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12n can only end with digits 0 or 5 when it baseprime factors include 2 and 5 as its factors.

but 12 has not 2 and 5 as its prime factors . so it is impossible for 12n to end with the digit 0 or 5

Let a square ABCD in which L,M,N&O are the midpoints .

in triangle AML and triangle CNO AM = CN ( AB = DC and M and O are the midpoints ) AL = CM ( AD = BC and L and N are the midpoints ) angle MAL = angle NCO ( all angles of a square = 90 degree ) by AAS critaria triangle AML CONGRUENT totriangle CNO therefore ML = ON ( CPCT )similarly in triangle MBN CONGRUENT toLDO and AND triangle AML is CONGRUENT to trianglenow , in Triangle AML ,angle AML = angle ALM ( AM = AL ) = 45 degree similarly in triangle LDO angle DLO = 45 degreethere fore ,angle MLO = 90 degree

by the properties of SQUAREall sides are equal and angles are 90 degree

sum of angles of 2 adjacent sides are supplementary (i.e =180°

x+20°+x-30°=1802x-10=1802x=190x=95°

the question is--2-(-8)=8-2

we know that --=+-2+8=8-26 = 6

solved

lhs=-2-(-8) =-2+8 =6rhs=8-2 =6so,lhs=rhs showed

5power -2 +t power4 =2

note: tan(inverse) is also known as arctan

px^2+3x+q=0when x= -1p(-1)^2+3(-1)+q=0p+q= 3---(1)when x= -2px^2+3x+q=0p(-2)^2+3(-2)+q=04p+q= 6---(2).p+q= 3---(1)by subtracting 3p= 3p= 3/3=1p+q=31+q= 3q= 3-1=2so q- p= 2-1=1

the smallest root is 1

3x=2[x+5] is the answer

I can't understand what you wrote

let,

5 - 2√3 be a rational number.

.°. 5 - 2√3 = p/q [ where p and q are integer , q ≠ 0 and q and p are co- prime number ]

=> - 2√3 = p/q -5

=>- 2√3 = p - 5q / q

=> √3 = p - 5q / - 2q

we know that p/q is a rational number.

.°. √3 is also a rational number.

This contradicts our assumption

5 - 2√3 is an irrational number

Like my answer if you find it useful!

Taking A(12, 8), B(-2, 6) and C(6, 0)

AB² = (12--2)² + (8-6)²= 196+4=200

AC² = (12-6)² + (8-0)²= 36+64=100

BC² = (-2-6)² + (6-0)²=64+36=100

By Pythogoras theorem

Since AB² = AC² + BC², the points (12, 8), (-2, 6) and (6, 0) are vertices of a right angledtriangle.

AB is the hypotenuse.

Mid-point of AB = [(12+-2)/2 , (8+6)/2] = (5, 7)Let the mid-point be M (5, 7)

AM =√(12-5)² + (8-7)²=√49+1=√50= 5√2

MB =√(5- -2)² + (7-6)²=√49+1= 5√2

AM = MB = 5√2

This proves thatthe midpoint of the hypotenuse is equidistant from the angular points.

thanks dear

HCF of 468 and 222

468 = (222 x 2) + 24222 = (24 x 9) + 624 = (6 x 4) + 0

Therefore, HCF = 6

6 = 222 - (24 x 9)= 222 - {(468 – 222 x 2) x 9 [where 468 = 222 x 2 + 24]= 222 - {468 x 9 – 222 x 2 x 9}= 222 - (468 x 9) + (222 x 18)= 222 + (222 x 18) - (468 x 9)= 222[1 + 18] – 468 x 9= 222 x 19 – 468 x 9= 468 x -9 + 222 x 19

Hence, HCF of 468 and 222 in the form of 468x + 222y is 468 x -9 + 222 x 19.

thanks

For x = - 1p*(-1)(-1) + 3(-1) +q = 0p - 3 + q = 0p + q = 3

For x=-2p*(-2)(-2)

Remaining solutionFor x = - 2p*(-2)(-2) + 3(-2) + q = 04p + q = 6

From first equationq = 3-p

4p + 3 - p = 63p = 3p = 1

q = 3-1 = 2

q-p = 2-1 = 1

x=-1so p-3+q=0so p+q=3x=-24p-6+q=94p+q=64p+q-p-q=6-3=3so 3p=3 so p=1so q=2so p-q=1-2=-1

(9x-7)(2x-5)-(3x-8)(5x-3)[9x(2x-5)-7(2x-5)]-[3x(5x-3)-8(5x-3)][18x^2-45x-14x+35]-[15x^2-9x-40x+24]3x^2+(-45-14+9+40)x+(35-24)3x^2-10x+11

For equation of the formax^2 + bx + c = 0

Discriminant = b^2 - 4ac

Then,For equation (x+5)^2 = 2(5x - 3)(x^2 + 25 + 10x) = 10x - 6x^2 + 31 = 0

Discriminant = 0 - 4*1*31= - 124

x^2+25+10x= 10x-6x^2+31= 0D= -4ac-4*31= -124

for 140

x + 2y = - 1...... (1)2x - 3y = 12..... (2)

Multiply eq(1) by 22x + 4y = - 2.....(3)

Subtract eq(2) from eq(3)

4y + 3y = - 2 - 127y = - 14y = - 14/7 = - 2

Put value of y in eq(1)x + 2*-2 = - 1x - 4 = - 1x = - 1 + 4 = 3

1+(-1)+1+(-1)2-20 as i square=-1

i² = - 1

So, i^6 = i² × i² × i² = 1

1 + i² + i⁴ + i^6

1 + ( -1) + 1 + 12

By using :a = bq + r 81 = 63 * 1 + 1863 = 18 * 3 + 918 = 9 * 2 + 0H.C.F = 9

= 63 - ( 18 * 3 )= 63 - 3*18= 63 - 3 [ 81 - ( 63 * 1 ) ]= 63 - 3 [ 81 - 63 ]= 63 - 81(3) + 63(3)= 63(4) + 81(-3)

So, In63x + 81yx = 4andy = -3Answer

Length of first wire = 25.85 mLength of second wire = 18.29 mTotal length of wire= 25.85 + 18.29= 44.14 m

how did you converted the units sir plz tell me

plz tell me

100cm = 1m85cm =.85m25m 85cm = 25m +. 85 m= 25.85 m

If we show for one element it will works.

So, given f(x) = x²,

Domain is R

So, Consider x = 1 and x = -1

f(1) = 1² = 1

f(-1) = (-1)² = 1

So, two different elements have same image. Therefore function is many one.

For into, consider -1 from rangeAs there exists no x such that

x² = - 1 so function is into

Equation will have equal rootsat b^2= 4ac

a cosx - b sinx=c squaring both the sides => (acosx - bsinx)² =c² = a²cos²x+b²sin²x-2absinxcosx----------(1)now let's find out the square of (asin x +b cosx)=> (asinx + b cos x)²= a²sin²x+b²cox²x+2absinxcosx --------(2)adding (1) and (2)=> (acosx-bsinx)²+(asinx+bcosx)²=a²cos²x+b²sin²x-2absinxcosx+a²sin²x+b²cox²x+2absinxcosx

=>c²+(asinx+bcosx)²=a²cos²x+b²sin²x-2absinxcosx+a²sin²x+b²cox²x+2absinxcosx [using (1)]

=> c²+(asinx+bcosx)²=b²cos²x+b²sin²x+a²sin²x+a²cox²x

=> c²+(asinx+bcosx)²= b²(sin²x +cos²x)+ a²(sin²x+cos²x)

=>c²+(asinx+bcosx)²= b²+ a²

=> ( asinx+bcosx)²= b²+ a²-c²=> ( asinx+bcosx) = √(b²+ a²-c²)

x2+x-5=0x2+5x-x-5=0x(x+5)-1(x+5)=0x+5=0,x-1=0x=-5,x=1then the product of its zeroes is -5×1=-5

x²+x-5=0a=1,b=1,c=-5product of roots=c/a=-5/1=-5