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33201.

Geih

Answer»
33202.

sin A (1+ tan A)+ cos A(1 + cot A) =(sec A+ cosec A)

Answer»
33203.

What is a pair of linear equation in two variables

Answer» ax + by+ c = 0
33204.

Slove x and y 139x+56y=641;56x+139y=724

Answer»
33205.

If one root of the equation x square +7x+k=0 is -2,then find the value of k and the other root

Answer» Given equation,x^2 + 7x + k = 0 , given root -2Substituting value of x = -2 in the equation4 - 14 + k = 0k = 10Now the equation is x^2 +7 x + 10 = 0Or x^2 + 2x + 5x + 10\xa0= x(x + 2) + 5( x + 2)= (x + 5)(x + 2)x= -5, -2,so, other value of x is -5
33206.

Solve this : (*= square)12y* - 12 √3y -72

Answer» Y*-root3-6Y*(-2root3y+3y)-6Y(Y-2root3)root3(Y-2root3)Y=2root3 and Y=-root3
33207.

Explain chapter 8 example 2 page no 210

Answer» Which book
33208.

Find HCF of 5309 and 3072 and express in form of linear combination

Answer»
33209.

Prove that x2+4x+5 has no real zeroes

Answer» f(x) = x4\xa0+ 4x2\xa0+ 5= (x2)2\xa0+ 4x2\xa0+ 5Let x2\xa0=n,Then, f(x) = n2\xa0+ 4n + 5,Here a=1,b=4,c=5The discriminant(D) = {tex}\\text{b}^2-4\\mathrm{ac}=\\;(4)^2-4\\times1\\times5=16-20=-4{/tex}Since the discriminant is negative so this polynomial has no zerosHence, f(x) = x4\xa0+ 4x2\xa0+ 5\xa0has no zero.
33210.

Ghhhffghh

Answer» swatantra
Jahjhsjojabavvaa
????
33211.

Two rails are represented by the equations x+2y-4=0and 2x+4y-12=0.

Answer» We have,{tex}x + 2y - 4=0{/tex}Putting {tex}y = 0{/tex}, we get{tex}x + 0 - 4 = 0{/tex}{tex} \\Rightarrow {/tex}\xa0{tex}x = 4{/tex}Putting x = 0, we get{tex}0 + 2y - 4 = 0{/tex}{tex} \\Rightarrow {/tex}\xa0{tex}y = 2{/tex}Thus, two solutions of equation {tex}x + 2y - 4 = 0{/tex} are:\tx40y02\tWe have,\xa0{tex}2x + 4y - 12 = 0{/tex}Putting {tex}x = 0{/tex}, we get{tex}0 + 4y - 12 = 0{/tex}{tex} \\Rightarrow {/tex}\xa0{tex}y = 3{/tex}Putting {tex}y = 0{/tex}, we get{tex}2x + 0(12) = 0{/tex}{tex} \\Rightarrow {/tex}\xa0x = 6Thus, two solutions of equation {tex}2x + 4y - 12 = 0{/tex} are:\tx06y30\tNow, we plot the points A (4, 0) and B (0, 2) and draw a line passing through these two points to get the graph of the line represented by the equations (i).We also plot the points P (0, 3) and Q (6, 0) and draw a line passing through these two points to get the graph of the line represented by the equation (ii).We observe that the lines are parallel and they do not intersect any where.REMARK The graphical representation of the above pair of linear equations provides us a pair of parallel lines.Let us write the pair of linear equations,{tex}x + 2y - 4 = 0{/tex}{tex}2x + 4y -12 = 0{/tex}as {tex}a_1x + b_1y + c_1=0{/tex}{tex}a_2x + b_2y +c_2 =0{/tex}where {tex}a_1=1, b_1= 2, c_1\xa0= -4{/tex},{tex}a_2\xa0= 2, b_2\xa0= 4\\ and \\ c_2\xa0= -12{/tex}{tex}\\frac { a _ { 1 } } { a _ { 2 } } = \\frac { 1 } { 2 } , \\frac { b _ { 1 } } { b _ { 2 } } = \\frac { 2 } { 4 } = \\frac { 1 } { 2 } \\text { and } \\frac { c _ { 1 } } { c _ { 2 } } = \\frac { - 4 } { - 12 } = \\frac { 1 } { 3 }{/tex}{tex}\\therefore \\quad \\frac { a _ { 1 } } { a _ { 2 } } = \\frac { b _ { 1 } } { b _ { 2 } } \\neq \\frac { c _ { 1 } } { c _ { 2 } }{/tex}will represent parallel lines, if{tex}\\frac { a _ { 1 } } { a _ { 2 } } = \\frac { b _ { 1 } } { b _ { 2 } } \\neq \\frac { c _ { 1 } } { c _ { 2 } }{/tex}The converse is also true for any pair of linear equations.It follows from the above examples that the pair of linear equations{tex}a_1x + b_1y + c_1 = 0{/tex}{tex}a_2x + b_2y +c_2=0{/tex}will represent:\tintersecting lines, if\xa0{tex}\\frac { a _ { 1 } } { a _ { 2 } } \\neq \\frac { b _ { 1 } } { b _ { 2 } }{/tex}\tcoincident lines, if\xa0{tex}\\frac { a _ { 1 } } { a _ { 2 } } = \\frac { b _ { 1 } } { b _ { 2 } } = \\frac { c _ { 1 } } { c _ { 2 } }{/tex}\tparallel lines, if\xa0{tex}\\frac { a _ { 1 } } { a _ { 2 } } = \\frac { b _ { 1 } } { b _ { 2 } } \\neq \\frac { c _ { 1 } } { c _ { 2 } }{/tex}
33212.

Prove that (3+2√5)2whole bracket square is irrational

Answer» Aap sharukh khan h kya
30
Not posible
33213.

Show that √6+√2 is irrational

Answer»
33214.

1/2+√3 is an irrational number

Answer»
33215.

If tanA+sinA=p and tanA-sinA=q, then prove that p square minus Q square =4 root p q

Answer» (x-2)+1=2x-3
33216.

X=1\\2+1\\2_1\\2+3x

Answer»
33217.

What is the no. 2

Answer»
33218.

Add 52+45

Answer» Hii shriya from where r u ???
97
97
97
5+2=7 and 5+4=9 so its 97
97
33219.

Estimate the given products 61×47

Answer»
33220.

If sinA +cosA=m, secA+cosecA=n, then prove that n(m×m-1)=2m

Answer» SecA+cosecA(sinA+cosA-1)(sinA+cosA+1)=(sin^2A+cos^A+2sinAcosA-1)(secA+cosecA)=(1+2sinAcosA-1)(secA+ cosecA)={2sinAcosA×(1/cosA)}+{2sinAcosA×(1/cosA)}=2sinA+2cosA=2(sinA+cosA)=2m
33221.

Hii V.K.

Answer» 10 bje online aana okk
33222.

prove that :sec6ϴ=tan6ϴ+3tan2ϴsec2ϴ+1

Answer»
33223.

Ncrt

Answer» Its NCERT
33224.

2coscos - 1/sin*cos

Answer» What is coscos
33225.

Find the hcf of 2 and 4

Answer» 2
33226.

Find the h.c.f of 176 and 38220

Answer» Given numbers are 176 and 38220.Here, 38220 > 17By using Euclid\'s division lemma, we get\xa0a = bq + r, where 0<_r < b. Here a as dividend, b as divisor, q as quotient and r as remainderDividend = divisor {tex}\\times{/tex}\xa0quotient + remainderdividend = divisor {tex}\\times{/tex}\xa0quotient + remainder38220 = (176 {tex}\\times{/tex}\xa0217) + 28 Here r = 28\xa0{tex}\\ne{/tex}\xa00 and b = 176On taking 176 as the new dividend and 28 as\xa0the new divisor and then apply Euclid\'s division lemma, we get176 = (28\xa0{tex}\\times{/tex}\xa06) + 8Here remainder = 8\xa0{tex}\\ne{/tex}\xa00So, on taking 28 as dividend and 8 as the divisor and then apply Euclid\'s division lemma, we get28 = (8 {tex}\\times{/tex}\xa03) + 4Again, remainder = 4\xa0{tex}\\ne{/tex}\xa00On taking 8 as the dividend and 4 as the divisor and then apply Euclid\'s division lemma, we get 8 = ( 4 {tex}\\times{/tex}\xa02) + 0\xa0Here, remainder = 0 and last divisor\xa0is 4.Hence, HCF of 176 and 38220 is 4.
33227.

x+√y=7√x+y=11

Answer» 9 or 4
33228.

solve 2x+3y=11&2x-4y=-24

Answer» 2x+3y=112x-4y=-24. Y=132x+3(13)=112x+39=112x=-28X=-14
33229.

State whether 35÷50 wil have a terminating decimal expansion or a non terminating repeating decimal

Answer» Its a terminating decimal
33230.

8 men and 12 women can do a piece of work in 10 days while

Answer» Write whole question
33231.

34578021567÷3×4

Answer» Using bodmas
46104 0 2 8 7 566
46104028756
33232.

Hii V.S

Answer» Hlw
Sorry V.K.
33233.

In triangle PQR angel Qis 90 and SIN R IS 3/5write the value of cos P

Answer»
33234.

Set of all tallest persons in your classroom

Answer»
33235.

Exercise 3b question no. 33

Answer»
33236.

If two zeroes of a polynomial p(x)=x^3 -4x^2 -3x +12 are √3 and √-3, then find its 3rd zero.

Answer»
33237.

Find a quadratic polynomial whose zeroes are -6 and -2/3.

Answer» X square -6x- 2/3
33238.

Textbook solution

Answer»
33239.

Represent √13 on the number line

Answer»
33240.

Value of root2

Answer» 1.414213562
1.4142135623730950488
1.414
1.414
1.414
33241.

If a and b be two zero of the quadratic p( x)= 2x

Answer»
33242.

Kitne marks ka ncert me ayega

Answer» Reality me kitne parcent ata ha
95percent
60%ncert ayaga
Full 80 marks
80 marks ka exam hota h
100 % ncert hota hai
33243.

Who is the father of geometry

Answer» Euclids
Euclids
Euclids
33244.

Chapter 6 ex 6d question 25

Answer»
33245.

If a and b are the zeroes of the polynomial p(x)=3x^2-5x+6,find:(i)(a/b)+(b/a)and (ii)a^3+b^3

Answer»
33246.

What is the absolute value of -/-6/

Answer» 1/6
-/-6/=-6
6
6.
6
33247.

Hello DV

Answer» Okk
Ni yrr time ni rhta
Brainly nhi chalate ab
Hey Vishu...
33248.

Find zeroes of√3x^2-8x+4√3 and verify relationship between zeroes and coefficients

Answer»
33249.

Find k for which the given to is solution of the equation x2+3ax+k=0,x=-a

Answer» K=2a+3a^2
which part of black sheep have wool
33250.

Show that √3 and 5 + 3√2 are irrational number.

Answer» ...