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f(x)=4x^3-6antiderivative F(x)=4x^4/4-6x+cwhere c is a constantnow F(0)=3c=3so antiderivative = x^4-6x+3

This is a physics question, please post only Mathematics question here.

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part 2

angle 2 = 55° ( vertically opposite angle)angle 1 = 68° ( corresponding angle)angle 1 + angle 2 + angle 3 = 180° ( sum of angle of triangle)68° + 55° + angle 3 = 180°123° + angle 3 = 180°angle 3 = 180° - 123° = 57°angle 4 = angle 3 = 57° ( corresponding angle)angle 4 + angle x = 180° ( linear pair)57° + angle x = 180°angle x = 123°

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Answer of the integral will bexy+y+chenceput in y=6x^26x^3+6x^2+c

if 3 is a zero hence3)^2+2(3)+a=09+6+a=0hence a=-15

(a) Coefficient of x^3 is 5(b) Coefficient of x^3 is 27(c) Coefficient of x^3 is 2/5(d) Coefficient of x^3 is 2(e) Coefficient of x^3 is 1(f) Coefficient of x^3 is sqrt(5)(g) Coefficient of x^3 is - 16(h) Coefficient of x^3 is - 4

f(0)= 1 after putting 0 in the place of xf(-1)= 1+4+3+2+1= 11f(2)= 16-14(8)+3(4)-2(2)+116-112+12-4+1= -87as we can see f(0) f(-1) = 11 that is not equal to -87no it is not correct

(50/100)*(x-y) =(30/100)*(x+y)

=>5x-5y=3x+3y

=>2x=8y

=>y=x/4

=>y=(25/100)x

Therefore y is 25% of x.

option E should be the correct answer.

This is the definition of limit to exist at some point a

EUCLID'S fifth axiom states – The whole is greater than a part.

In the given figure clearly line AB is the whole while AC is a part of it hence AB>AC.

If thenumberyou areroundingis followed by 5, 6, 7, 8, or 9,roundthenumber up. Example: 38roundedto thenearest tenis 40. If thenumberyou areroundingis followed by 0, 1, 2, 3, or 4,roundthenumberdown. Example: 33roundedto thenearest tenis 30.

So,89=90415=4203951=3950

thanks

He is allowing 50 percentage on 150 percentagethen he will not have loss if he allow upto 50 percentagethen 50 percentage of 15050/150*100= 33.3 percentage

x^2 - 8x - 1280 = 0

x^2 - 40x + 32x - 1280 = 0

x(x - 40) + 32(x - 40) = 0

(x + 32)(x - 40) = 0

x = - 32, 40

ye ans Kaise aaya

2^x'-1+x+1=1280,; 2^2x=2^320; 2x=320; x=320/2=160

2^x-1 + 2^x+1=1280; 2^x/2 +2^2.2=1280, d=a^x. d/2+ 2y=1280;; 5d=1280×2; d=1280×2/5==128×2×2=2^9; 2^×=a%3; x=3

We have PQ = QR hence ∠QPR = ∠QRP

Also ∠PQR + ∠QPR + ∠QRP = ∠180º (sum of interior angles of a triangle).

∠PQR = 128º

∠QPR = ∠QRP = 180 - 128 / 2 = 26º

We can say that ∠QSR and ∠QPR are angles in the same segment, then:

∠QSR = ∠QPR = 26º

Now in triangle QRS, since QR = RS hence ∠QRS = ∠SQR = 26º

Also ∠QRS = 180º - 26 - 26º = 128º

Then ∠ROS = 2∠SQR = 2 x 26º = 52º

Also we can say that ∠QRS + ∠QTS = 180º, then ∠QTS = 180º - 128º = 52º

From the figure we also know that: ∠PQS + ∠SQR = ∠PQR

Then:

∠PQS = 128º - 26º = 102º

Now in cycic quadrilateral PQST, ∠PQS + ∠PTS = 180º

Then:

∠PTS = 180º - 102º = 78º

Now we can say that

∠PTQ + ∠QTS = ∠PTS

∠PTQ = 78º - 52º = 26º

Hence the angles measurement is 26º

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C.P.=1499loss of 12%S.P.=1499*0.78=1169.22

i) 1280+15*1280/100

= 1472

ii) 1280-10*1280/100

= 1152

SP = 1320

Loss = 12%

Let CP be x

So, we get equation

x - 12% of x = 1320

x - 12/100 × x = 1320

88x/100 = 1320

x = (1320 × 100)/88

x = 1500

So, CP is 1500

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a^x=b => x=logab

b^y=c=> y=logbc

c^z=a=> z=logca

xyz=(logab.logbc).logca

=logac.logca

=logaa

=1

Given equation is x+y/xyz if x=-4 y=2*1/2 z=1*1/2 = 3/2.

tan-1x+tan-1y+tan-1z=π -: tan-1(x+y/1-xy)+z=π. [tan-1x+tan-1y =(x+y/1-xy)] -: (x+y/1-xy)+z=tanπ -: (x+y/1-xy)+z=0 -: (x+y/1-xy)=-z -: x+y=-z(1-xy) -: x+y=-z+xyz Hence, x+y+z=xyz