332354 + Interview Questions in Secondary School in Math

1.	4-x2=5x standard form and values of a b c
Answer» Answer: HOPE IT WILL HELP YOU Step-by-step EXPLANATION: PLEASE MARK AS BRAINLIST ANSWER AND FOLLOW ME

1.

4-x2=5x standard form and values of a b c

Answer»

Answer:

HOPE IT WILL HELP YOU

Step-by-step EXPLANATION:

PLEASE MARK AS BRAINLIST ANSWER AND FOLLOW ME

Discussion

2.	What is the central tendency????? sigma??????
Answer» Answer: the TENDENCY for the values of a random VARIABLE to CLUSTER ROUND its mean, mode, or median. . . . HOPE it helps you. .

2.

What is the central tendency????? sigma??????

Answer»

Answer:

the TENDENCY for the values of a random VARIABLE to CLUSTER ROUND its mean, mode, or median.

.

HOPE it helps you. .

Discussion

3.	5) Find the quadratic equation, if the sum of the roots is 24 andthe roots is 24 and the product of the roots is 140.
Answer» ANSWER: Let the quadratic equation be = ax^2 + B x + c = The product of ROOTS is given by = c / a = 140. c = 140 a. = The sum of roots is given by = b / a = 24. b = 24 a. = a x^2 + b x + c = 0 = a x^2 + 24 BAX + 140 a = 0 = a ( x^2 + 24 x + 140) = 0 = x^2 + 24 x + 140 = 0.

3.

5) Find the quadratic equation, if the sum of the roots is 24 andthe roots is 24 and the product of the roots is 140.

Answer»

ANSWER:

Let the quadratic equation be = ax^2 + B x + c

= The product of ROOTS is given by = c / a = 140. c = 140 a.

= The sum of roots is given by = b / a = 24. b = 24 a.

= a x^2 + b x + c = 0

= a x^2 + 24 BAX + 140 a = 0

= a ( x^2 + 24 x + 140) = 0

= x^2 + 24 x + 140 = 0.

Discussion

4.	Find four numbers in G.P.such that their product is I and the sum of the middle two 10 numbers 1S
Answer» ANSWER: Last three of the FOUR NUMBERS are in A.P. and hence they may be chosen as a−d,a,a+d Also the first number is same as the last one i.e. a+d. Therefore the four numbers are a+d,a−d,a,a+d. The first three of the above four are in G.P. ∴(a−d) 2 =a(a+d). But d=6 given ∴(a−6) 2 =a(a+6) or a 2 −12a+36=a 2 +6a or 18a=36 ∴a=2. Putting for a and d the four numbers are 8,−4,2,8, which satisfy the given conditions.

4.

Find four numbers in G.P.such that their product is I and the sum of the middle two 10 numbers 1S

Answer»

ANSWER:

Last three of the FOUR NUMBERS are in A.P. and hence they may be chosen as a−d,a,a+d

Also the first number is same as the last one i.e. a+d.

Therefore the four numbers are a+d,a−d,a,a+d. The first three of the above four are in G.P.

∴(a−d)

2

=a(a+d). But d=6 given

∴(a−6)

2

=a(a+6)

or a

2

−12a+36=a

2

+6a

or 18a=36 ∴a=2.

Putting for a and d the four numbers are 8,−4,2,8, which satisfy the given conditions.

Discussion

5.	Use existential and universal quantifiers to express the statement "Everyone has exactly twobiological parents" using the propositional function P(x, y), which represents "x is the biologicalparent of y."
Answer» ANSWER: Here is your answer Please MARK me as the BRINALIST

5.

Use existential and universal quantifiers to express the statement "Everyone has exactly twobiological parents" using the propositional function P(x, y), which represents "x is the biologicalparent of y."

Answer»

ANSWER:

Here is your answer

Please MARK me as the BRINALIST

Discussion

6.	21. BE and CF are two equal altitudes of a triangle ABC. Prove thata) triangle ABC congruent to triangle ACF.b) AB = AC ; ABC is an isosceles triangle..
Answer» ANSWER: in triangle BAC and QPR Ange A =angle P Angle B=angle Q by AA similarly triangle BAC~triangle QPR 6/h=4/28 6X28=h h=42cm.

6.

21. BE and CF are two equal altitudes of a triangle ABC. Prove thata) triangle ABC congruent to triangle ACF.b) AB = AC ; ABC is an isosceles triangle..

Answer»

ANSWER:

in triangle BAC and QPR

Ange A =angle P

Angle B=angle Q

by AA similarly

triangle BAC~triangle QPR

6/h=4/28

6X28=h

h=42cm.

Discussion

7.	In ∆ABC,if
Answer» ANSWER: ANGLE A is 55°. Step-by-step EXPLANATION: SUM of all ANGLES is 180°. so, - 75° +50°+

7.

In ∆ABC,if

Answer»

ANSWER:

ANGLE A is 55°.

Step-by-step EXPLANATION:

SUM of all ANGLES is 180°.

so,

- 75° +50°+

Discussion

8.	A rectangular hall has its dimensions12 metre, 8 metre and 6 metres.(i) Determine the area of four walls ofthe hall(ii) Calculate the expenses incurred to paintthe wall at 30 per square metre.
Answer» ANSWER of FIRST PART is 14

Discussion

9.	Fill.in the blanks1. Integer _____ is the identity under addition
Answer» Answer: The additive IDENTITY of any INTEGER a is a NUMBER b which when added with a, leaves it unchanged, i.e. b is called as the additive identity of any integer a if a + b = a. Now, when we add 0 with any of the integers a we get a + 0 = a = 0 + a So, 0 is the additive identity for integers.

9.

Fill.in the blanks1. Integer _____ is the identity under addition

Answer»

Answer:

The additive IDENTITY of any INTEGER a is a NUMBER b which when added with a, leaves it unchanged, i.e. b is called as the additive identity of any integer a if a + b = a. Now, when we add 0 with any of the integers a we get a + 0 = a = 0 + a So, 0 is the additive identity for integers.

Discussion

10.	Crammers Rule method
Answer» Answer: in TRIANGLE BAC and QPR Ange A =ANGLE P Angle B=angle Q by AA similarly triangle BAC~triangle QPR 6/h=4/28 6X28=h h=42cm.

10.

Crammers Rule method

Answer»

Answer:

in TRIANGLE BAC and QPR

Ange A =ANGLE P

Angle B=angle Q

by AA similarly

triangle BAC~triangle QPR

6/h=4/28

6X28=h

h=42cm.

Discussion

11.	I have been practising morning
Answer» Step-by-step EXPLANATION: WHATS your QUESTION ASK perfectlyi am not UNDERSTANDING this

11.

I have been practising morning

Answer»

Step-by-step EXPLANATION:

WHATS your QUESTION

ASK perfectlyi am not UNDERSTANDING this

Discussion

12.	Please send this question answer.....
Answer» ANSWER: I THINK...... Step-by-step explanation: 3) answer is RIGHT

12.

Please send this question answer.....

Answer»

ANSWER:

I THINK......

Step-by-step explanation:

3) answer is RIGHT

Discussion

13.	__+__= 10 and __x__= -11 (HAS TO BE SAME SET OF NUMBERS)
Answer» ANSWER: Qkqkakaoa98a9aqiqiqo

13.

+= 10 and x= -11 (HAS TO BE SAME SET OF NUMBERS)

Answer»

ANSWER:

Qkqkakaoa98a9aqiqiqo

Discussion

14.	How many diagonals are there in a quadrilateral?
Answer» Answer: The two DIAGONALS of a CONVEX quadrilateral are the line segments that connect opposite VERTICES. The two bimedians of a convex quadrilateral are the line segments that connect the midpoints of opposite SIDES.

14.

How many diagonals are there in a quadrilateral?

Answer»

Answer:

The two DIAGONALS of a CONVEX quadrilateral are the line segments that connect opposite VERTICES. The two bimedians of a convex quadrilateral are the line segments that connect the midpoints of opposite SIDES.

Discussion

15.	Solve this problem .heat conduction Solve: 3 + 2 = 0.
Answer» HOPE it can HELP you...... PLZ MARK me as a BRAINLIST.....

15.

Solve this problem .heat conduction Solve: 3 + 2 = 0.

Answer»

HOPE it can HELP you......

PLZ MARK me as a BRAINLIST.....

Discussion

16.	[11/11, 1:09 pm] .: Hello............. ^_^Q. (6)Let Principal be 'P'and Rate of interest be 'R'Time (n) = 2 yearsSimple Interest (S.I) = Rs. 400Compound Interest (C.I) = Rs. 410S.I=100P×R×T=100P×R×T=400=100PR×2=400=50PR=400=PR=400×50=PR=20000C.I=P((1+100R)n−1)=P((1+100R)n−1)=410=P((1+100R)2−1)=410=P((1)2+2×1×100R+(100R)2−1)=410=P(1−1×50R+(100R)2)=410=P(100R+10000R2)=410(takecommonR)=PR(501+10000R)=410(putthevalueofPR=20000)=20000(501+10000R)=410=20000(10000200+R)=410=10000R+200=20000410=10000R+200=200041=R+200=200041×10000=R+200=41×5=R+200=205=R=205−200=R=5Rate of interest = 5%Now Principal,as PRNowPrincipal,asP×R=20000=>P×5=20000=>P=520000=>P=4000Principal = Rs. 4000Q. (7)A man invested Rs. 1000 for 3 years at 11% Simple Interest .....................(CASE 1)Principal (P) = Rs. 1000Rate of interest (R) = 11% per annumTime (n) = 3 yearsS.I=100P×R×T=1001000×11×3=10×11×3=330S.I = Rs. 330He also invested Rs. 1000 at 10% compound interest per annum compounded annually for 3 years .........................( CASE 2 )Principal (P) = Rs. 1000Rate of interest (R) = 10% per annumTime (n) = 3 yearsC.I=P((1+100R)n−1)=1000((1+10010)3−1)=1000((1+101)3−1)=1000((1010+1)3−1)=1000((1011)3−1)=1000×(10001331−1)=1000×(10001331−1000)=1000×1000331=331C.I = Rs. 331(S.I =330 < C.I = 331)(CASE 1) < (CASE 2)S.I < C.ISo, investment at compound Interest is better.Q. (8)Principal (P) = Rs. 400000Rate of interest = 16% per annum= 16/2 = 8% per half yearly (R)Time (n) = 1 year = 2 half yearsAmount=P(1+100R)n=400000(1+1008)2=400000(1+252)2=400000(2525+2)2=400000×(2527)2=400000×25×2527×27=400000×625729=640×729=466560So, they earn Rs. 466560...................^_^[11/11, 1:32 pm] .: Hello............. ^_^
Answer» <P>Step-by-step explanation: [11/11, 1:09 pm] .: Hello............. ^_^ Q. (6) LET Principal be 'P' and Rate of INTEREST be 'R' Time (N) = 2 years Simple Interest (S.I) = Rs. 400 Compound Interest (C.I) = Rs. 410 $\begin{gathered}S.I = \frac{P \times R \times T}{100} \\ \\ = \frac{P \times R \times T}{100} = 400 \\ \\ = \frac{PR \times 2}{100} = 400 \\ \\ = \frac{PR}{50} = 400 \\ \\ = PR = 400 \times 50 \\ = PR = 20000\end{gathered}$ S.I=100P×R×T=100P×R×T=400=100PR×2=400=50PR=400=PR=400×50=PR=20000 $\begin{gathered}C.I = P( \: \: ( {1 + \frac{R}{100} })^{n} - 1) \\ \\ = P( \: \: ( {1 + \frac{R}{100} })^{n} - 1) = 410 \\ \\ = P( \: \: ( {1 + \frac{R}{100} })^{2} - 1) = 410 \\ \\ = P( \: \: {(1)}^{2} + 2 \times 1 \times \frac{R}{100} + { (\frac{R}{100}) }^{2} - 1 ) = 410 \\ \\ = P( \: \: 1 - 1 \times \frac{R}{50} + { (\frac{R}{100}) }^{2} \: \: \: ) = 410 \\ \\ = P( \: \frac{R}{100} + \frac{ {R}^{2} }{10000} \: \: ) = 410 \\ \\ (take \: \: common \: R \: ) \\ \\ = PR ( \frac{1}{50} + \frac{R}{10000} ) = 410 \\ (put \: the \: value \: of \: PR \: = 20000) \\ \\ = 20000( \frac{1}{50} + \frac{R}{10000} ) = 410 \\ \\ = 20000( \frac{200 + R }{10000} ) = 410 \\ \\ = \frac{R + 200}{10000} = \frac{410}{20000 } \\ \\ = \frac{R + 200}{10000} = \frac{41}{2000 } \\ \\ = R + 200 = \frac{41}{2000} \times 10000 \\ \\ = R + 200 = 41 \times 5 \\ = R + 200 = 205 \\ = R = 205 - 200 \\ = R = 5\end{gathered}$ C.I=P((1+100R)n−1)=P((1+100R)n−1)=410=P((1+100R)2−1)=410=P((1)2+2×1×100R+(100R)2−1)=410=P(1−1×50R+(100R)2)=410=P(100R+10000R2)=410(takecommonR)=PR(501+10000R)=410(putthevalueofPR=20000)=20000(501+10000R)=410=20000(10000200+R)=410=10000R+200=20000410=10000R+200=200041=R+200=200041×10000=R+200=41×5=R+200=205=R=205−200=R=5 Rate of interest = 5% Now Principal, as PR $\begin{gathered}Now Principal, \\ as P \times R = 20000 \\ = > P \times 5 = 20000 \\ = > P = \frac{20000}{5} \\ \\ = > P = 4000\end{gathered}$ NowPrincipal,asP×R=20000=>P×5=20000=>P=520000=>P=4000 Principal = Rs. 4000 Q. (7) A man invested Rs. 1000 for 3 years at 11% Simple Interest .....................(CASE 1) Principal (P) = Rs. 1000 Rate of interest (R) = 11% per annum Time (n) = 3 years

16.

[11/11, 1:09 pm] .: Hello............. ^_^Q. (6)Let Principal be 'P'and Rate of interest be 'R'Time (n) = 2 yearsSimple Interest (S.I) = Rs. 400Compound Interest (C.I) = Rs. 410S.I=100P×R×T=100P×R×T=400=100PR×2=400=50PR=400=PR=400×50=PR=20000C.I=P((1+100R)n−1)=P((1+100R)n−1)=410=P((1+100R)2−1)=410=P((1)2+2×1×100R+(100R)2−1)=410=P(1−1×50R+(100R)2)=410=P(100R+10000R2)=410(takecommonR)=PR(501+10000R)=410(putthevalueofPR=20000)=20000(501+10000R)=410=20000(10000200+R)=410=10000R+200=20000410=10000R+200=200041=R+200=200041×10000=R+200=41×5=R+200=205=R=205−200=R=5Rate of interest = 5%Now Principal,as PRNowPrincipal,asP×R=20000=>P×5=20000=>P=520000=>P=4000Principal = Rs. 4000Q. (7)A man invested Rs. 1000 for 3 years at 11% Simple Interest .....................(CASE 1)Principal (P) = Rs. 1000Rate of interest (R) = 11% per annumTime (n) = 3 yearsS.I=100P×R×T=1001000×11×3=10×11×3=330S.I = Rs. 330He also invested Rs. 1000 at 10% compound interest per annum compounded annually for 3 years .........................( CASE 2 )Principal (P) = Rs. 1000Rate of interest (R) = 10% per annumTime (n) = 3 yearsC.I=P((1+100R)n−1)=1000((1+10010)3−1)=1000((1+101)3−1)=1000((1010+1)3−1)=1000((1011)3−1)=1000×(10001331−1)=1000×(10001331−1000)=1000×1000331=331C.I = Rs. 331(S.I =330 < C.I = 331)(CASE 1) < (CASE 2)S.I < C.ISo, investment at compound Interest is better.Q. (8)Principal (P) = Rs. 400000Rate of interest = 16% per annum= 16/2 = 8% per half yearly (R)Time (n) = 1 year = 2 half yearsAmount=P(1+100R)n=400000(1+1008)2=400000(1+252)2=400000(2525+2)2=400000×(2527)2=400000×25×2527×27=400000×625729=640×729=466560So, they earn Rs. 466560...................^_^[11/11, 1:32 pm] .: Hello............. ^_^

Answer»

<P>Step-by-step explanation:

[11/11, 1:09 pm] .: Hello............. ^_^

Q. (6)

LET Principal be 'P'

and Rate of INTEREST be 'R'

Time (N) = 2 years

Simple Interest (S.I) = Rs. 400

Compound Interest (C.I) = Rs. 410

$\begin{gathered}S.I = \frac{P \times R \times T}{100} \\ \\ = \frac{P \times R \times T}{100} = 400 \\ \\ = \frac{PR \times 2}{100} = 400 \\ \\ = \frac{PR}{50} = 400 \\ \\ = PR = 400 \times 50 \\ = PR = 20000\end{gathered}$ S.I=100P×R×T=100P×R×T=400=100PR×2=400=50PR=400=PR=400×50=PR=20000

$\begin{gathered}C.I = P( \: \: ( {1 + \frac{R}{100} })^{n} - 1) \\ \\ = P( \: \: ( {1 + \frac{R}{100} })^{n} - 1) = 410 \\ \\ = P( \: \: ( {1 + \frac{R}{100} })^{2} - 1) = 410 \\ \\ = P( \: \: {(1)}^{2} + 2 \times 1 \times \frac{R}{100} + { (\frac{R}{100}) }^{2} - 1 ) = 410 \\ \\ = P( \: \: 1 - 1 \times \frac{R}{50} + { (\frac{R}{100}) }^{2} \: \: \: ) = 410 \\ \\ = P( \: \frac{R}{100} + \frac{ {R}^{2} }{10000} \: \: ) = 410 \\ \\ (take \: \: common \: R \: ) \\ \\ = PR ( \frac{1}{50} + \frac{R}{10000} ) = 410 \\ (put \: the \: value \: of \: PR \: = 20000) \\ \\ = 20000( \frac{1}{50} + \frac{R}{10000} ) = 410 \\ \\ = 20000( \frac{200 + R }{10000} ) = 410 \\ \\ = \frac{R + 200}{10000} = \frac{410}{20000 } \\ \\ = \frac{R + 200}{10000} = \frac{41}{2000 } \\ \\ = R + 200 = \frac{41}{2000} \times 10000 \\ \\ = R + 200 = 41 \times 5 \\ = R + 200 = 205 \\ = R = 205 - 200 \\ = R = 5\end{gathered}$ C.I=P((1+100R)n−1)=P((1+100R)n−1)=410=P((1+100R)2−1)=410=P((1)2+2×1×100R+(100R)2−1)=410=P(1−1×50R+(100R)2)=410=P(100R+10000R2)=410(takecommonR)=PR(501+10000R)=410(putthevalueofPR=20000)=20000(501+10000R)=410=20000(10000200+R)=410=10000R+200=20000410=10000R+200=200041=R+200=200041×10000=R+200=41×5=R+200=205=R=205−200=R=5

Rate of interest = 5%

Now Principal,

as PR

$\begin{gathered}Now Principal, \\ as P \times R = 20000 \\ = > P \times 5 = 20000 \\ = > P = \frac{20000}{5} \\ \\ = > P = 4000\end{gathered}$ NowPrincipal,asP×R=20000=>P×5=20000=>P=520000=>P=4000

Principal = Rs. 4000

Q. (7)

A man invested Rs. 1000 for 3 years at 11% Simple Interest .....................(CASE 1)

Principal (P) = Rs. 1000

Rate of interest (R) = 11% per annum

Time (n) = 3 years

Discussion

17.	__+__=3 and __x__=-40 (HAS TO BE SAME SET OF NUMBERS)
Answer» ANSWER: ONE no.is 8 and other is (-5) Step-by-step explanation: Mark this as BRAINLIEST....

17.

+=3 and x=-40 (HAS TO BE SAME SET OF NUMBERS)

Answer»

ANSWER:

ONE no.is 8 and other is (-5)

Step-by-step explanation:

Mark this as BRAINLIEST....

Discussion

18.	Prove that angle substanded by an arc at the center is double the angel substanded by it any point on the remaining part of the circle(according to class 9)please answer
Answer» Step-by-step explanation: To prove : ∠POQ=2∠PAQ To prove this theorem we consider the arc AB in three different situations, minor arc AB, major arc AB and semi-circle AB. Construction : Join the line AO extended to B. Proof : ∠BOQ=∠OAQ+∠AQO .....(1) Also, in △ OAQ, OA=OQ [Radii of a circle] Therefore, ∠OAQ=∠OQA [Angles opposite to equal sides are equal] ∠BOQ=2∠OAQ .......(2) SIMILARLY, BOP=2∠OAP ........(3) Adding 2 & 3, we get, ∠BOP+∠BOQ=2(∠OAP+∠OAQ) ∠POQ=2∠PAQ .......(4) HOPE you got it.... For the CASE 3, where PQ is the major arc, equation 4 is replaced by Reflex angle, ∠POQ=2∠PAQ

18.

Prove that angle substanded by an arc at the center is double the angel substanded by it any point on the remaining part of the circle(according to class 9)please answer

Answer»

Step-by-step explanation:

To prove : ∠POQ=2∠PAQ

To prove this theorem we consider the arc AB in three different situations, minor arc AB, major arc AB and semi-circle AB.

Construction :

Join the line AO extended to B.

Proof :

∠BOQ=∠OAQ+∠AQO .....(1)

Also, in △ OAQ,

OA=OQ [Radii of a circle]

Therefore,

∠OAQ=∠OQA [Angles opposite to equal sides are equal]

∠BOQ=2∠OAQ .......(2)

SIMILARLY, BOP=2∠OAP ........(3)

Adding 2 & 3, we get,

∠BOP+∠BOQ=2(∠OAP+∠OAQ)

∠POQ=2∠PAQ .......(4)

HOPE you got it....

For the CASE 3, where PQ is the major arc, equation 4 is replaced by

Reflex angle, ∠POQ=2∠PAQ

Discussion

19.	What is "Hexamethylene diamine and adipic acid for class 8
Answer» NYLON -6,6 is synthesized by polycondensation of hexamethylenediamine and ADIPIC acid. Equivalent amounts of hexamethylenediamine and adipic acid are COMBINED with water in a reactor. ... The nylon salt GOES into a reaction vessel where polymerization process takes place either in BATCHES or continuously.

Discussion

20.	What is microwave and microwave communication and its application ?
Answer» Answer: Microwaves refer to the electromagnetic rays with frequencies between 300MHz and 300GHZ in the electromagnetic spectrum. Microwaves are small when compared with the waves used in radio BROADCASTING. ... Microwaves are most commonly used in satellite communications, RADAR signals, phones, and navigational APPLICATIONS

20.

What is microwave and microwave communication and its application ?

Answer»

Answer:

Microwaves refer to the electromagnetic rays with frequencies between 300MHz and 300GHZ in the electromagnetic spectrum. Microwaves are small when compared with the waves used in radio BROADCASTING. ... Microwaves are most commonly used in satellite communications, RADAR signals, phones, and navigational APPLICATIONS

Discussion

21.	Perimeter of an equilateral triangle is 36 cm. Find its area using Heron's formula
Answer» ANSWER: let one side be "a" a + a + a = 36 3a = 36 a = 36/3 a = 12 cm. Now, if we APPLY Herons Formula, 12 + 12 + 12/2 = 18. ( 18 is the semi - perimeter) (18- 12) (18-12) (18-12) 6 X 6 X 6 2X3X2X3X2X3 2X2X3X3 root 2X3 (We consider both pairs of 2X2 as one) 6 root 6 is your answer. Please rate my answer and MARK me as brainliest! THANKS!

21.

Perimeter of an equilateral triangle is 36 cm. Find its area using Heron's formula

Answer»

ANSWER:

let one side be "a"

a + a + a = 36

3a = 36

a = 36/3

a = 12 cm.

Now, if we APPLY Herons Formula,

12 + 12 + 12/2 = 18. ( 18 is the semi - perimeter)

(18- 12) (18-12) (18-12)

6 X 6 X 6

2X3X2X3X2X3

2X2X3X3 root 2X3

(We consider both pairs of 2X2 as one)

6 root 6 is your answer.

Please rate my answer and MARK me as brainliest!

THANKS!

Discussion

22.	If the simple interest on a sum of money for 2 years at 5% per annum is Rs. 60, what is the compound interest on the same at the samerate and for the same time?dSelect one:a. Rs. 63.5O b. Rs. 62O C. Rs.61.5O d. Rs. 64
Answer» Answer: 62 Step-by-step EXPLANATION: please MAKE it BRAINLIEST ANSWERAND follow me

22.

If the simple interest on a sum of money for 2 years at 5% per annum is Rs. 60, what is the compound interest on the same at the samerate and for the same time?dSelect one:a. Rs. 63.5O b. Rs. 62O C. Rs.61.5O d. Rs. 64

Answer»

Answer:

62

Step-by-step EXPLANATION:

please MAKE it BRAINLIEST ANSWERAND follow me

Discussion

23.	Сделать как сказано в заданиях и без тупух вопрос и понятно написать в крации
Answer» Answer: CX go DIFF I HTTP an gf if NG

23.

Сделать как сказано в заданиях и без тупух вопрос и понятно написать в крации

Answer»

Answer:

CX go DIFF I HTTP an gf if NG

Discussion

24.	Give the absolute value of -13/-15
Answer» ANSWER: the EXACT VALUE of -13/-15= 1.5384615.....

24.

Give the absolute value of -13/-15

Answer»

ANSWER:

the EXACT VALUE of -13/-15= 1.5384615.....

Discussion

25.	Find the ratio of 4 hours to 300 minuts
Answer» Step-by-step explanation: 1 HOUR = 60 min 4 HOURS = 60×4 = 240 min ratio = 240:300 = 4:5

25.

Find the ratio of 4 hours to 300 minuts

Answer»

Step-by-step explanation:

1 HOUR = 60 min

4 HOURS = 60×4 = 240 min

ratio = 240:300

= 4:5

Discussion

26.	Sin20+cos20=✔2sinx how x?
Answer» ANSWER: I'm SORRY PLEASE GIVE me BRAINLIESTS

26.

Sin20+cos20=✔2sinx how x?

Answer»

ANSWER:

I'm SORRY PLEASE GIVE me BRAINLIESTS

Discussion

27.	Find the measure of x in the following figure (according to class 9)please answer
Answer» Answer: 40° Step-by-step EXPLANATION: PLEASE MAKE it BRAINLIEST answerand FOLLOW me

27.

Find the measure of x in the following figure (according to class 9)please answer

Answer»

Answer:

40°

Step-by-step EXPLANATION:

PLEASE MAKE it BRAINLIEST answerand FOLLOW me

Discussion

28.	What is HELL is carbonization
Answer» Answer: Carbonization is the term for the conversion of an organic SUBSTANCE into carbon or a carbon-containing residue through pyrolysis or destructive distillation. When biomaterial is exposed to sudden SEARING HEAT, it can be carbonized extremely quickly, TURNING it into SOLID carbon.

28.

What is HELL is carbonization

Answer»

Answer:

Carbonization is the term for the conversion of an organic SUBSTANCE into carbon or a carbon-containing residue through pyrolysis or destructive distillation. When biomaterial is exposed to sudden SEARING HEAT, it can be carbonized extremely quickly, TURNING it into SOLID carbon.

Discussion

29.	Integral (f(x) + f'(x)) dc
Answer» Answer: USE INTEGRATION website in Google u can DIFFERENTIATE ANYTHING

29.

Integral (f(x) + f'(x)) dc

Answer»

Answer:

USE INTEGRATION website in Google

u can DIFFERENTIATE ANYTHING

Discussion

30.	__+__ =-6 and __x__=8 (HAS TO BE THE SAME SET OF NUMBERS)
Answer» ANSWER: -4+-2and -4×-2 Step-by-step EXPLANATION: PLS VOTE me

30.

+ =-6 and x=8 (HAS TO BE THE SAME SET OF NUMBERS)

Answer»

ANSWER:

-4+-2and -4×-2

Step-by-step EXPLANATION:

PLS VOTE me

Discussion

31.	- Geometric Application of determinants: (Curves and surfaces through given points)The idea is to get an equation from the vanishing of the determinant of a homogeneous linearsystem as the condition for a nontrivial solution. We explain the trick for obtainingsuch a system for the case of a line L through two given points P1 : (x1 yı)and P2 : (x2.yz). The unknown line is ax +by+c=0, say. We write it as ax +by+c.1=0. To get a nontrivial solution for a,b,c, the determinant of the"coefficients” x,y, 1 must be zero. The system isaxtby +c. 1 =0 (Line L)axı +by. +0.1=0 (P1 on L)... (1)ax2 +by2 +0.1=0 (P2 on L)(a) Line through two given points: Derive from D=0 in (1) the familiarformula(2-1))(1-x)=O'-1)Ya-D(b) Plane : Find the analog of (1) for a plane through three given points. Apply it when thepoints are(1,1,1), (3,2,6), (5,0,5)(c) Circle : Find a similar formula for a circle in the plane through three given points. Findand sketch the circle through (2,6), (6,3), (7,1).(d) Sphere: Find the analog of the formula in (C) for a sphere through four given points. Findthe sphere through (0,0,5), (4,0,1), (0,4,1), (0,0,-3) by the formula or byinspection
Answer» ANSWER: KO shnskslam skskakanab bakakama a jakakaksossm

31.

- Geometric Application of determinants: (Curves and surfaces through given points)The idea is to get an equation from the vanishing of the determinant of a homogeneous linearsystem as the condition for a nontrivial solution. We explain the trick for obtainingsuch a system for the case of a line L through two given points P1 : (x1 yı)and P2 : (x2.yz). The unknown line is ax +by+c=0, say. We write it as ax +by+c.1=0. To get a nontrivial solution for a,b,c, the determinant of the"coefficients” x,y, 1 must be zero. The system isaxtby +c. 1 =0 (Line L)axı +by. +0.1=0 (P1 on L)... (1)ax2 +by2 +0.1=0 (P2 on L)(a) Line through two given points: Derive from D=0 in (1) the familiarformula(2-1))(1-x)=O'-1)Ya-D(b) Plane : Find the analog of (1) for a plane through three given points. Apply it when thepoints are(1,1,1), (3,2,6), (5,0,5)(c) Circle : Find a similar formula for a circle in the plane through three given points. Findand sketch the circle through (2,6), (6,3), (7,1).(d) Sphere: Find the analog of the formula in (C) for a sphere through four given points. Findthe sphere through (0,0,5), (4,0,1), (0,4,1), (0,0,-3) by the formula or byinspection

Answer»

ANSWER:

KO shnskslam skskakanab

bakakama a

jakakaksossm

Discussion

32.	2. A certain pipe can fill up a tank 2 hours faster than another pipe. It takes 3hours for both pipes to fill up the same tank. In how many hourswould the first pipe fill up the tank?
Answer» 3 DIVIDED by 2 =10 10-5=5ans

32.

2. A certain pipe can fill up a tank 2 hours faster than another pipe. It takes 3hours for both pipes to fill up the same tank. In how many hourswould the first pipe fill up the tank?

Answer»

3 DIVIDED by 2 =10

10-5=5ans

Discussion

33.	Find out the square root of following by division method
Answer» ANSWER: DUDE where is the NUMBER? We NEED a number to SOLVE your question.

33.

Find out the square root of following by division method

Answer»

ANSWER:

DUDE where is the NUMBER?

We NEED a number to SOLVE your question.

Discussion

34.	Please please pleasesolve this bengali maths
Answer» ANSWER: SORRY I AM NOT KNOWING BENGALI LANGUAGE

34.

Please please pleasesolve this bengali maths

Answer»

ANSWER:

SORRY I AM NOT KNOWING BENGALI LANGUAGE

Discussion

35.	Successor of a+b-1 is
Answer» a-1+b-1-0 ANSWER.......................

Discussion

36.	What is physics?explain
Answer» Answer: the branch of SCIENCE concerned with the nature and properties of matter and energy. The SUBJECT matter of physics includes mechanics, heat, light and other RADIATION, SOUND, electricity, magnetism, and the structure of atoms.

36.

What is physics?explain

Answer»

Answer:

the branch of SCIENCE concerned with the nature and properties of matter and energy. The SUBJECT matter of physics includes mechanics, heat, light and other RADIATION, SOUND, electricity, magnetism, and the structure of atoms.

Discussion

37.	What is called congruency ??
Answer» Step-by-step EXPLANATION: ɪɴ ɢᴇᴏᴍᴇᴛʀʏ,ᴛᴡᴏ ғɪɢᴜʀᴇs ᴏʀ ᴏʙᴊᴇᴄᴛs ᴀʀᴇ ᴄᴏɴɢʀᴜᴇɴᴛ ɪғ ᴛʜᴇʏ ʜᴀᴠᴇ ᴛɢᴇ sᴀᴍᴇ sʜᴀᴘᴇ ᴀɴᴅ sɪᴢᴇ, ᴏʀ ɪғ ᴏɴᴇ ʜᴀs ᴛʜᴇ sᴀᴍᴇ sʜᴀᴘᴇ ᴀɴᴅ sɪᴢᴇ ᴀs ᴛʜᴇ ᴍɪʀʀᴏʀ ɪᴍᴀɢᴇ ᴏғ ᴛʜᴇ ᴏᴛʜᴇʀ. ᴘʟᴇᴀsᴇ ᴍᴀʀᴋ ᴍᴇ ᴀs ʙʀᴀɪɴʟɪsᴛ ᴀɴᴅ ғᴏʟʟᴏᴡ ᴍᴇ.....

37.

What is called congruency ??

Answer»

Step-by-step EXPLANATION:

ɪɴ ɢᴇᴏᴍᴇᴛʀʏ,ᴛᴡᴏ ғɪɢᴜʀᴇs ᴏʀ ᴏʙᴊᴇᴄᴛs ᴀʀᴇ ᴄᴏɴɢʀᴜᴇɴᴛ ɪғ ᴛʜᴇʏ ʜᴀᴠᴇ ᴛɢᴇ sᴀᴍᴇ sʜᴀᴘᴇ ᴀɴᴅ sɪᴢᴇ, ᴏʀ ɪғ ᴏɴᴇ ʜᴀs ᴛʜᴇ sᴀᴍᴇ sʜᴀᴘᴇ ᴀɴᴅ sɪᴢᴇ ᴀs ᴛʜᴇ ᴍɪʀʀᴏʀ ɪᴍᴀɢᴇ ᴏғ ᴛʜᴇ ᴏᴛʜᴇʀ.

ᴘʟᴇᴀsᴇ ᴍᴀʀᴋ ᴍᴇ ᴀs ʙʀᴀɪɴʟɪsᴛ ᴀɴᴅ ғᴏʟʟᴏᴡ ᴍᴇ.....

Discussion

38.	2. In the given figure O is the centre of a circle. Find themeasure of
Answer» ANSWER: BOC =120° Step-by-step explanation: here angle BOC= 2 angle ABC if this HELPS you then you can follow me

38.

2. In the given figure O is the centre of a circle. Find themeasure of

Answer»

ANSWER:

BOC =120°

Step-by-step explanation:

here angle BOC= 2 angle ABC

if this HELPS you then you can follow me

Discussion

39.	Increasing the growth of algae.they Occupy the morecalled alisafication.the more Oxygen;lakes and pondsareDrow the line segment 8. 1cm anddivick in to the ration 5:4 Constructthis and find what is the ration ofАс: с.
Answer» 1 ANS is LAKES and PONDS becomes GREEN

Discussion

40.	Which of the following options is correct if the centre of xy connecting the following points x and y in of P (2,3)?
Answer» ANSWER: Question Plot the FOLLOWING points in the Cartesian plane whose x,y CO - ordinates are given.x23 - 10 - 9 - 4y - 3 - 34110 - 6(x,y) Answer · 0 votes X = 2 3 - 1 0 - 9 - 4 Y = - 3 - 3 4 11 0 - 6 (x,y) = A B C D E F More

40.

Which of the following options is correct if the centre of xy connecting the following points x and y in of P (2,3)?

Answer»

ANSWER:

Question

Plot the FOLLOWING points in the Cartesian plane whose x,y CO - ordinates are given.x23 - 10 - 9 - 4y - 3 - 34110 - 6(x,y)

Answer · 0 votes

X = 2 3 - 1 0 - 9 - 4 Y = - 3 - 3 4 11 0 - 6 (x,y) = A B C D E F

More

Discussion

41.	V) 1:15 pm -11:30 am = কত ঘন্টা কত মিনিট ?
Answer» Answer: 3 HOUR and 15 minutes Step-by-step EXPLANATION: PLEASE MAKE it brainliest answerand follow me okkkkkkkkkkkkkkkkkkkkkkkkkkkkkk okkkkkkkkkkkkkkkkkkkkkkkkkkkkkk okkkkkkkkkkkkkkkkkkkkkkkkkkkkkk okkkkkkkkkkkkkkkkkkkkkkkkkkkkkk okkkkkkkkkkkkkkkkkkkkkkkkkkkkkk okkkkkkkkkkkkkkkkkkkkkkkkkkkkkk okkkkkkkkkkkkkkkkkkkkkkkkkkkkkk okkkkkkkkkkkkkkkkkkkkkkkkkkkkkk okkkkkkkkkkkkkkkkkkkkkkkkkkkkkk okkkkkkkkkkkkkkkkkkkkkkkkkkkkkk okkkkkkkkkkkkkkkkkkkkkkkkkkkkkk okkkkkkkkkkkkkkkkkkkkkkkkkkkkkk okkkkkkkkkkkkkkkkkkkkkkkkkkkkkk okkkkkkkkkkkkkkkkkkkkkkkkkkkkkk okkkkkkkkkkkkkkkkkkkkkkkkkkkkkk okkkkkkkkkkkkkkkkkkkkkkkkkkkkkk okkkkkkkkkkkkkkkkkkkkkkkkkkkkkk okkkkkkkkkkkkkkkkkkkkkkkkkkkkkk okkkkkkkkkkkkkkkkkkkkkkkkkkkkkk

41.

V) 1:15 pm -11:30 am = কত ঘন্টা কত মিনিট ?

Answer»

Answer:

3 HOUR and 15 minutes

Step-by-step EXPLANATION:

PLEASE MAKE it brainliest answerand follow me

okkkkkkkkkkkkkkkkkkkkkkkkkkkkkk okkkkkkkkkkkkkkkkkkkkkkkkkkkkkk okkkkkkkkkkkkkkkkkkkkkkkkkkkkkk okkkkkkkkkkkkkkkkkkkkkkkkkkkkkk okkkkkkkkkkkkkkkkkkkkkkkkkkkkkk okkkkkkkkkkkkkkkkkkkkkkkkkkkkkk okkkkkkkkkkkkkkkkkkkkkkkkkkkkkk okkkkkkkkkkkkkkkkkkkkkkkkkkkkkk okkkkkkkkkkkkkkkkkkkkkkkkkkkkkk okkkkkkkkkkkkkkkkkkkkkkkkkkkkkk okkkkkkkkkkkkkkkkkkkkkkkkkkkkkk okkkkkkkkkkkkkkkkkkkkkkkkkkkkkk okkkkkkkkkkkkkkkkkkkkkkkkkkkkkk okkkkkkkkkkkkkkkkkkkkkkkkkkkkkk okkkkkkkkkkkkkkkkkkkkkkkkkkkkkk okkkkkkkkkkkkkkkkkkkkkkkkkkkkkk okkkkkkkkkkkkkkkkkkkkkkkkkkkkkk okkkkkkkkkkkkkkkkkkkkkkkkkkkkkk okkkkkkkkkkkkkkkkkkkkkkkkkkkkkk

Discussion

42.	What is the difference between the sum of first 50 terms ofAs12,20,28..and22,32,42..?
Answer» Step-by-step EXPLANATION: ${HOPE HELPS YOU}$

42.

What is the difference between the sum of first 50 terms ofAs12,20,28..and22,32,42..?

Answer»

Step-by-step EXPLANATION:

${HOPE HELPS YOU}$

Discussion

43.	Track(ii) How much will you subtract from the scale to get the correct length of object?(111) motion of2. A 20 cm scale has a broken end and begins from 1.5 cm.(1) Which mark of this scale will you use to measure length of an object?
Answer» ANSWER: ksslndkd snenannanakskoshahkahahhahejejjskejehehiischsicjeofheofjeofje9ugrg0eife0jfienglejforof4ug0rig0rgrvirlfhogurofjrijforigg0r0grggr9uroug9rig0rig

43.

Track(ii) How much will you subtract from the scale to get the correct length of object?(111) motion of2. A 20 cm scale has a broken end and begins from 1.5 cm.(1) Which mark of this scale will you use to measure length of an object?

Answer»

ANSWER:

ksslndkd snenannanakskoshahkahahhahejejjskejehehiischsicjeofheofjeofje9ugrg0eife0jfienglejforof4ug0rig0rgrvirlfhogurofjrijforigg0r0grggr9uroug9rig0rig

Discussion

44.	Answer this correctly to get brainliest answer
Answer» ANSWER is 3 Mark my BRAINLIEST

44.

Answer this correctly to get brainliest answer

Answer»

ANSWER is 3

Mark my BRAINLIEST

Discussion

45.	Find the ratio of following and write it in lowest form 150 cm to 1.5 m
Answer» Answer: 1:1....................................

45.

Find the ratio of following and write it in lowest form 150 cm to 1.5 m

Answer»

Answer:

1:1....................................

Discussion

46.	441 ملمs
Answer» Answer: 441 ملمs $solve \: x + 9 = 11$

46.

441 ملمs

Answer»

Answer:

441 ملمs

$solve \: x + 9 = 11$

Discussion

47.	How to find length + breadths of any given area........
Answer» Answer: To find length or breadth when Area of a Rectangle is given When we NEED to find length of a rectangle we need to DIVIDE area by breadth. Length of a rectangle= $area \div breadth$ ℓ = A ÷ b. Similarly when we need to find breadth of a rectangle we need to divide area by length. Breadth of a rectangle =Area ÷ length. b = A ÷ ℓ Step-by-step explanation: Hope this HELPS

47.

How to find length + breadths of any given area........

Answer»

Answer:

To find length or breadth when Area of a Rectangle is given

When we NEED to find length of a rectangle we need to DIVIDE area by breadth.
Length of a rectangle=
$area \div breadth$
ℓ = A ÷ b.
Similarly when we need to find breadth of a rectangle we need to divide area by length.
Breadth of a rectangle =Area ÷ length.
b = A ÷ ℓ

Step-by-step explanation:

Hope this HELPS

Discussion

48.	Find the value of sec45/ cos45+ cosec45/sin45
Answer» ANSWER: ${\huge{\boxed{\orange{\text{Correct Answer Is <klux>4</klux>}}}}}$ Step-by-step EXPLANATION: ${\red{\text{sec45°/cos45° + cosec45°/sin45°}}}$ ⇒ 1/cos45°•cos45° + 1/sin45°•sin45° ⇒ 1/[1/√2•1/√2] + 1/[1/√2•1/√2] ⇒ 2 + 2 ⇒ 4

48.

Find the value of sec45/ cos45+ cosec45/sin45

Answer»

ANSWER:

${\huge{\boxed{\orange{\text{Correct Answer Is <klux>4</klux>}}}}}$

Step-by-step EXPLANATION:

${\red{\text{sec45°/cos45° + cosec45°/sin45°}}}$

⇒ 1/cos45°•cos45° + 1/sin45°•sin45°

⇒ 1/[1/√2•1/√2] + 1/[1/√2•1/√2]

⇒ 2 + 2

⇒ 4

Discussion

49.	(11) ZBEC(iii) ZBAC(2014)Find :(i) ZCAD(ii) ZCBD(iii) ZADC(2008)(b) In the figure (ii) given below, AB is parallel to DC, ZBCE = 80° and ZBAC = 25°
Answer» Answer: so it equals letters Step-by-step EXPLANATION: i CANT do math THANKS for the POINTS

49.

(11) ZBEC(iii) ZBAC(2014)Find :(i) ZCAD(ii) ZCBD(iii) ZADC(2008)(b) In the figure (ii) given below, AB is parallel to DC, ZBCE = 80° and ZBAC = 25°

Answer»

Answer:

so it equals letters

Step-by-step EXPLANATION:

i CANT do math THANKS for the POINTS

Discussion

50.	898420 divided by 385
Answer» ANSWER: 233355844155 is the answer

50.

898420 divided by 385

Answer»

ANSWER:

233355844155 is the answer

Discussion

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