled in 48 hours?6.

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48 containers in 6hrs In 1hr it will fill 48/6 = 8 containers∴ In 48hrs it will fill 48 × 8 = 384 containers.

68456.

6/ What is the area of a square park whose side is 150 metres? Findthe total cost of planting grass in it at 10 per square metre.

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like the answer if it helps you

68457.

Tanya is planting grass in her rectangular backyard what formula could Tanya use to find how many square feet of grass she will need?

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To find the area of rectangle,

area = length*breadth in square units

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68458.

Points A(-1,y) and B(5,7) yFind the values of y. Hence, find the radius of the circle. (cBSE 201lie on a circle with centre O(2,-3

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Since both points lie on the circle, so, length AO is equal to length BO. since both are the radius of given circle.By applying distance formula,(2-(-1))² + (y-(-3y))²= (5-2)² + (7-(-3y))²⇒9+16y² =9 +(7+3y)²⇒16y²=49+9y²+42y⇒7y²-42y-49=0⇒y²-6y-7=0⇒y²-7y+y-7=0⇒y(y-7)+1(y-7)=0(y-7)=0 or (y+1)=0y=7 or -1

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68459.

CBSE 201is 168, find the value sCBSE 201347. The sum of first n terms of an A.P. is 5n2+3n. If its mth termm. Also, find the 20th term of this A.P

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sum of n termsn/2 (2a+(n-1)d)=5n^2 +3ndivide by na+(n-1)d/2=5n+3a-d/2+nd/2=5n+3comparing both sidesa-d/2=3d/2=5d=10a=8now,a+(m-1)d=168.........2put value of a and d8+(m-1)10=168(m-1)10=160m-1=16m=1720th term will bea+19d=8+19×10=198

68460.

20. Let Cbe the circle with centre (0,0) and radius 3 units. The equation of the locus of the mid points of the2Tchords of the circle C that subtend an angle of at its centre is(A)x+yx2 2744

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1

2

3 Hence, option (D) is correct.

how did the value od OP came

68461.

if a and b are roots of a quadratic equation t^2-2t+2=0 then find the value of k if (k+a)^n-(k+b)^n/a-b os equal to sin(nu)/sin(u)^n

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oi pagal question to thik se pahle

matlab kuch bhi question padhiye pahle

68462.

6. The time t, of a complete oscillation of a simplependulam of lenght ' is given by the equationT 2T,la where "g' is a gravitational constantfind approximate percentage error in 'T whenthe percentage of error in .1, is 1%

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T = 2π√1/g

→Δl/l× 100 = 1 (Given) Here we are using Percentage error formula dy/y×100 (OR) dx/x×100 → dl/l× 100 = 1

→ T = 2π√1/g

→ T = 2π(l/g)¹/²

Now, Apply Log on Both sides

→ logT = log2π(l/g)¹/²

→ logT - log2π + log(l/g)¹/²

→ logT = log2π + 1/2log(l/g)

→ logT = log2π + 1/2[log l - log g ]

→ log T = log 2π + 1/2log l - 1/2log g

→ 1/T× dt = 0 + 1/2× 1/l - 1/2× (0)

→ 1/T× dt = 1/2 dl/l

→ dT/T = 1/2 dl/l

→ dT/T× 100 = 1/2 dl/l× 100 { Apply Percentage error formula }

→ dT/T× 100 = 1/2 % { FINAL ANSWER }

68463.

How many integers between 10 and 100 are divisible by 3? Solve by a formula.

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68464.

Find how many integers between 200 and 500 are divisible by 8LTS30/1/2

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200, 208, 216........... 496

a = 200

d = 8

L = 500

=> a +(n-1)d = 496

=> 200 + (n-1)8 = 496

=> (n-1)8 = 296

=> n-1 = 37

=> n = 38

No. of integer between 200 and 500 that is divisible by 8 is 38

68465.

T IN CIO Ti of the A P 0.27,24,21, 16,llow many integers between 10 and 100 are divisible by 37 Bolve by a formula

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This means that there are30 numbers between 10 and 100 that are divisible by 3. 100 divided by 2 is 50. This means that there are50 numbersbetween 1 and 100 that are divisible by 2. However, some of these numbers are counted twice (e.g. 6 is divisible by both 3 and 2)

68466.

a) HillWhich is required as a raw material in photosynthesis in addition to water-a) CO2i NAD(d) Mineral elements

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CARBON Dioxide.The answer is Option a.

The raw materials of photosynthesis,waterandcarbon dioxide, enter the cells of the leaf, and the products of photosynthesis, sugar and oxygen, leave the leaf.Waterenters the root and is transported up to the leaves through specialized plant cells known as xylem vessels.

68467.

CBSE20127. Find the sum of first 16 terms of the AlCBSE 20110, 6, 2,

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68468.

8. FHUFind the solution of the equation 2x + 5y = 1; 2x + 3y = 3.

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68469.

1,If the sum of 7 terms of an A.P. is 49 and that of 17 terms is 289, find the sum of n terms[CBSE 2013, NCERT

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thanks for the answer

68470.

.8.Solve the quadratic equation:2x+1,

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like if u find it useful

Let 4x-3/2x+1 = yand 2x+1/4x-3 =1/y since its reciprocalso, y-10(1/y) = 3y-10/y = 3y²-10 = 3yy² - 3y - 10 = 0y² - 5y +2y - 10 = 0y(y-5) + 2(y-5) = 0(y-5)(y+2) = 0y = 5 and y = -2

For y = 5(4x-3)/(2x + 1) = 54x - 3 = 10x + 56x = - 8x = - 4/3

For y = - 2(4x-3)/(2x + 1) = - 24x - 3 = - 4x - 28x = 1x = 1/8

68471.

Find how many integers between 200 and 500 are divisible by 8CBSE

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Integers between 200 and 500 divisible by 8 are

208, 216........... 496 its an AP

a = 208

d = 8

L = 500

=> a +(n-1)d = 496

=> 208 + (n-1)8 = 496

=> (n-1)8 = 288

=> n-1 = 36

=> n = 37

No. of integer between 200 and 500 that is divisible by 8 is 37

68472.

22. If the sum of 7 terms of an A.P. is 49 and that of17termsis289,findthesumofnterms.[CBSE 2013, 2016, NCERT]

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68473.

ar lh AP and their product is 49. The sum of three consecutive terms of an AP is 21 and the sum of thesquares of these terms is 165. Find these terms.

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68474.

16. Find the curved surface area of the right circular cone whose radius is 3cm and the height is4cm. (Take t 3.14)

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68475.

16. Find the cu4cm. (Take-3.14)rved surface area of the right circular cone whose radius is 3cm and the height is

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tqu

68476.

arethere,andwhy?If the point ( 3, 4 ) lies on the graph of the equation 3y-ax + 7、find the value of a.

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super thank you

68477.

16. Find the curved surface area of the right circular cone whose radius is 3cm and the height iscm. (Take T 3.14)

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Given: Radius=3cmHeight= 4cmFormula used: Curved surface area of right circular cone=pi*r*LSolution:L=sqroot of h^2+r^2=sqoot(16+9)=5cmArea=3.14*5*3=47.1Area=47.1cmsq

68478.

13. AD is an altitude of an equilateral triangle ABC. On AD as base anothercquilateral riangle ADE is constructed. Prove that ar(AMDE) ar(AABC)-3:4

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68479.

Drawn the graph of the equation 2x+3y-12

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68480.

1. Determine k so that (3k -2), (4k-6) and (k +2) are three consecutieVeCBSE 200terms of an AP.

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68481.

b-calIf, ,- are in A.P.s then provea' b'ca+c a+bare also in A.P.、

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68482.

5, Determine k so that k2, 4k - 6 and 3k 2 are the three consecutive terms of anА.Đ.is 41 Find the 40th term.

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68483.

Which of the following statements is correct?a) A rectangle is also a parallelogram.b) A rhombus is also a parallelogram.c) A trapezium is also a parallelogram.d) A rhombus is also a kite

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Except (c) all are correct

Explanation:A rectangle is considered a special case of a parallelogram because: A parallelogram is a quadrilateral with 2 pairs of opposite, equal and parallel sides. A rectangle is a quadrilateral with 2 pairs of opposite, equal and parallel sides BUT ALSO forms right angles between adjacent sides.A rhombus is a parallelogram with four congruent sides.A trapezium is NOT a parallelogram because a parallelogram has 2 pairs of parallel sides. But a trapezium only has 1 pair of parallel sides.A kite is a quadrilateral with two adjacent pairs of sides of equal length. A rhombus is a quadrilateral with all sides of equal length. So a rhombus does have two pairs of adjacent sides of equal length and is, therefore, a kite.

68484.

3. Determine k so that k2 +4k+8,2k2 + 3k6 and 3k2 + 4k + 4 are three consecutiveterms of an A.P.

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It is given that (3k− 2),(4k− 6) and(k+ 2) are three consecutive terms of an AP.

∴(4k− 6)−(3k− 2) =(k+ 2)−(4k− 6)

⇒ 4k− 6− 3k+ 2 =k+ 2− 4k+ 6

⇒k− 4 =−3k+ 8

⇒k+ 3k=8 + 4

⇒ 4k=12

⇒k=3

Hence, the value ofkis 3.

68485.

7 Find the perimeter of the figurezem22semsem24em220m

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Perimeter of whole figure=4×(4+2+4) cm=4×10 cm=40 cm

68486.

ox. Prove thatcot A+ cosecA-I = It cosaCOLA – COSECAt sina.

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LHS:Cosec A + cot A - 1 / cot A - cosec A + 1

we know that,cosec ² A - cot ² A = 1substituting this in the numerator

= cosec A + cot A - (cosec ² A - cot ² A) / (cot A - cosec A + 1)x²-y²= (x+y)(x-y)

= cosec A + cot A - (cosec A + cot A) (cosec A - cot A) / (cot A - cosec A + 1)

= (cosec A + cot A)(1-cosec A + cot A) / (cot A - cosec A + 1)

= cosec A + cot A

= 1/sin A + cos A/sin A

= (1+cos A) / sin A

= RHS

Hence proved

68487.

1o. Find the curved surface area of the right circular cone whose radius is 3em and the height is4em. (Take 3.14)

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Curved surface area of cone = 3.14xrxlWhere r = radius of basel = Slant height of coneTo find slant height use pythagoras theoramlxl = rxr + hxhlxl = 9 + 16lxl = 25l = 5 cm

CSA of cone = 3.14x3x5= 47.1 cm^2

68488.

13. AD is an altitude of an equilateral AABC. On AD as base, another equilateralAADEisconstructed. Prove that area(AADE): area(AABC) 3:4

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68489.

what is molecular orbital

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A molecular orbital (MO) is an allowed spatial distribution of electrons in a molecule that is associated with a particular orbital energy. Unlike an atomic orbital (AO), which is centered on a single atom, a molecular orbital extends over all the atoms in a molecule or ion.

68490.

If the point (3,4) lues on the graph of the equation 3y=ax+7,find the value of a.

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3y=ax+7

pass throu 3,4

substituting,

12=3a+7

5=3a=>a=5/3

68491.

x 1, y 2 is a solution of theuation ax+ay 3, then find thelues of a:2

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Given :

a²x + ay = 3

So, x = 1 and y = 2

2a² + a - 3 = 0

2a² - 2a + 3a - 3 = 0

2a( a - 1) + 3 ( a - 1) = 0

( 2a + 3) ( a - 1) = 0

So, a = 1 and a = -3/2

68492.

8:88) रेखाओं 7x-15y-2= 0 तथा 6x + 12y-18= 0का प्रतिच्छेदन बिन्दु है :1.(D) इनमें से कोई नहीं29

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Hindi me nhii kripaya kar kk English

A is the right answer plz like

a is the right answer

68493.

(2)If α and β are the roots of ar-bx + c = 00), then calculate α + β.

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thank you

68494.

, Ifa + b + c = 12, a2 + b2 + c2-90, find the value of a3 + b3 + c3-3abc.

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thnks

68495.

28)Froman aeroplane vertically above a straight horizontal plane, the angles of depression of. two consecutive kilometer stones on the opposite sides of the aeroplane are found to be α and βShow that the height of the aeroplane istan α tan βtan α + tan β

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68496.

815. In the given figure, ABCD is aparallelogram in which AC is a diagonal.G, E and F are the mid-points of AB, BCGand AC respectively. If AGEF is anequilateral triangle, then prove thatparallelogram ABCD is a rhombus notsquare.

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68497.

Diagonal AC of a parallelogram ABCD bisectsE A (seeFig. 8.19). Show that0) it'biseets C also,(u) ABCD is a rhombus.

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68498.

Determine K, so thatF(x)=\left\{\begin{array}{cc}{\frac{x^{2}-36}{x-6}} & {\text { ; } f x \neq 6} \\ {K} & { ; x=6}\end{array}\right.F(x) is continuous at x=6

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please answer rest of the questions

68499.

Find the height of the parallelogram whose area is 54 cm2 and the base is 15 cm

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68500.

find the equality of parallelogram whose area is 52. Metre square and the base is 250 cm.

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