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every point on the number represents one and one only real number

Total= 840Chocolate flavour=(2/3)*840=560

Given f(x) = 27x^3and g(x) = x^1/ 3(gof) (x) = g [f(x)] = g [27x^3] = [27x^3]^1/ 3= 3x

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(23x^2y^3z)*(-15x^3yz^2)= 23*(-15)(x^[2+3]y^[3+1]z^[1+2])= -345(x^5y^4z^3)

it was wrong

Perimeter=32cmSum of two sides = 11cm+8cm=19cmThird side=32-19=13cm

kx(x-2) + 6 = 0kx^2 - 2kx + 6 = 0for equal rootsD=0b^2 -4ac=04k^2 -24k=04k(k-6)=0k=0,6

x = 24000/125 = 192

(1 + i)/(1 - i)}^m = 1

multiply (1 + i) numberator as well as denominator .

{(1 + i)(1 + i)/(1 - i)(1 + i)}^m = 1

{(1 + i)²/(1² - (i)²)}^m = 1

{(1 + i² +2i)/2 }^m = 1

{( 2i)/2}^m = 1

{i}^m = 1

we know, i^4n = 1 where , n is an integer.

so, m = 4n where n is an integerse.g m = 4 { because least positive integer 1 } hence, m = 4 ( answer )

5+(2/10)+(7/100)=5+0.2+0.07=5.27

As givenIn 1 kg = 4 packets are being madethen in 6 kgs = 4*6= 24 packets of 250 gm each so Answer is 24 packets

sum of 7 terms=7/2(2a+6d)=497(a+3d)=49so a+3d=7sum of 17 terms=17/2(2a+16d)=28917(a+8d)=289so a+8d=17a+8d-a-3d=17-7=105d=10so d=2a+6=7 so a=1sum of n terms=n/2(2+(n-1)2)=n/2(2+2n-2) =n^2

A) yes, because in both the triangle Base and hypotenuse are equal B) yes, it is parallel because distance between both lines are 4cm

A)yes because in both the triangle base and hypotenuse are equalB) yes it is parallel because distance between both lines of 4 c.m

S6 = 6/2(2a + 5d)S10 = 10/2(2a + 9d)

Now,S6/S10 = 3/73(2a + 5d)/5(2a + 9d) = 3/721(2a + 5d) = 15(2a + 9d)14a + 35d = 10a + 45d4a = 10dd = 2a/5........(1)

Given,a16 = 3a - 4a + 15d = 3a - 4

Put value of d from eq (1)a + 15*2a/5 = 3a - 4a + 6a = 3a - 44a = - 4a = - 1

d = 2*-1/5 = - 2/5

Therefore, AP isa, a + d, a + 2d........-1, (-1 + - 2/5), (-1 + - 4/5).....-1, - 7/5, - 9/5.....

4 terms is the correct answer .

ax³+bx²+cx+d

your options are certainly wrong

sn=n/2(2a+(n-1)d)n=16,a=10,d=-4s16=16/2(2(10)+(16-1)-4)s16=8(20-60)s16=8(-40)s16=-320

Let the first term be a and common difference be dnth term = a+(n-1)d

Given Third Term + Seventh term = (a+2d)+(a+6d) = 6 ==> a+4d = 3hence, a= 3-4d

Third Term * Seventh term = (a+2d)*(a+6d) = 8(3-4d+2d)*(3-4d+6d) = 8==> (3-2d)*(3+2d) = 8i.e. 9-4d^2 = 8==> d^2 = (9-8)/4 = 0.25==> d = 0.5 or -0.5

Now to check which is correct d...Substitute and find

Case (a): d= 0.5a+4d = 3==> a=3-4d = 3-4(0.5)=13rd term = a+2d= 1+2*0.5 = 27th term = a+6d= 1+6*0.5 = 4Sum = 6 and Product = 8Case (b): d= -0.5a+4d = 3==> a=3-4d = 3-4(-0.5) = 3+2 = 53rd term = a+2d= 5+2*(-0.5) = 47th term = a+6d= 5+6*(-0.5) = 2Sum = 6 and Product = 8

Since both are matching, we will go with bothvalues

Sum of first 16 terms = n*(2a+(n-1)d)/2 = 16*(2a+15d)/2= 8*(2a+15d)

Case (a): d= 0.5Sum = 8*(2*1+15*0.5)=76

Case (b): d= 0.5Sum = 8*(2*5+15*(-0.5))=20

we know that tan(a+b) = (tan(a)+tan(b))/(1-tan(a)tan(b)) = [m/(m+1) + 1/(2m+1)]/[1-m/(m+1)(2m+1)] = [2m²+m+m+1/(m+1)(2m+1)]/[(2m²+3m+1-m)/(m+1)(2m+1)]= 2m²+2m+1/2m²+2m+1= 1

so. tan(a+b) = 1=> a+b = tan-¹(1) = π/4.

we know that tan(a+b) = (tan(a)+tan(b))/(1-tan(a)tan(b)) = [m/(m+1) + 1/(2m+1)]/[1-m/(m+1)(2m+1)] = [2m²+m+m+1/(m+1)(2m+1)]/[(2m²+3m+1-m)/(m+1)(2m+1)]= 2m²+2m+1/2m²+2m+1= 1

so. tan(a+b) = 1=> a+b = tan-¹(1) = π/4..

Hi,

Here is the answer to your question.

find 11th term of Aap 3 7 11..

Any quadratic polynomial with zeroes a and b can be written as

( x - a) ( x - b) = x² - ( a + b) x + ab

Here zeroes are -4 and 3.

So, a + b = -4 + 3 = - 1

ab = (-4) × 3 = -12

So, quadratic equation is

x² - x -12

lim ( 3x² - 8x + 15)x-->2

lim ( 3 × 2² - 8 × 2 + 15)x-->2

lim ( 12 - 16 + 15) = 12 - 1 = 11x-->2

option 4 is the correct answer