332354 + Interview Questions in Secondary School in Math

1.	1.\(c) 48(a) 18(b) 21(f) 124(d) 63)(e) 100Write all the factors of the given numbers:
Answer» PLEASE MARK me as BRAINLIEST

1.

1.\(c) 48(a) 18(b) 21(f) 124(d) 63)(e) 100Write all the factors of the given numbers:

Answer»

PLEASE MARK me as BRAINLIEST

Discussion

2.	Successor of the given number.10, 00, 001
Answer» Answer: Successor of a GIVEN number is 1 more than the given number. ... For example, 9,99,99,999 is predecessor of 10,00,00,000 or we can ALSO SAY 10,00,00,000 is the successor of 9,99,99,999.

2.

Successor of the given number.10, 00, 001

Answer»

Answer:

Successor of a GIVEN number is 1 more than the given number. ... For example, 9,99,99,999 is predecessor of 10,00,00,000 or we can ALSO SAY 10,00,00,000 is the successor of 9,99,99,999.

Discussion

3.	In a quad. ABCD ANGLE A + ANGLE D =90 DEGREE. THEN PROVE THAT AC^2 + BD^2=BC^2 + AD^2.
Answer» ANSWER: klmklmklmklmklmklmklmk

3.

In a quad. ABCD ANGLE A + ANGLE D =90 DEGREE. THEN PROVE THAT AC^2 + BD^2=BC^2 + AD^2.

Answer»

ANSWER:

klmklmklmklmklmklmklmk

Discussion

4.	The ratio of the present ages of Anjali and Ashu is 2:3 Five year hence , the ratio of their ages will be 3:4 find their present ages. pls answer me
Answer» Let Anjali and Anshu 's present ages be 2x yrs and 3X yrs respectively ATQ, 2x+5 / 3x+5 = 3/4 or, 4(2x+5) = 3( 3x+5) or, 8x +20= 9x +15 or, -15+20= 9x-8x or, 5=x Therefore, Anjali's present AGE= 2x = 2×5= 10 yrs Anshu's present age = 3x= 3×5= 15 yrs pls mark me as the BRAINLIEST!! Hope it helps

4.

The ratio of the present ages of Anjali and Ashu is 2:3 Five year hence , the ratio of their ages will be 3:4 find their present ages. pls answer me

Answer»

Let Anjali and Anshu 's present ages be 2x yrs and 3X yrs respectively

ATQ, 2x+5 / 3x+5 = 3/4

or, 4(2x+5) = 3( 3x+5)

or, 8x +20= 9x +15

or, -15+20= 9x-8x

or, 5=x

Therefore,

Anjali's present AGE= 2x = 2×5= 10 yrs

Anshu's present age = 3x= 3×5= 15 yrs

pls mark me as the BRAINLIEST!!

Hope it helps

Discussion

5.	The pave of equation y=3 and y=-5havesolution
Answer» Answer: -5 - 3 = -5 - 3= -2 Step-by-step explanation: I HOPE THIS WILL HELP YOU!!

5.

The pave of equation y=3 and y=-5havesolution

Answer»

Answer:

-5 - 3 =

-5 - 3= -2

Step-by-step explanation:

I HOPE THIS WILL HELP YOU!!

Discussion

6.	Find the roots of the following quadratic equations by the factorisation method:
Answer» Answer:1)2/3,-3/2 2)3,-1/2 3)√2,-1/3√2 4)√5/3,-2√5 5)1/21,1/21 Step-by-step EXPLANATION:

6.

Find the roots of the following quadratic equations by the factorisation method:

Answer»

Answer:1)2/3,-3/2

2)3,-1/2

3)√2,-1/3√2

4)√5/3,-2√5

5)1/21,1/21

Step-by-step EXPLANATION:

Discussion

7.	The ratio of the present ages of Anjali and Ashu is 2:3 Five year hence , the ratio of their ages will be 3:4 find their present ages.
Answer» EXPLANATION. To find the present age. The ratio of the present age of ANJALI and Ashu = 2:3 Let the age of Anjali = 2x Let the age of ashu be = 3x 5 years hence, Let the age of Anjali = 2x + 5 Let the age of Ashu = 3x + 5 The ratio of their age will be = 3:4 => 2x + 5 / 3x + 5 = 3/4 => 4 ( 2x + 5 ) = 3 ( 3x + 5 ) => 8x + 20 = 9x + 15 => x = 5 THEREFORE, present age of Anjali = 2x => 2 X 5 = 10 years. Present age of ashu = 3x => 3 X 5 = 15 years.

7.

The ratio of the present ages of Anjali and Ashu is 2:3 Five year hence , the ratio of their ages will be 3:4 find their present ages.

Answer»

EXPLANATION.

To find the present age.

The ratio of the present age of ANJALI and

Ashu = 2:3

Let the age of Anjali = 2x

Let the age of ashu be = 3x

5 years hence,

Let the age of Anjali = 2x + 5

Let the age of Ashu = 3x + 5

The ratio of their age will be = 3:4

=> 2x + 5 / 3x + 5 = 3/4

=> 4 ( 2x + 5 ) = 3 ( 3x + 5 )

=> 8x + 20 = 9x + 15

=> x = 5

present age of Anjali = 2x

=> 2 X 5 = 10 years.

Present age of ashu = 3x

=> 3 X 5 = 15 years.

Discussion

8.	the average of 25 obsevations is 27. if one observation is included the average still remains 27 .find the included observation
Answer» Answer: 0 hope it HELPS you

8.

the average of 25 obsevations is 27. if one observation is included the average still remains 27 .find the included observation

Answer»

Answer:

0

hope it HELPS you

Discussion

9.	Q2 LCM of two numbers is a multipleof their HCF
Answer» Answer: Yes ... this STATEMENT is true The LCM of two NUMBERS is a MULTIPLE of their HCF

9.

Q2 LCM of two numbers is a multipleof their HCF

Answer»

Answer:

Yes ... this STATEMENT is true

The LCM of two NUMBERS is a MULTIPLE of their HCF

Discussion

10.	Represent 5 / 11 on the number line
Answer» Answer: Make this answer brainliest Step-by-step EXPLANATION: HOPE this HELPS you Solution is ATTACHED

10.

Represent 5 / 11 on the number line

Answer»

Answer:

Make this answer brainliest

Step-by-step EXPLANATION:

HOPE this HELPS you

Solution is ATTACHED

Discussion

11.	1 + SinA/cosA=1+sinA+cosA/cosA+1-sinA
Answer» (SINA + 1 - cosA) / (cosA - 1 + sinA) Divide each term by cosA (TANA + SECA - 1) / (1 - secA + tanA) =[tanA + secA - ( sec2A−tan2A )] / ( 1 - secA + tanA) = (tanA+secA)( 1 - secA + tanA )/(1 - secA + tanA) = tanA + secA = (sinA / cosA )+ 1 / cosA = (sinA + 1 )/ cosA Hence Proved

11.

1 + SinA/cosA=1+sinA+cosA/cosA+1-sinA

Answer»

(SINA + 1 - cosA) / (cosA - 1 + sinA)

Divide each term by cosA

(TANA + SECA - 1) / (1 - secA + tanA)

=[tanA + secA - ( sec2A−tan2A )] /

( 1 - secA + tanA)

= (tanA+secA)( 1 - secA + tanA )/(1 - secA + tanA)

= tanA + secA

= (sinA / cosA )+ 1 / cosA

= (sinA + 1 )/ cosA

Hence Proved

Discussion

12.	In an arthamatic progression Tn=3n-1 then t is
Answer» ANSWER: you SEE in pic Step-by-step EXPLANATION: MARK as a BRAINILIST

12.

In an arthamatic progression Tn=3n-1 then t is

Answer»

ANSWER:

you SEE in pic

Step-by-step EXPLANATION:

13.	Go sa gu 160,165,305
Answer» ANSWER: SORRY MATE which SUB is THOS

Go sa gu 160,165,305

Answer»

ANSWER:

SORRY MATE which SUB is THOS

Discussion

14.	Find the value of θ in each of the following. (θ is an acute angle.) (i) sec 2θ = 2 √3 (ii) 4 cot 3θ - 4 = 0 (iii) 3 tan2 θ + 2 = 3
Answer» Find the VALUE of θ in each of the following. (θ is an ACUTE angle.) (i) SEC 2θ = 2/√3 (ii) 4 COT 3θ - 4 = 0 (iii) 3 tan² θ + 2 = 3 [correct question] Given: (i) sec 2θ = 2/√3 (ii) 4 cot 3θ - 4 = 0 (iii) 3 tan² θ + 2 = 3 To find: The value of the θ. Solution: (i) sec 2θ = 2/√3 sec 2θ = sec 30° 2θ = 30° θ = 15° (ii) 4 cot 3θ - 4 = 0 4 cot 3θ = 4 cot 3θ = 1 cot 3θ = cot 45° 3θ = 45° θ = 15° (iii) 3 tan² θ + 2 = 3 3 tan² θ = 3-2 3 tan² θ =1 tan² θ = 1/3 tanθ = √1/3 tanθ = 1/√3 tanθ = tan 30° θ = 30°

14.

Find the value of θ in each of the following. (θ is an acute angle.) (i) sec 2θ = 2 √3 (ii) 4 cot 3θ - 4 = 0 (iii) 3 tan2 θ + 2 = 3

Answer»

Find the VALUE of θ in each of the following. (θ is an ACUTE angle.) (i) SEC 2θ = 2/√3 (ii) 4 COT 3θ - 4 = 0 (iii) 3 tan² θ + 2 = 3 [correct question]

Given:

(i) sec 2θ = 2/√3

(ii) 4 cot 3θ - 4 = 0

(iii) 3 tan² θ + 2 = 3

To find:

The value of the θ.

Solution:

(i) sec 2θ = 2/√3

sec 2θ = sec 30°

2θ = 30°

θ = 15°

(ii) 4 cot 3θ - 4 = 0

4 cot 3θ = 4

cot 3θ = 1

cot 3θ = cot 45°

3θ = 45°

θ = 15°

(iii) 3 tan² θ + 2 = 3

3 tan² θ = 3-2

3 tan² θ =1

tan² θ = 1/3

tanθ = √1/3

tanθ = 1/√3

tanθ = tan 30°

θ = 30°

Discussion

15.	If A x : x N and x 100      and B = {y : y  N and y is a multiple of 2 less than 100}, find A B
Answer» ANSWER: PLEASE mark me BRAINLIST and follow me in brainly

15.

If A x : x N and x 100      and B = {y : y  N and y is a multiple of 2 less than 100}, find A B

Answer»

ANSWER:

PLEASE mark me BRAINLIST and follow me in brainly

Discussion

16.	X=_/3 ,y=6_/2 z=4_/7if x is equals to root 3 Y is equals to 62 to Z is equals to 4 root 7
Answer» ANSWER: ASK GOOGLE Step-by-step EXPLANATION: UNDERSTOOD

16.

X=_/3 ,y=6_/2 z=4_/7if x is equals to root 3 Y is equals to 62 to Z is equals to 4 root 7

Answer»

ANSWER:

ASK GOOGLE

Step-by-step EXPLANATION:

UNDERSTOOD

Discussion

17.	Please tell me this answer I make you brainlist
Answer» Answer: thank you for your time and consideration and I will let you PLEASE let us do you want a ride with us if you have a good Morning Everyone you have any idea of your schedule this video was your age batao of my life with 2 chapter 2 ke liya Kya h yo you can get it again and I look good Morning Everyone you have a chance could you send krne ke pass it on that yo want a refund I want or I can just like tik Tok you please give me and I can get you a check or CASH nhi aa rha h kya you have gone or not SURE how are things at the meeting tonight or suna of your schedule for the week of your thoughts and if so CUTE I look good in this video was your night and consideration for your call at the meeting tomorrow m and m not getting any questions or KOI

17.

Please tell me this answer I make you brainlist

Answer»

Answer:

thank you for your time and consideration and I will let you PLEASE let us do you want a ride with us if you have a good Morning Everyone you have any idea of your schedule this video was your age batao of my life with 2 chapter 2 ke liya Kya h yo you can get it again and I look good Morning Everyone you have a chance could you send krne ke pass it on that yo want a refund I want or I can just like tik Tok you please give me and I can get you a check or CASH nhi aa rha h kya you have gone or not SURE how are things at the meeting tonight or suna of your schedule for the week of your thoughts and if so CUTE I look good in this video was your night and consideration for your call at the meeting tomorrow m and m not getting any questions or KOI

Discussion

18.	(a+b)^3-a-b factorize that question
Answer» ANSWER: answer for the GIVEN PROBLEM is given

18.

(a+b)^3-a-b factorize that question

Answer»

ANSWER:

answer for the GIVEN PROBLEM is given

Discussion

19.	Multiply with reciprocal of -8 by 9 is ____
Answer» HERE IS UR ANSWER DEAR, reciprocal of -8= -1/8 *MULTIPLICATION*: =-(-1/8) × 9 = (-9/8) = 1.125 $#MISS PHENOMENOL❤❤$

19.

Multiply with reciprocal of -8 by 9 is ____

Answer»

HERE IS UR ANSWER DEAR,

reciprocal of -8= -1/8

MULTIPLICATION:

=-(-1/8) × 9

= (-9/8)

= 1.125

$#MISS PHENOMENOL❤❤$

Discussion

20.	Box 1 contains three cards bearing numbers 1, 2, 3; box 2 contains five cards bearing numbers 1, 2, 3, 4, 5; and box 3 contains seven cards bearing numbers 1, 2, 3, 4, 5, 6, 7. A card is drawn from each of the boxes. Let xi be the number on the card drawn from 'i' th box i=1,2,3. a, The probability that x1+x2+x3 is odd is. b, The probability that x1, x2 and x3 are in the arithmetic progression, is
Answer» Answer: Let be the number on the card drawn from the ith box, i = 1, 2, 3. The PROBABILITY that is ODD is The probability that are in an aritmetic progression is Step-by-step explanation: follow!!!!!!!!!

20.

Box 1 contains three cards bearing numbers 1, 2, 3; box 2 contains five cards bearing numbers 1, 2, 3, 4, 5; and box 3 contains seven cards bearing numbers 1, 2, 3, 4, 5, 6, 7. A card is drawn from each of the boxes. Let xi be the number on the card drawn from 'i' th box i=1,2,3. a, The probability that x1+x2+x3 is odd is. b, The probability that x1, x2 and x3 are in the arithmetic progression, is

Answer»

Answer:

Let

be the number on the card drawn from the ith box, i = 1, 2, 3. The PROBABILITY that

is ODD is The probability that

are in an aritmetic progression is

Step-by-step explanation:

follow!!!!!!!!!

Discussion

21.	Give me some codes for stylish text i am REQUESTING to everyone .please help
Answer» ANSWER: $\huge{\orange{\boxed{\boxed{\purple{\underline{\underline{\red{\mathfrak{1}}}}}}}}}$ $\huge\boxed{\fcolorbox{black}{pink}{<klux>2</klux>}}$ I'm having only 2 CODES

21.

Give me some codes for stylish text i am REQUESTING to everyone .please help

Answer»

ANSWER:

$\huge{\orange{\boxed{\boxed{\purple{\underline{\underline{\red{\mathfrak{1}}}}}}}}}$

$\huge\boxed{\fcolorbox{black}{pink}{<klux>2</klux>}}$

I'm having only 2 CODES

Discussion

22.	ଗୋଟିଏ ଆୟତାକୃତି ବିଶିଷ୍ଟ କ୍ଷେତ୍ରର ଦୈର୍ଘ 1 ମିଟର ଓପ୍ରସୁ 40 ସେ.ମି. । ଏହାର କ୍ଷେତ୍ରଫଳ ଓ ପରିସୀମା ସ୍ଥିରକର ।
Answer» ANSWER: OZ.. ..Ifhdd FD hxvkhlxfjzjgxgjd

22.

ଗୋଟିଏ ଆୟତାକୃତି ବିଶିଷ୍ଟ କ୍ଷେତ୍ରର ଦୈର୍ଘ 1 ମିଟର ଓପ୍ରସୁ 40 ସେ.ମି. । ଏହାର କ୍ଷେତ୍ରଫଳ ଓ ପରିସୀମା ସ୍ଥିରକର ।

Answer»

ANSWER:

OZ.. ..Ifhdd FD hxvkhlxfjzjgxgjd

Discussion

23.	The mean of the 5 smallest numbers from a group is 15 while the mean of all the numbers of the group taken together is 17. If the mean of the number leaving the smallest five out is 18.25, how many numbers were there in the group in all
Answer» Answer: 15 is ANS bro Step-by-step EXPLANATION: CN j vvbbjn jnn

23.

The mean of the 5 smallest numbers from a group is 15 while the mean of all the numbers of the group taken together is 17. If the mean of the number leaving the smallest five out is 18.25, how many numbers were there in the group in all

Answer»

Answer:

15 is ANS bro

Step-by-step EXPLANATION:

CN j

vvbbjn

jnn

Discussion

24.	If sinA = 1/3 , then 2cot2A + 2 = .............
Answer» If sinA=1/3 Then 2cot²A+2 =2(cot²A+1) =2cosec²A. [` . ` cot²A+1=cosec²A] =2(1/sinA)² [' . ' cosecA=1/sinA] =2×3²/1 = 18 PLEASE FOLLOW ME THANKS MY ANSWER

24.

If sinA = 1/3 , then 2cot2A + 2 = .............

Answer»

If sinA=1/3

Then 2cot²A+2

=2(cot²A+1)

=2cosec²A. [` . ` cot²A+1=cosec²A] =2(1/sinA)² [' . ' cosecA=1/sinA]

=2×3²/1

= 18

PLEASE FOLLOW ME THANKS MY ANSWER

Discussion

25.	The denominator of a fraction is twice the numerator.When both are increased by 2 the fraction becomes (3)/(4). Find the fraction.
Answer» ANSWER: 1/2 is the answer MAKE me BRAINLIEST PLEASEEEEEEEEEE

25.

The denominator of a fraction is twice the numerator.When both are increased by 2 the fraction becomes (3)/(4). Find the fraction.

Answer»

ANSWER:

1/2 is the answer

MAKE me BRAINLIEST PLEASEEEEEEEEEE

Discussion

26.	A car increases its speed from 60km/h to 70km/h in 10 second. It's acceleration is
Answer» Answer: INITIAL velocity = u = 20km/hr = 20 x 5/18 = 5.56 m/s Final velocity = V = 50 km/hr = 50 x 5/18 = 13.89 m/s Time = t = 10 sec Acceleration = a a = (v- u) / t = (13.89 - 5.56) / 10 = 0.833 m/s² Step-by-step explanation:

26.

A car increases its speed from 60km/h to 70km/h in 10 second. It's acceleration is

Answer»

Answer:

INITIAL velocity = u = 20km/hr = 20 x 5/18 = 5.56 m/s

Final velocity = V = 50 km/hr = 50 x 5/18 = 13.89 m/s

Time = t = 10 sec

Acceleration = a

a = (v- u) / t

= (13.89 - 5.56) / 10

= 0.833 m/s²

Step-by-step explanation:

Discussion

27.	Find the smallest number by each of the following number must be divide to make a perfect cube 1125
Answer» ANSWER: Answer:do yourself and MAKE me BRAINLIEST

27.

Find the smallest number by each of the following number must be divide to make a perfect cube 1125

Answer»

ANSWER:

Answer:do yourself and MAKE me BRAINLIEST

Discussion

28.	-1 , -3/2 , -2 ,..... are these terms in ap ? justify
Answer» ANSWER: if they are in a.p d MUST be same d1= -3/2+1 = -1/2 d2= -2+3/2 = -1/2 d1=d2 so, they are in AP

28.

-1 , -3/2 , -2 ,..... are these terms in ap ? justify

Answer»

ANSWER:

if they are in a.p d MUST be same

d1= -3/2+1 = -1/2

d2= -2+3/2 = -1/2

d1=d2

so, they are in AP

Discussion

29.	Find smallest perfect square which is divisible by 4 ,9,10each
Answer» Answer: The SMALLEST square number that is DIVISIBLE by each of the numbers 4,9 and 10 is 900 Step-by-step EXPLANATION:

29.

Find smallest perfect square which is divisible by 4 ,9,10each

Answer»

Answer:

The SMALLEST square number that is DIVISIBLE by each of the numbers 4,9 and 10 is 900

Step-by-step EXPLANATION:

Discussion

30.	38. The number of trees in different parks of a city are 33, 38, 48, 33, 4, 34, 33 and 24. The mode and range of this data is:
Answer» Answer: Mode = most common Mode = 33 Range = BIGGEST number - smallest number Range = 48 - 4 Range = 44

30.

38. The number of trees in different parks of a city are 33, 38, 48, 33, 4, 34, 33 and 24. The mode and range of this data is:

Answer»

Answer:

Mode = most common

Mode = 33

Range = BIGGEST number - smallest number

Range = 48 - 4

Range = 44

Discussion

31.	The product of given integers is(_3) × (+5)
Answer» ANSWER:- the PRODUCT of GIVEN INTEGERS is 15

31.

The product of given integers is(_3) × (+5)

Answer»

ANSWER:-

the PRODUCT of GIVEN INTEGERS is 15

Discussion

32.	Fractorisation of Xsquare +1=
Answer» ANSWER: x^2+1= x^2+(1)^2 = (x+1)^2 = x^2 +x+1

32.

Fractorisation of Xsquare +1=

Answer»

ANSWER:

x^2+1= x^2+(1)^2

= (x+1)^2

= x^2 +x+1

Discussion

33.	The reciprocal of positive rational number is _________
Answer» ANSWER: The RECIPROCAL of POSITIVE RATIONAL NUMBER is positive number.

33.

The reciprocal of positive rational number is _________

Answer»

ANSWER:

The RECIPROCAL of POSITIVE RATIONAL NUMBER is positive number.

Discussion

34.	Write an irrational no between √3 and √7
Answer» Answer: 9/49 is a CORRECT answer MATE please 10 *thanks*

34.

Write an irrational no between √3 and √7

Answer»

Answer:

9/49 is a CORRECT answer MATE

please 10 thanks

Discussion

35.	If U={x:x€N and x
Answer» ANSWER: Given: U = { x:x BELONGS to N and x< 10} A= { 1,2,3,5,8} B= { 2,5,6,7,9} So, n(u)= 9 n(A)= 5 n(B)= 5 (AnB)= { 2,5} n(AnB)=2 n(AuB)= n(A)+ n(B) -n(AnB) = 5+5-2 =8 n{(AuB)`}= n(U)- n(AUB)= 9-8=1 so, the answer is 1 please mark it as BRAINLIST if it helps ✨✨

35.

If U={x:x€N and x

Answer»

ANSWER:

Given:

U = { x:x BELONGS to N and x< 10}

A= { 1,2,3,5,8}

B= { 2,5,6,7,9}

So,

n(u)= 9

n(A)= 5

n(B)= 5

(AnB)= { 2,5}

n(AnB)=2

n(AuB)= n(A)+ n(B) -n(AnB)

= 5+5-2

=8

n{(AuB)`}= n(U)- n(AUB)= 9-8=1

so, the answer is 1

please mark it as BRAINLIST if it helps ✨✨

Discussion

36.	2y + 3 ≤ 11 what is the largest value that y could be? x ≤ 0 give an example of what value of x could be
Answer» ANSWER: 2y+3=11 2y=8 y=4 x=0or-1 please MARK as BRAINLIEST answer

36.

2y + 3 ≤ 11 what is the largest value that y could be? x ≤ 0 give an example of what value of x could be

Answer»

ANSWER:

2y+3=11

2y=8

y=4

x=0or-1

please MARK as BRAINLIEST answer

Discussion

37.	Answer with solutionfind the square root of the following rational number
Answer» ANSWER: under ROOT 49/16=7/4 Step-by-step EXPLANATION:

37.

Answer with solutionfind the square root of the following rational number

Answer»

ANSWER:

under ROOT 49/16=7/4

Step-by-step EXPLANATION:

Discussion

38.	Integration of (1/2-3x)+1/√(3x-2) dx
Answer» ANSWER: dx1+3x2 SUBSTITUTE: x=13√tanu dx=13√sec2udu ∫13√sec2udu1+3(13√tanu)2 Pull out that constant: 13–√∫sec2udu1+3(13tan2u) Do the math: 13–√∫sec2udu1+tan2u Trigonometric IDENTITY: 1+tan2u=sec2u: 13–√∫sec2udusec2u Cancel: 13–√∫du That LOOKS PRETTY easy: 13–√u+C Undo the substitution: 13–√arctan(3–√x)+C Check the answer on desmos:

38.

Integration of (1/2-3x)+1/√(3x-2) dx

Answer»

ANSWER:

dx1+3x2

SUBSTITUTE:

x=13√tanu

dx=13√sec2udu

∫13√sec2udu1+3(13√tanu)2

Pull out that constant:

13–√∫sec2udu1+3(13tan2u)

Do the math:

13–√∫sec2udu1+tan2u

Trigonometric IDENTITY: 1+tan2u=sec2u:

13–√∫sec2udusec2u

Cancel:

13–√∫du

That LOOKS PRETTY easy:

13–√u+C

Undo the substitution:

13–√arctan(3–√x)+C

Check the answer on desmos:

Discussion

39.	Find the 13th term of an arithmetic sequence if 5 times 5th term is equal to 8 times 8 th term
Answer» Step-by-step explanation: 5a 5 =8A 8 5(a+4d)=8(a+7d) ⇒3(a+12d)=0 ⇒a 13 =0

39.

Find the 13th term of an arithmetic sequence if 5 times 5th term is equal to 8 times 8 th term

Answer»

Step-by-step explanation:

5a

5

=8A

8

5(a+4d)=8(a+7d)

⇒3(a+12d)=0

⇒a

13

=0

Discussion

40.	6. If P= { ∶ ℰ ℕ 1 < < 11}, Q = { = − 2, ℰ ℕ < 6} andR = {4, 6, 8, 9, 10, 12}, then verify P - (Q ⋂ R) = (P – Q) U(P – R)
Answer» ANSWER: answer for the GIVEN PROBLEM is given

40.

6. If P= { ∶ ℰ ℕ 1 < < 11}, Q = { = − 2, ℰ ℕ < 6} andR = {4, 6, 8, 9, 10, 12}, then verify P - (Q ⋂ R) = (P – Q) U(P – R)

Answer»

ANSWER:

answer for the GIVEN PROBLEM is given

Discussion

41.	Find the value of xx, 12, 16
Answer» Answer: answer is 8 Step-by-step explanation: SINCE the DIFFERENCE between any consecutive nunber is 4 so 12-4=8 hope it helps mark as BRAINLIEST

41.

Find the value of xx, 12, 16

Answer»

Answer:

answer is 8

Step-by-step explanation:

SINCE the DIFFERENCE between any consecutive nunber is 4 so 12-4=8

hope it helps mark as BRAINLIEST

Discussion

42.	Find the equation of line containing the points (1,2)(1,-2)
Answer» Given , The equation of line passing through the POINTS (1,2) and (1,-2) We know that , the two point FORM is given by $\boxed{ \sf{y - y_{1} = \frac{ y_{2} - y_{1} }{x_{2} - x_{1} } (x - x_{1})}}$ Thus , y - 2 = (-2 - 2)/(1 - 1) × (x - 1) y - 2 = -4/0 × (x - 1) y - 2 = 1/0 × (x - 1) 0 = x - 1 x = 1 ★ The equation of the line is x = 1

42.

Find the equation of line containing the points (1,2)(1,-2)

Answer»

Given ,

The equation of line passing through the POINTS (1,2) and (1,-2)

We know that , the two point FORM is given by

$\boxed{ \sf{y - y_{1} = \frac{ y_{2} - y_{1} }{x_{2} - x_{1} } (x - x_{1})}}$

Thus ,

y - 2 = (-2 - 2)/(1 - 1) × (x - 1)

y - 2 = -4/0 × (x - 1)

y - 2 = 1/0 × (x - 1)

0 = x - 1

x = 1

★ The equation of the line is x = 1

Discussion

43.	What is the vaue of tan 15
Answer» ANSWER: $\sqrt{3} - 1$ ...........

43.

What is the vaue of tan 15

Answer»

ANSWER:

$\sqrt{3} - 1$

...........

Discussion

44.	Please tell me the answer....it is interpretation questions .....if u know so plz tell
Answer» Answer: HELLO!!... Is it STH asked by UR FRND??...

44.

Please tell me the answer....it is interpretation questions .....if u know so plz tell

Answer»

Answer:

HELLO!!...

Is it STH asked by UR FRND??...

Discussion

45.	Factirize ( x2 - 4x - 21)
Answer» ANSWER: Quadratic equations whose roots are integers can be solved by factoring. To factor, find two NUMBERS whose sum is 4 (the coefficient of the MIDDLE term) and whose product is -21 (the constant term). ... Solving a Quadratic Equation By Factoring. x2 + 4x - 21 = 0 x2 + 4x - 21 = 0 0 = 0 0 = 0 So, the roots are 3 and -7

45.

Factirize ( x2 - 4x - 21)

Answer»

ANSWER:

Quadratic equations whose roots are integers can be solved by factoring. To factor, find two NUMBERS whose sum is 4 (the coefficient of the MIDDLE term) and whose product is -21 (the constant term).

...

Solving a Quadratic Equation By Factoring.

x2 + 4x - 21 = 0 x2 + 4x - 21 = 0

0 = 0 0 = 0

So, the roots are 3 and -7

Discussion

46.	Please pleasefast fast Please tell 1st one answer
Answer» Step-by-step EXPLANATION: I HOPE this will be HELP you.....

46.

Please pleasefast fast Please tell 1st one answer

Answer»

Step-by-step EXPLANATION:

I HOPE this will be HELP you.....

Discussion

47.	6. Dividend=? Divisor quotient +remainderDivisor remainder +quotientDivisor +remainder *quotientDivisor +quotient remainder
Answer» ANSWER: DIVISOR*quotient+remainder

47.

6. Dividend=? Divisor quotient +remainderDivisor remainder +quotientDivisor +remainder *quotientDivisor +quotient remainder

Answer»

ANSWER:

DIVISOR*quotient+remainder

Discussion

48.	☣️☣️☣️☣️☣️☣️☣️☣️☣️☣️☣️☣️☣️☣️☣️☣️☣️☣️☣️☣️☣️☣️☣️☣️☣️☣️ 6. The sides of a triangle are in the ratio 13 : 14 : 15 and its perimeter is84 cm. Find the area of the triangle.7. Find the area of A ABC in which BC =8 cm, AC = 15 cm and AB = 17 cm. Find the length of altitude drawn on AB. 8. An isosceles triangle has perimeter 30 cm and each of its equal sides is12 cm. Find the area of the triangle. 9. The perimeter of an isosceles triangle is 32 cm. The ratio of one of the equal sides to its base is 3:2. Find the area of the triangle.
Answer» Answer: Step-by-step EXPLANATION: (6) Let the sides of the triangle be 13x,14x and 15x. Therefore, 13x+14x+15x=84 42x=84 x=2 Hence, sides of the triangle are 26 cm , 28 cm & 30 cm. Area of the triangle =√s(s-a)(s-b)(s-c). [Heron's Formula] s=a+b+c/2=26+28+30/2= 84/2=42. a=26cm Area=√42(42-26)(42-28)(42-30). b=28 cm Area=√42* 161412. c=30 cm Area=√422688 Area=√112896 Area=336 cm² ■■■■■■■■■■■■■■■■■■■■ (7) Let the sides be 13x, 14x and 15x Perimeter = Sum of all sides 84 = 13x + 14x + 15x 42x = 84 x = 2 So, the sides are a = 26 cm, b = 28 cm and c = 30 cm s = p/2 = 84/2 = 42 cm Area = √s(s - a)(s - b)(s - c) = √42(42 - 26)(42 - 28)(42 - 30) = √42161412 = 4√141436 = 4146 = 336 cm^2 So, area of triangle = 336 cm^2 ■■■■■■■■■■■■■■■■■■■■■■■ (8) Semi perimeter = s s =( AB +BC + AC)/2 s = (17 + 8 + 15)/2 = 40/2 = 20cm By Heron's formula, area of triangle ABC, Area = √s(s-a)(s-b)(s-c) = √20 (20 - 17) (20 - 8) (20 - 15) = √20 (3 x 12 x 5) = √20 x 180 = √3600 = 60cm² Again area of triangle = 1/2 base x height\ 60 = 1/2 x 17 x h h = 120/17 h = 7.05cm ■■■■■■■■■■■■■■■■■■■■■■ (9) Let The THIRD Side Be xcm. Now Perimeter = 12+12+x 24+x = 30 x = 6 Now Find The Area Using Herons Formula Semiperimeter = 30/2 = 15 Sq Root(15(15-12)(15-12)(15-6) ) = Sq Root(15339) = Sq Root(1215) = 34.85 cm^2 So Area is 34.85cm^2 ■■■■■■■■■■■■■■■■■■■■■ the perimeter of isosceles triangle = 32cm p of isosceles triangle = 2 equal side + base let length of equal side be 3x and base be 2x then 2 * 3x + 2x = 32 cm 6x + 2x = 32cm 8x = 32 cm x= 4 cm length of equal side = 34= 12cm length of base = 24=8cm area of isosceles triangle=b1/4√4a2-b2 = 81/4√4144-64 =2√512 =2*22.62 = 45.25cm2 ■■■■■■■■■■■■■■■■■■■■■ Hope it helps :)

48.

☣️☣️☣️☣️☣️☣️☣️☣️☣️☣️☣️☣️☣️☣️☣️☣️☣️☣️☣️☣️☣️☣️☣️☣️☣️☣️ 6. The sides of a triangle are in the ratio 13 : 14 : 15 and its perimeter is84 cm. Find the area of the triangle.7. Find the area of A ABC in which BC =8 cm, AC = 15 cm and AB = 17 cm. Find the length of altitude drawn on AB. 8. An isosceles triangle has perimeter 30 cm and each of its equal sides is12 cm. Find the area of the triangle. 9. The perimeter of an isosceles triangle is 32 cm. The ratio of one of the equal sides to its base is 3:2. Find the area of the triangle.

Answer»

Answer:

Step-by-step EXPLANATION:

(6) Let the sides of the triangle be 13x,14x and 15x.

Therefore, 13x+14x+15x=84

42x=84

x=2

Hence, sides of the triangle are 26 cm , 28 cm & 30 cm.

Area of the triangle =√s*(s-a)*(s-b)*(s-c).

[Heron's Formula]

s=a+b+c/2=26+28+30/2= 84/2=42.

a=26cm

Area=√42*(42-26)*(42-28)*(42-30).

b=28 cm

Area=√42* 16*14*12. c=30 cm

Area=√42*2688

Area=√112896

Area=336 cm²

■■■■■■■■■■■■■■■■■■■■

(7) Let the sides be 13x, 14x and 15x

Perimeter = Sum of all sides

84 = 13x + 14x + 15x

42x = 84

x = 2

So, the sides are a = 26 cm, b = 28 cm and c = 30 cm

s = p/2 = 84/2 = 42 cm

Area = √s(s - a)(s - b)(s - c)

= √42(42 - 26)(42 - 28)(42 - 30)

= √42*16*14*12

= 4√14*14*36

= 4*14*6

= 336 cm^2

So, area of triangle = 336 cm^2

■■■■■■■■■■■■■■■■■■■■■■■

(8) Semi perimeter = s

s =( AB +BC + AC)/2

s = (17 + 8 + 15)/2 = 40/2 = 20cm

By Heron's formula, area of triangle ABC,

Area = √s(s-a)(s-b)(s-c)

= √20 (20 - 17) (20 - 8) (20 - 15)

= √20 (3 x 12 x 5)

= √20 x 180

= √3600 = 60cm²

Again area of triangle = 1/2 base x height\

60 = 1/2 x 17 x h

h = 120/17

h = 7.05cm

■■■■■■■■■■■■■■■■■■■■■■

(9) Let The THIRD Side Be xcm.

Now Perimeter = 12+12+x

24+x = 30

x = 6

Now Find The Area Using Herons Formula

Semiperimeter = 30/2 = 15

Sq Root(15(15-12)(15-12)(15-6) ) = Sq Root(15*3*3*9) = Sq Root(1215) = 34.85 cm^2

So Area is 34.85cm^2

■■■■■■■■■■■■■■■■■■■■■

the perimeter of isosceles triangle = 32cm

p of isosceles triangle = 2 * equal side + base

let length of equal side be 3x and base be 2x

then

2 * 3x + 2x = 32 cm

6x + 2x = 32cm

8x = 32 cm

x= 4 cm

length of equal side = 3*4= 12cm

length of base = 2*4=8cm

area of isosceles triangle=b*1/4*√4a2-b2

= 8*1/4√4*144-64

=2√512

=2*22.62

= 45.25cm2

■■■■■■■■■■■■■■■■■■■■■

Hope it helps

:)

Discussion

49.	:) The picture shows a rectangle divided intotwo partsInstead of the broken line separating these 5parts, crawa straight line to divide the rect-angle into two other parts of the same areaCalculate the areas of these parts.
Answer» ANSWER: f     2 2 2 a b a b    2 , FIND the value of a b  .

49.

:) The picture shows a rectangle divided intotwo partsInstead of the broken line separating these 5parts, crawa straight line to divide the rect-angle into two other parts of the same areaCalculate the areas of these parts.

Answer»

ANSWER:

f

   

2 2 2 a b a b    2

, FIND the value of

a b  .

Discussion

50.	) The number of students of Std 6th and Std 7th who went to visit theTadoba Tiger Project at Chandrapur was 140 and 196 respectively. The students of each class are to be divided into groups of the same number of students. Each group can have a paid guide. What is the maximum number of students there can be in each group?
Answer» $\huge\underline\bold\red{AnSweR}$ An oncologist is a doctor who TREATS cancer and provides medical care for a PERSON diagnosed with cancer. The FIELD of oncology has three major areas: medical, SURGICAL, and radiation. A medical oncologist treats cancer using chemotherapy or other medications, such as targeted therapy or immunotherapy. Treatments: Chemotherapy; Radiation therapy Includes DISEASES: Ovarian cancer Associated Diseases or Conditions: Cancer

50.

) The number of students of Std 6th and Std 7th who went to visit theTadoba Tiger Project at Chandrapur was 140 and 196 respectively. The students of each class are to be divided into groups of the same number of students. Each group can have a paid guide. What is the maximum number of students there can be in each group?

Answer»

$\huge\underline\bold\red{AnSweR}$

An oncologist is a doctor who TREATS cancer and provides medical care for a PERSON diagnosed with cancer. The FIELD of oncology has three major areas: medical, SURGICAL, and radiation. A medical oncologist treats cancer using chemotherapy or other medications, such as targeted therapy or immunotherapy.

Treatments: Chemotherapy; Radiation therapy

Includes DISEASES: Ovarian cancer

Associated Diseases or Conditions: Cancer

Discussion

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THEREFORE,

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