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Let, x²+3xy+x+my-m=(ax+by+c)(ex+fy+g)Since the given expression does not contain y² then one of the linear factors must not contain the term containing y. Therefore,x²+3xy+x+my-m=(ax+by+c)(ex+g)or, x²+3xy+x+my-m=aex²+bexy+cex+agx+bgy+cgEquating the coefficients from both sides,ae=1 ----------------------(1)be=3 ----------------------(2)ce+ag=1 -----------------(3)bg=m ---------------------(4)cg=-m --------------------(5)Since all the coefficients are integers then from (1),a=e=1∴, from (2), b=3Putting in(3),c+g=1 ----------------(6)Now dividing (4) by (5),b/c=-1or, c=-bor, c=-3∴, from (6),-3+g=1or, g=4Putting in (5),-m=-3×4or, m=12 Ans.

Let, x²+3xy+x+my-m=(ax+by+c)(ex+fy+g)Since the given expression does not contain y² then one of the linear factors must not contain the term containing y. Therefore,x²+3xy+x+my-m=(ax+by+c)(ex+g)or, x²+3xy+x+my-m=aex²+bexy+cex+agx+bgy+cgEquating the coefficients from both sides,ae=1 ----------------------(1)be=3 ----------------------(2)ce+ag=1 -----------------(3)bg=m ---------------------(4)cg=-m --------------------(5)Since all the coefficients are integers then from (1),a=e=1∴, from (2), b=3Putting in(3),c+g=1 ----------------(6)Now dividing (4) by (5),b/c=-1or, c=-bor, c=-3∴, from (6),-3+g=1or, g=4Putting in (5),-m=-3×4or, m=12 Ans.

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A =√(24(24 - 21)(24 - b)(24 - c))

A =√(2332(24 - b)(24 - c))

A = 6√(2(24 - b)(24 - c))

Let's see if we can figure out what b and c would have to be from here:

>b and c have to be integers

>b and c have to add to 27 (because the perimeter is 48)

>2(24 - b)(24 - c) must be a perfect square (because A has to be an integer)

Given ratio of sum of n terms of two AP’s = (7n+1):(4n+27)We can consider the 9th term as the m th term.Let’s consider the ratio these two AP’s m th terms as am: a’m→(2)Recall the nth term of AP formula, an= a + (n – 1)dHence equation (2) becomes,am: a’m= a + (m – 1)d : a’ + (m – 1)d’On multiplying by 2, we getam: a’m= [2a + 2(m – 1)d] : [2a’ + 2(m – 1)d’]= [2a + {(2m – 1) – 1}d] : [2a’ + {(2m – 1) – 1}d’]= S2m – 1: S’2m – 1= [7(2m – 1) + 1] : [4(2m – 1) +27] [from (1)]= [14m – 7 +1] : [8m – 4 + 27]= [14m – 6] : [8m + 23]Thus the ratio of mth terms of two AP’s is [14m – 6] : [8m + 23].now substitute the value of m as 9so the answer becomes120/95

As per Pascal's Triangle

coefficients will be1 4 6 4 1

(3x+2y)^4=1⋅(3x)^4(2y)^0+4(3x^3)(2y)+6(3y)^2(2y)^2+4(3x)(2y)^3+1(3x)0(2y)^4

81x^4+216x^3y+216x^2y^2+96xy^3+16y^4

3/2x-1=5(3x-2)=5*23x=2+103x=12x=12/3x=4

Distance between these coordinates is √(4-2)²+(3-3)²+(1-5)²= √2²+4²= √20 units

The unit digit of a two-digit number is 3u = 3and seven times the sum of the digits is the number itself.7(t+u) = 10t+u

Substitute 3 for u:

7(t+3) = 10t+37t+21 = 10t+3-3t = -18t = 6The tens digit is 6 and the units digit is 3, so the number is 63.

7(3+x)=10x+321+7n=10x+310=3xx=6number=6X10+3=63

76/4=19 in each roll 19members get it19*8=152

76/4=19; 19 x 8 = 152 rolls

rolls of cloth=8each rolls have =76mtotal cloths=76*8=608meach boy given =4m1roll given to 76÷4=19 boystotal cloths given to boys= 608÷4=152boys

1rupee= 100 paisa

121rupee=12100 paisa

no. of boys is 12100÷25=484

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We have the word MISSISSIPPI

In this word there are 4 I’s, 4 S’s, 2 P’s and 1 M.

No two S should be together, hence we can place S at these places

_ M _ I _ I _ I _ I _ P _ P _ .

Therefore, the possible number of words is given by8C4. 7! / 4!2!.

=8C4. 7 .6! / 4. 2!

= 7.8C4 .6C4

the area becomes4times the original area

when l,b doubles then l=2a,b=2b,new area=2ax2b=4ab.so 4times the original area

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the total no. of distinct zeros are 2. as it is a quadratic equation.

Why we multiple and what is the meaning A

103×107(100+3)(100+7)(x)^2 + x (a+b) + (a×b)

(100)^2 + 100 (3+7) + (3×7)

10000 + 1000 + 21

11021Ans:--11021

January month has 31 daysThus no. of days starting from 8th Januaryto 31 January = 31 - 7 = 24

1 Feb to 12 Feb no. Of days = 12

No. Of days between 7 January and 13 February = 24 + 12 = 36

1.2) answer is zero(0) which is Greater than 11 and -11 and its it's difference is also -11

4.5 cm,0.045m,0.000045km is the correct answer of the given question

45mm=4.5cm=.045m=.000045km

The right answer is 4.5cm,.045m,. 000045km.

45mm=4.5cm=.04m=.000045km

45 mm in cm,m and km are respectively 4.5 cm,0.045m and 0.000045 km.

A+B+C= 180DIVIDED BY 2 BOTH SIDEA+B+C/2=180/2B+C/2+A/2=90B+C/2=90-A/2cosec^2B+C/2-tan^2A/2cosec^2(90-A/2)- tan^A/2sec^2A/2-tan^2A/2let A/2= psec^2p-tan^2p=1

A+B+C=180; B+C=180-A___(1); cosecx^2( b+c/2)-tan^2(a/2); Substitutiing 1 in cosecx^2(b+c/2) = cosecx^2(180/2-a/2)-tan^2(a/2)= cosex^2(90-a/2)-tan^2(a/2); identity cosecx(90-x)=secx;=sec^2(a/2)-tan^2(a/2); identity secx^2- tanx^2=1;sec^2(a/2)-tan^2(a/2)=1

gfsf Girish fyh rheostats Sharma Diana

Machine languageis the onlylanguagea computer is capable of understanding. ... Computer programs are written in one or moreprogramming languages, like C++, Java, or Visual Basic. A computer cannot directly understand theprogramming languagesused to create computer programs, so theprogramcode must be compiled.

if a+b+c=0 , a^3+b^3+c^3=3abca^3+b^3+c^3/abc=3abc/abc=3 answer

hexagon has 6 sides make each side with scale

∫ ( x + sin(x)) dx /(1 + cos(x)

= ∫ ( x + 2sin(x/2)cos(x/2)) dx /(1 + 2cos^2(x/2) - 1)

= ∫ ( x + 2sin(x/2)cos(x/2)) dx / 2cos^2(x/2)

= ∫ x dx / 2 cos^2(x/2) + ∫ tan(x/2) dx

= 1/2∫ x sec^2(x/2) dx + ∫ tan(x/2) dx --------------------(1)

integrate first integral by partsu = xdu = dx

dv = (1/2)sec^2(x/2) dxv = tan(x/2)

1/2∫ x sec^2(x/2) dx = x tan(x/2) - ∫ tan(x/2) dx

substitute in (1)

1/2∫ x sec^2(x/2) dx + ∫ tan(x/2) dx = x tan(x/2) - ∫ tan(x/2) dx + ∫ tan(x/2) dx

= x tan(x/2) + C

send pictures

When ashapeis rotated about its centre, if it comes to rest in a position and looks exactly like the original, then it has rotational symmetry. rotational symmetry order). This means that a regular hexagon has 6 sides, 6 lines of symmetry and an order of rotational symmetry of 6.Angle of rotation = 360/6 = 60.

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The sum of the interior angles of ahexagon is 720dgrees. You can have 6 triangles in ahexagonif you join vertices to the centre. So on account six triangles 6*180degreesminus the angle at the centre which is 360. or 180*6-360 =720 degreesis the sum of the interor angles of a six sided polygon orhexagon

The formula for finding the area of a hexagon is Area = (3√3 s2)/ 2 where s is the length of a side of the regular hexagon.therefore after substituting s=4 cm,A=41.5sqcm

When ashapeis rotated about its centre, if it comes to rest in a position and looks exactly like the original, then it has rotational symmetry. rotational symmetry order). This means that a regular hexagon has 6 sides, 6 lines of symmetry and an order of rotational symmetry of 6.Angle of rotation = 360/order= 360/6= 60.

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take photos clera because this is not a good poc

Theformulaforfindingthesumof the measure of the interioranglesis (n - 2) ×180.So, here n = 6 (6-2)× 180° = 4× 180° = 720°As it is a regular hexagon so all angles are equal. Let one angle be x6x = 720°x = 720°/6 = 120°x = 120°

no this is not be correct answer

According to the trigonometric identity, tan70=tan(20+50) tan70=(tan20+tan50)/1-tan20tan50 tan70-tan20tan50tan70=tan20+tan50 Also tan70tan20=tan70cot70=1 Hence tan70-tan50=tan20+tan50 So tan70=tan20+2tan50

LHS = ( cosA-sinA+1)/(cosA+sinA-1)

divide numerator and denominator with

sinA , we get

= ( cotA - 1 + cosecA)/( cotA+1 - cosecA )

= [cotA+cosecA- 1 ]/[ 1 - cosecA + cotA ]

=[(cotA+cosecA)-(cosec²A-cot²A)]/(1-cosecA+cotA)

=[(cotA+cosecA)-(cosecA+cotA)(cosecA-cotA)]/(1-cosecA+cotA)

= [(cosecA+cotA)(1-cosecA+cotA)]/(1-cosecA+cotA)

= cosecA + cotA

= RHS

Sum of all angles inside a quadrilateral is equal to 360.

15x-3+11x+2+14x-4+85=36040x+80=36040x=360-8040x=280x=280/40x=7

Angle B=14x-4=(14*7)-4=98-4=94

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