81279 + Interview Questions in Mathematics QA in Mathematics

1.	12. The sum of two numbers is 55. If one number exceeds other by 5. Find the numbers.f 5 tables andchairs ÄŻs1560 Tf a table costs60 more than a chair cost find the cos
Answer» let the two numbers be X and and X+5henceX+X+5=552x=50X=25

Discussion

2.	The H. C. F. of two numbersis 27 and L. C. Mis 16? Ifona nf th
Answer» One number is 81 So let the other number be y We know that the product of two numbers will be equal to the product of their LCM and hcf. 81×y=27×162 y=27×162/81 y=54 The other number is 54

2.

The H. C. F. of two numbersis 27 and L. C. Mis 16? Ifona nf th

Answer»

One number is 81

So let the other number be y

We know that the product of two numbers will be equal to the product of their LCM and hcf.

81×y=27×162

y=27×162/81

y=54

The other number is 54

Discussion

3.	2. The H. C. F. and the L. C. M. of two numbers are 36 and 60900 respectivelylfone of the numbers is 1044, find the other number.
Answer»

Discussion

4.	12. If the difference between two numbers is 26, and one number is three times the other. Fthem.
Answer» Let first number be xThen other number = 3x As per given condition3x - x = 262x = 26x = 26/2 = 13 Numbers are 13 and 39

4.

12. If the difference between two numbers is 26, and one number is three times the other. Fthem.

Answer»

Let first number be xThen other number = 3x

As per given condition3x - x = 262x = 26x = 26/2 = 13

Numbers are 13 and 39

Discussion

5.	each1. If A is a square matrix of order 3 and 12A\|-kAăwrite the value of k
Answer»

Discussion

6.	which two numbers in tenths place on the number line al(b) 0.45 (c) 0.19 (d) 0.66 (e) 0.92 (fe?
Answer»

Discussion

7.	The H. C. F. of two numbers is 27 and L.C. M is 162. If one of the number is 81, findthe other.
Answer»

Discussion

8.	\frac { \operatorname { cos } A - \operatorname { sin } A } { \operatorname { cos } A + \operatorname { sin } A } - \frac { \operatorname { cos } A + \operatorname { sin } A } { \operatorname { cos } A - \operatorname { sin } A } = \frac { 4 \operatorname { tan } A } { \operatorname { tan } ^ { 2 } A - 1 }
Answer»

Discussion

9.	\left. \begin array l \operatorname sin 60 ^ \circ \operatorname cos 30 ^ \circ %2B \operatorname sin 30 ^ \circ \operatorname cos 60 ^ \circ \\ \frac \operatorname cos 45 ^ \circ \operatorname sec 30 ^ \circ %2B \operatorname cosec 30 ^ \circ \quad \text (iv) \frac \operatorname sin 30 ^ \circ %2B \operatorname tan 45 ^ \circ - \operatorname cosec 6 \operatorname sec 30 ^ \circ %2B \operatorname cos 60 ^ \circ %2B \operatorname cosec 6 \\ \frac 5 \operatorname cos ^ 2 60 ^ \circ %2B 4 \operatorname sec ^ 2 30 ^ \circ - \operatorname tan ^ 2 45 ^ \circ \operatorname sin ^ 2 30 ^ \circ %2B \operatorname cos ^ 2 30 ^ \circ \end array \right.
Answer» 5cos^2 60 + 4sec^2 30 - tan^2 45 / sin^2 30 + cos^2 30 = 5*(1/4) + 4(4/3) - 1/1 = 5/4 + 16/3 - 1 = (15 + 64 - 12)/12 = 67/12

9.

\left. \begin array l \operatorname sin 60 ^ \circ \operatorname cos 30 ^ \circ %2B \operatorname sin 30 ^ \circ \operatorname cos 60 ^ \circ \\ \frac \operatorname cos 45 ^ \circ \operatorname sec 30 ^ \circ %2B \operatorname cosec 30 ^ \circ \quad \text (iv) \frac \operatorname sin 30 ^ \circ %2B \operatorname tan 45 ^ \circ - \operatorname cosec 6 \operatorname sec 30 ^ \circ %2B \operatorname cos 60 ^ \circ %2B \operatorname cosec 6 \\ \frac 5 \operatorname cos ^ 2 60 ^ \circ %2B 4 \operatorname sec ^ 2 30 ^ \circ - \operatorname tan ^ 2 45 ^ \circ \operatorname sin ^ 2 30 ^ \circ %2B \operatorname cos ^ 2 30 ^ \circ \end array \right.

Answer»

5cos^2 60 + 4sec^2 30 - tan^2 45 / sin^2 30 + cos^2 30

= 5*(1/4) + 4(4/3) - 1/1

= 5/4 + 16/3 - 1

= (15 + 64 - 12)/12

= 67/12

Discussion

10.	\frac { \operatorname { cos } A } { \operatorname { sin } B \operatorname { sin } C } + \frac { \operatorname { cos } B } { \operatorname { sin } C \operatorname { sin } A } + \frac { \operatorname { cos } C } { \operatorname { sin } A \operatorname { sin } B } = 2
Answer»

Discussion

11.	\frac { \operatorname { sin } ( A - B ) } { \operatorname { cos } A \operatorname { cos } B } + \frac { \operatorname { sin } ( B - C ) } { \operatorname { cos } B \operatorname { cos } C } + \frac { \operatorname { sin } ( C - A ) } { \operatorname { cos } C \operatorname { cos } A } = 0
Answer» vinesh bro can u explain it

Discussion

12.	\frac { \operatorname { sin } ( B - C ) } { \operatorname { cos } B \operatorname { cos } C } + \frac { \operatorname { sin } ( C - A ) } { \operatorname { cos } C \operatorname { cos } A } + \frac { \operatorname { sin } ( A - B ) } { \operatorname { cos } A \operatorname { cos } B } = 0
Answer» Given, sin(a-b)/cos a cos b + sin(b-c)/cos b cos c + sin(c-a)/cos c cos a = sin a cos b - cos a sin b/cos a cos b + sin b cos c - cos b sin c/cos b cos c + sin c cos a - cos c sin a/cos c cos a = sin a cos b/cos a cos b - cos a sin b/cos a cos b + sin b cos c/cos b cos c - cos b sinc/cos b cos c + sin c cos a/cos c cos a - cos c sin a/cos c cos a = tan a - tan b + tan b - tan c + tan c - tan a = 0.

12.

\frac { \operatorname { sin } ( B - C ) } { \operatorname { cos } B \operatorname { cos } C } + \frac { \operatorname { sin } ( C - A ) } { \operatorname { cos } C \operatorname { cos } A } + \frac { \operatorname { sin } ( A - B ) } { \operatorname { cos } A \operatorname { cos } B } = 0

Answer»

Given, sin(a-b)/cos a cos b + sin(b-c)/cos b cos c + sin(c-a)/cos c cos a

= sin a cos b - cos a sin b/cos a cos b + sin b cos c - cos b sin c/cos b cos c + sin c cos a - cos c sin a/cos c cos a

= sin a cos b/cos a cos b - cos a sin b/cos a cos b + sin b cos c/cos b cos c - cos b sinc/cos b cos c + sin c cos a/cos c cos a - cos c sin a/cos c cos a

= tan a - tan b + tan b - tan c + tan c - tan a

= 0.

Discussion

13.	\frac { \operatorname { sin } A + \operatorname { sin } 3 A + \operatorname { sin } 5 A + \operatorname { sin } 7 A } { \operatorname { cos } A + \operatorname { cos } 3 A + \operatorname { cos } 5 A + \operatorname { cos } 7 A } = \operatorname { tan } 4 A
Answer» SinA+sin3A+sin5A+sin7A/cosA+cos3A+cos5A+cos7A=(sinA+sin7A)+(sin3A+sin5A)/(cosA+cos7A)+(cos3A+cos5A)=(2sin4Acos3A+2sin4AcosA)/(2cos4Acos3A+2cos4AcosA)=2sin4A(cos3A+cosA)/2cos4A(cos3A+cosA)=sin4A/cos4A=tan4A (Proved) thanks you

13.

\frac { \operatorname { sin } A + \operatorname { sin } 3 A + \operatorname { sin } 5 A + \operatorname { sin } 7 A } { \operatorname { cos } A + \operatorname { cos } 3 A + \operatorname { cos } 5 A + \operatorname { cos } 7 A } = \operatorname { tan } 4 A

Answer»

SinA+sin3A+sin5A+sin7A/cosA+cos3A+cos5A+cos7A=(sinA+sin7A)+(sin3A+sin5A)/(cosA+cos7A)+(cos3A+cos5A)=(2sin4Acos3A+2sin4AcosA)/(2cos4Acos3A+2cos4AcosA)=2sin4A(cos3A+cosA)/2cos4A(cos3A+cosA)=sin4A/cos4A=tan4A (Proved)

thanks you

Discussion

14.	18) Evaluate (2+1)-(2-1)
Answer» Denote √2 as "a" (a+1)^5-(a-1)^5 =(1+a)^5+(1-a)^5 =2(1+10a²+5a^4) =2(1+102+54) =2(1+20+20) =2(41) =82

14.

18) Evaluate (2+1)-(2-1)

Answer»

Denote √2 as "a"

(a+1)^5-(a-1)^5

=(1+a)^5+(1-a)^5

=2(1+10a²+5a^4)

=2(1+10*2+5*4)

=2(1+20+20)

=2(41)

=82

Discussion

15.	1. Evaluate.2
Answer»

Discussion

16.	Bisectors of interior B and exterior ACD of a A ABC intersect at the point T. Prove that BTCBAC2
Answer» thank you for giving solution for me

Discussion

17.	44.If a + b + c0, find the value of(b+e (cta) (a+by0bcсаab
Answer» given that a+b+c=0 there replace all numerator b+c= -a and c+a= -b and a+b= -c now solve as follows

Discussion

18.	Evaluate 2(5•+1•)
Answer» According to Distributive property25=10and21=210+2=12

Discussion

19.	(b) Find the value of (TRAPPCT:)a+bh+c\cta
Answer»

Discussion

20.	li a 6tc.A=11 b ctaI C at6_estz
Answer» C3>C1+C2+C3a+b+c+1 take outthen 2 columns are same so the value of derivative becomes 0hence Ans: 0 c2= c3+ç2take a+b+c common from second coloumn.Now in the det we get two coloumns are same. by properties of determiners.. If any two rows or coloumns are same in the det then the value of detreminent is 0 ..so the answer is =0

20.

li a 6tc.A=11 b ctaI C at6_estz

Answer»

C3>C1+C2+C3a+b+c+1 take outthen 2 columns are same so the value of derivative becomes 0hence Ans: 0

c2= c3+ç2take a+b+c common from second coloumn.Now in the det we get two coloumns are same. by properties of determiners.. If any two rows or coloumns are same in the det then the value of detreminent is 0 ..so the answer is =0

Discussion

21.	1. Evaluate-2
Answer»

Discussion

22.	(3x+y)(3x-y)-(3x+y)(3x+y)
Answer» in simply matter the answer is −3xy2−y3−3xy−y2 = 3x+3x-3x+3x+y-y+y+y= 6x-6x+y-3y= -2y answer (3x-y)-(3x-y)(3x-y)-3x+y -2y² is the answer for this equation in simply the solution of this question is -3xy2-y3-3xy-y3 -2y(3x+y) is correct -2y(y+3x) ans ho ga hence proved -2y(y+3x) is the answer.because the question is very easy -2y(y+3x) is your answer on ho go go for for those those those hi both both (3x+y)(3x-y)-(3x+y)(3x+y)=9x^2-y^2-(3x+y)^2=9x^2-y^2-9x^2-6xy+y^2=-6xy -2y(y+3x) is right answer according to me (3x+y)[(3x-y)-(3x+y)](3x+y)[3x-y-3x-y](3x+y)[-y-y](3x+y)(-2y)-2y(3x+y) 0,...................................... sbse phle brekit ki ek dusre m multiple kr lege.fir jo bhi cat pit k bchega vhi hmara answer hoga .for as the answer of question is -6xy

22.

(3x+y)(3x-y)-(3x+y)(3x+y)

Answer»

in simply matter the answer is −3xy2−y3−3xy−y2

= 3x+3x-3x+3x+y-y+y+y= 6x-6x+y-3y= -2y answer

(3x-y)-(3x-y)(3x-y)-3x+y

-2y² is the answer for this equation

in simply the solution of this question is -3xy2-y3-3xy-y3

-2y(3x+y) is correct

-2y(y+3x) ans ho ga hence proved

-2y(y+3x) is the answer.because the question is very easy

-2y(y+3x) is your answer

on ho go go for for those those those hi both both

(3x+y)(3x-y)-(3x+y)(3x+y)=9x^2-y^2-(3x+y)^2=9x^2-y^2-9x^2-6xy+y^2=-6xy

-2y(y+3x) is right answer according to me

(3x+y)[(3x-y)-(3x+y)](3x+y)[3x-y-3x-y](3x+y)[-y-y](3x+y)(-2y)-2y(3x+y)

0,......................................

sbse phle brekit ki ek dusre m multiple kr lege.fir jo bhi cat pit k bchega vhi hmara answer hoga .for as

the answer of question is -6xy

Discussion

23.	7 \frac{1}{3}-2 \frac{1}{4}+\left(4 \frac{1}{3}\right)-\left(2 \frac{1}{5}-\frac{3}{4}\right)
Answer» 22/3-9/4+13/3-11/5+3/4=35/3-6/4-11/5=(700-90-132)/60=478/60=239/30

Discussion

24.	(C)(-34-4)8. Factors of 9tors of 9x2 - 2x -are-(3x + 7 ) (3x + 1) (B) (3x - 2) Berth(3x+3) (3x-1) D) (3x-4) (+1)(A)
Answer» option (c) is the right answer

Discussion

25.	15. Solve the following pairs of equations3x+y + 3x-y = 4 and 2(3x + y)ä¸2(3x-y) =-8
Answer» 1/(3x+y) + 1/(3x-y)=3/4....(1) 1/2(3x+y) + 1/2(3x-y)=1/81/(3x+y) + 1/(3x-y)=1/4....(2) Subtract 1 from 2 Slope same, Parallel linesno solution

25.

15. Solve the following pairs of equations3x+y + 3x-y = 4 and 2(3x + y)ä¸2(3x-y) =-8

Answer»

1/(3x+y) + 1/(3x-y)=3/4....(1)

1/2(3x+y) + 1/2(3x-y)=1/81/(3x+y) + 1/(3x-y)=1/4....(2)

Subtract 1 from 2

Slope same, Parallel linesno solution

Discussion

26.	3x y 3x-y42(3x + y)2(3x y) 8
Answer»

Discussion

27.	1/3x+y +1/3x-y =3/4, 1/2 (3x+y) - 1/2 (3x-y) = -1/8.
Answer»

Discussion

28.	18. Solve the following pairs of equations3+=_ and3x + y3x-y2(3x +y) 2(3x-y)4
Answer» hit like if you find it useful

Discussion

29.	ULICALION. VILICAResolve into partial fractions.com(2x-1)(3x-1)Options :2x-13x-12 3X-12x-1NA 2x-13x-1
Answer» option 2 is correctwe can solve easily. just check x terms in cross multiplication. if it is equal then x terms are going to cancel.

Discussion

30.	ULICALION. VILICAResolve into partial fractions or(2x-1)(3x-1)Options :2x-13x-12 3X-12x-1NA 2x-13x-1
Answer» .... hii hi hi hi

30.

ULICALION. VILICAResolve into partial fractions or(2x-1)(3x-1)Options :2x-13x-12 3X-12x-1NA 2x-13x-1

Answer»

....

hii hi hi hi

Discussion

31.	atb13. Prove that bucctabtccuaa+bctal aa + b = 2 bbtc cbcacab
Answer» dr. Robert and the money ah are making the money and I will

Discussion

32.	g3. Evaluate:gin® 63°+sm” 27z sin’ ¢ i 17०57 17% न 005 73
Answer» PLEASE LIKE THE SOLUTION

Discussion

33.	Simplify:a+bbtccta
Answer» (x^a/ x^b)^a+b* (x^b/x^c)^b+c* (xc/x^a)^c+a= (x^a-b)^a+b* (x^b-c)^b+c* (x^c-a)^c+a= x^a2 - b2 * x^b2 -c2 * x^c2 -a2= xa2 -a2 +b2-b2 +c2-c2= x^0= 1

Discussion

34.	Solve thatatbletabtcC-a
Answer» open the power in a same medium and cut the powers and gain answers also what

34.

Solve thatatbletabtcC-a

Answer»

open the power in a same medium and cut the powers and gain answers also

what

Discussion

35.	In1. Evaluate2(G) 3
Answer» thank you kritikumar larmar😊

Discussion

36.	Evaluate:er-G厂-Gf-(ジ)Evaluate:-\| ×\|4
Answer» send a clear answer

Discussion

37.	Evaluate Gay
Answer» (a/b)⁻¹ = (b/a) {(4/3)⁻¹ - (1/4)⁻¹}⁻¹= {(3/4) - (4)}⁻¹ = {(3 - 16)/4}⁻¹ = {-13/4}⁻¹ = -4/13

37.

Evaluate Gay

Answer»

(a/b)⁻¹ = (b/a)

{(4/3)⁻¹ - (1/4)⁻¹}⁻¹= {(3/4) - (4)}⁻¹ = {(3 - 16)/4}⁻¹ = {-13/4}⁻¹ = -4/13

Discussion

38.	△ABC, ifA = 72° and <B = 63°, find LC.. In a
Answer» C will be 45° because the sum of a triangle is 180° and A = 72° and B = 63° given .So we add A and B then subtract it from 180° . Then we find C as 45° .

Discussion

39.	ヅIn the given figure, <B-65° and LC = 45° in Δ ABC and DAE ll BCAUE .If <DAB = x° and <EAC = y®, find the values of x and y.6545
Answer»

Discussion

40.	a+c() btc(B) c=a+ba +1(T) b=-2
Answer» a, b, c samantar shreni hai to 2b = a + c Hota hai, b = ( a + c)/2 k) phela option sahi hai

40.

a+c() btc(B) c=a+ba +1(T) b=-2

Answer»

a, b, c samantar shreni hai to

2b = a + c Hota hai,

b = ( a + c)/2

k) phela option sahi hai

Discussion

41.	(iii) 1253สูรุ 1: (i) 642 (ii) 325(ii) 322(ii) ( )7TIgIT : (i) 93(iii) 16-112(iii) ll!2ET.:(i) 23.254
Answer»

Discussion

42.	Write the formula for the following coordination compounds.a) potassiumhexacyanidoferrate(II)b) pentacarbonyliron(0)c) pentaamminenitrito -N-cobalt(III)iond) hexaamminecobalt(III)sulphatee) sodiumtetrafluoridodihydroxidochromate(III)
Answer» option c is correct answer. option c is the right answer option c is the right answer option, c is the right answer option c is correct answer.

42.

Write the formula for the following coordination compounds.a) potassiumhexacyanidoferrate(II)b) pentacarbonyliron(0)c) pentaamminenitrito -N-cobalt(III)iond) hexaamminecobalt(III)sulphatee) sodiumtetrafluoridodihydroxidochromate(III)

Answer»

option c is correct answer.

option c is the right answer

option, c is the right answer

option c is correct answer.

Discussion

43.	7. Express 10 cm. in kilometers.8. Find the values of the followingi) 0.03 x 1000 ) 36.8 x 10part in the l。。。9. Shade theadjacent figure
Answer» 1] 1cm = 1/100mand 1m=1/1000kmso 1cm = 1/100× 1/1000km=1/100000km You question says to convert 10cm to m and km If 1cm =1/100mthen 10cm =1/100×10=10/100=1/10m=0.1m Now if 1cm = 1/100000km then 10cm = 1/100000×10km =10/100000 =1/10000km = 0.0001km2]0.03 * 1000 = 30 , 36.8 *10 = 368

43.

7. Express 10 cm. in kilometers.8. Find the values of the followingi) 0.03 x 1000 ) 36.8 x 10part in the l。。。9. Shade theadjacent figure

Answer»

1] 1cm = 1/100mand 1m=1/1000kmso 1cm = 1/100× 1/1000km=1/100000km

You question says to convert 10cm to m and km

If 1cm =1/100mthen 10cm =1/100×10=10/100=1/10m=0.1m

Now if 1cm = 1/100000km then 10cm = 1/100000×10km =10/100000 =1/10000km = 0.0001km2]0.03 * 1000 = 30 , 36.8 *10 = 368

Discussion

44.	11) Express as Kilometers using decimals.(a) 8 m(b) 88 m.
Answer» The first one is 0.008km and second one is 0.088 km

Discussion

45.	Question numbers 23 to 30 carry 4 marks each23. A passenger, while boarding the plane, slipped from the stairs and got hu:t. The pilot took the pÄąssengerin the emergency clinic at the airport for treatment. Due to this, plane got delayed by half an hour Toreach the destination 1500 km away in time, so that the passengers could catch the connecting light,d of the plane was increased by 250 km/hour than the usual speed. What is the usual speed ofthe plane
Answer»

Discussion

46.	28. A passenger, while boarding the plane, slipped from the stairs and got hurt. The pilot took thepassenger in the emergency clinic at the airport for treatment. Due to this, the plane gotdelayed by half an hour. To reach the destination 1500 km away in time, so that the passengerscould catch the connecting flight, the speed of the plane was increased bythe usual speed. Find the usual speed of the plane.250 km/hour than
Answer» Like my answer if you find it useful!

Discussion

47.	whileboarding the plane, slipped from the stairs and got hurt. The34. A passenger,ilot took the passenger in the emergencyclinic at the airport for treatment. Duethe plane got delayed by half an hour. To reach destination 1500 km awayin time, so that the passengers could catch the connecting flight, the speed of theplane was increased by 250 km/hour than the usual speed. Find the usual speed[CBSE 2016 (D.B.)]is,of the plane.
Answer»

Discussion

48.	Question numbers 23 to 30 carry 4 marks each23. A passenger, while boarding the plane, slipped from the stairs and got hurt. The pilot took the pissengerin the emergency clinic at the airport for treatment. Due to this, plane got delayed by half an hour Toreach the destination 1500 km away in time, so that the passengers could catch the connecting flightthe speed of the plane was increased by 250 km/hour than the usual speed. What is the usul speed cfthe plane?
Answer»

Discussion

49.	4.Add the product: ala-b), b(b-c), c(c-a)5
Answer» a(a-b) + b(b-c) + c(c-a)a² - ab + b² - bc + c² - aca² + b² + c² - ab - bc - ac

Discussion

50.	13. Given the function g(x) graphed here,a. Evaluate g(-3)b. Solve g(x)=35+ALA2+
Answer» So hereg(-3)= 2thanks so hereg(-3)=2thank you

50.

13. Given the function g(x) graphed here,a. Evaluate g(-3)b. Solve g(x)=35+ALA2+

Answer»

So hereg(-3)= 2thanks

so hereg(-3)=2thank you

Discussion

Explore topic-wise InterviewSolutions in Current Affairs.

12. The sum of two numbers is 55. If one number exceeds other by 5. Find the numbers.f 5 tables andchairs ÄŻs1560 Tf a table costs60 more than a chair cost find the cos

The H. C. F. of two numbersis 27 and L. C. Mis 16? Ifona nf th

2. The H. C. F. and the L. C. M. of two numbers are 36 and 60900 respectivelylfone of the numbers is 1044, find the other number.

12. If the difference between two numbers is 26, and one number is three times the other. Fthem.

each1. If A is a square matrix of order 3 and 12A|-kAăwrite the value of k

which two numbers in tenths place on the number line al(b) 0.45 (c) 0.19 (d) 0.66 (e) 0.92 (fe?

The H. C. F. of two numbers is 27 and L.C. M is 162. If one of the number is 81, findthe other.

\frac { \operatorname { cos } A } { \operatorname { sin } B \operatorname { sin } C } + \frac { \operatorname { cos } B } { \operatorname { sin } C \operatorname { sin } A } + \frac { \operatorname { cos } C } { \operatorname { sin } A \operatorname { sin } B } = 2

\frac { \operatorname { sin } ( A - B ) } { \operatorname { cos } A \operatorname { cos } B } + \frac { \operatorname { sin } ( B - C ) } { \operatorname { cos } B \operatorname { cos } C } + \frac { \operatorname { sin } ( C - A ) } { \operatorname { cos } C \operatorname { cos } A } = 0

\frac { \operatorname { sin } ( B - C ) } { \operatorname { cos } B \operatorname { cos } C } + \frac { \operatorname { sin } ( C - A ) } { \operatorname { cos } C \operatorname { cos } A } + \frac { \operatorname { sin } ( A - B ) } { \operatorname { cos } A \operatorname { cos } B } = 0

\frac { \operatorname { sin } A + \operatorname { sin } 3 A + \operatorname { sin } 5 A + \operatorname { sin } 7 A } { \operatorname { cos } A + \operatorname { cos } 3 A + \operatorname { cos } 5 A + \operatorname { cos } 7 A } = \operatorname { tan } 4 A

18) Evaluate (2+1)-(2-1)

1. Evaluate.2

Bisectors of interior B and exterior ACD of a A ABC intersect at the point T. Prove that BTCBAC2

44.If a + b + c0, find the value of(b+e (cta) (a+by0bcсаab

Evaluate 2(5•+1•)

(b) Find the value of (TRAPPCT:)a+bh+c\cta

li a 6tc.A=11 b ctaI C at6_estz

1. Evaluate-2

(3x+y)(3x-y)-(3x+y)(3x+y)

7 \frac{1}{3}-2 \frac{1}{4}+\left(4 \frac{1}{3}\right)-\left(2 \frac{1}{5}-\frac{3}{4}\right)

(C)(-34-4)8. Factors of 9tors of 9x2 - 2x -are-(3x + 7 ) (3x + 1) (B) (3x - 2) Berth(3x+3) (3x-1) D) (3x-4) (+1)(A)

15. Solve the following pairs of equations3x+y + 3x-y = 4 and 2(3x + y)ä¸2(3x-y) =-8

3x y 3x-y42(3x + y)2(3x y) 8

1/3x+y +1/3x-y =3/4, 1/2 (3x+y) - 1/2 (3x-y) = -1/8.

18. Solve the following pairs of equations3+=_ and3x + y3x-y2(3x +y) 2(3x-y)4

ULICALION. VILICAResolve into partial fractions.com(2x-1)(3x-1)Options :2x-13x-12 3X-12x-1NA 2x-13x-1

ULICALION. VILICAResolve into partial fractions or(2x-1)(3x-1)Options :2x-13x-12 3X-12x-1NA 2x-13x-1

atb13. Prove that bucctabtccuaa+bctal aa + b = 2 bbtc cbcacab

g3. Evaluate:gin® 63°+sm” 27z sin’ ¢ i 17०57 17% न 005 73

Simplify:a+bbtccta

Solve thatatbletabtcC-a

In1. Evaluate2(G) 3

Evaluate:er-G厂-Gf-(ジ)Evaluate:-| ×|4

Evaluate Gay

△ABC, ifA = 72° and &lt;B = 63°, find LC.. In a

ヅIn the given figure, &lt;B-65° and LC = 45° in Δ ABC and DAE ll BCAUE .If &lt;DAB = x° and &lt;EAC = y®, find the values of x and y.6545

a+c() btc(B) c=a+ba +1(T) b=-2

(iii) 1253สูรุ 1: (i) 642 (ii) 325(ii) 322(ii) ( )7TIgIT : (i) 93(iii) 16-112(iii) ll!2ET.:(i) 23.254

Write the formula for the following coordination compounds.a) potassiumhexacyanidoferrate(II)b) pentacarbonyliron(0)c) pentaamminenitrito -N-cobalt(III)iond) hexaamminecobalt(III)sulphatee) sodiumtetrafluoridodihydroxidochromate(III)

7. Express 10 cm. in kilometers.8. Find the values of the followingi) 0.03 x 1000 ) 36.8 x 10part in the l。。。9. Shade theadjacent figure

11) Express as Kilometers using decimals.(a) 8 m(b) 88 m.

4.Add the product: ala-b), b(b-c), c(c-a)5

13. Given the function g(x) graphed here,a. Evaluate g(-3)b. Solve g(x)=35+ALA2+

each1. If A is a square matrix of order 3 and 12A|-kAăwrite the value of k

15. Solve the following pairs of equations3x+y + 3x-y = 4 and 2(3x + y)ä¸2(3x-y) =-8

△ABC, ifA = 72° and <B = 63°, find LC.. In a

ヅIn the given figure, <B-65° and LC = 45° in Δ ABC and DAE ll BCAUE .If <DAB = x° and <EAC = y®, find the values of x and y.6545