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Therefore the statement is true.

let the amount of sand be 1.7x and amount of concrete be 1.9x so, total weight is (1.7+1.9)x = 31.104kg => 3.6x = 31.104 so, x = 31.104/3.6 = 8.64kg therefore , the amount of sand is 1.7x = 1.7*8.64 = 14.688 kg.

Percentage= 8.5/17*100= 50

1.6x²-√2x-2=06x²-3√2x+2√2x-2=03x(2x-√2)+√2(2x-√2)=0(2x-√2)(3x+√2)=02x-√2=0 or 3x+√2=0x=√2/2 or x=-√2/3

let the number be 'x'According to questionx+12=160 of 1/xx²+12x=160(multiply both sides by x)x²+12x-160=0x²+20x-8x-160=0x(x+20)-8(x+20)=0(x+20)(x-8)=0x+20=0 or x-8=0x=-20 or x=8

2.let the no be xincreased by 12=x+12160 times its reciprocal=160(1/x);160/xby equalling,x+12=160/x x*2+12x=160x*2+12x-160=0by factorizing,(x+20)(x-8)=0(x+20)=0 and (x-8)=0x=-20 and x=8

The general form of AP is a+(n-1)d11th term is a+10dwe need a and d a(first term)=-5d=t2-t1=(-5/2)-(-5)=(-5/2+5)d=5/211th term=a+10d=-5+10(5/2)=-5+(5)(5)=-5+25=20

In triangleAPB nd tri.BRQangleAPB=angleBSR (alternate angles)PB=BS(given-B is mid pt.)anglePBA=angleSBR(vertiacally opp. angles)triangleAPB and tri. BRQ are congruent by ASA.=>AP=SR (by cpct)__(1)PQRS is a parallelogram,=>PQ=SR (opp. sides of a parallelogram are equal)__(2)according to eq. (1) and (2)AP=SRPQ=SR=>AP=PQ.AP+PQ=AQPQ+PQ=AQ (AP=PQ)2PQ=AQHence proved!

total amount of cement = 45 kg 75 g converting the above quantity into kg75g = (75/1000) = 0.075kgtotal amount of cement = 45+.075 = 45.075kg

used cement = 26 kg 598 gconverting the above quantity into kg598g = (588/1000) = 0.598gused cement = 26+.598 = 26.598kg

remaining amount of cement= (total amount of cement - used cement)= 45.075 -26.598= 18.477kg

amount of cement left in the sack is 18.477kg

Let us assume 5√2 to be Rational

Then,5√2 = a / b [ Where, a & b are co-prime and b ≠ 0 ]

√2 = a / 5b

a / 5b is rational

But, we know √2 is Irrational

This contradiction arise due to our wrong supposition that 5√2 is Rational.

Hence,5√2 is Irrational

(a) false(b) true(c) false, in litre(d) true

LHS =ap +ap+2q =

{a+(p-1)d} + {a +(p+2q-1)d}

a + dp -d + a +dp +2dq-d

2a +2pd -2d +2dq=2(a+dp-d+dq)

RHS

{a + (p+q-1)d}2

(a+dp+dq-d)2

LHS= RHS

We know that lengths of tangents drawn to a circle from an external point are equal,so,AP=AQ , i.e option (A)

A)ap=aq is the answer

(1) 10^-1 * 10^2 * 10^-3 * 10^4 * 10^-5 * 10^6

= 10^[-1 + 2 - 3 + 4 - 5 + 6]

= 10^[12 - 9]

= 10^3

= 1000

(2) (3^4 * 2^3) = 81*8 = 648 (2^5 * 3^2) = 32*9 = 288

Therefore, (3^4 * 2^3) is larger

Angle OAP = 90 °

therefore,

OP² = OA² + AP²

OP² = 16²+30² = 1156

OP = x = √1156 = 34

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Apis mellifera ligustica is theItalian beewhich is a subspecies of the westernhoney bee(Apis mellifera)

Theembryo sacof date palm Polygonum type contained fourmature cellsfollowing the early degeneration of the three antipods. Theembryo sac contains twosynergidcells, an eggcelland a large centralcellwhichisthe future first endospermcell.

Option-A

the answer is 4 because it is -c/d

If red is 40 and in the pie chart it is 1/2 then the whole is 80.Then white is 80/8=10

when red part is 40 the half of the pie so the other three colours of the circle is the halfso 40 ×1/8= 5

A=90 AB=AC so abgles agaimst them are also same lets say those are xso 90+x+x=180so 2x=90so.x=45

ABC is an equilateral triangle , where D point on side BC in such a way that BD = BC/3 . Let E is the point on side BC in such a way that AE⊥BC .

Now, ∆ABE and ∆AEC ∠AEB = ∠ACE = 90° AE is common side of both triangles , AB = AC [ all sides of equilateral triangle are equal ]From R - H - S congruence rule , ∆ABE ≡ ∆ACE ∴ BE = EC = BC/2

Now, from Pythagoras theorem , ∆ADE is right angle triangle ∴ AD² = AE² + DE² ------(1)∆ABE is also a right angle triangle ∴ AB² = BE² + AE² ------(2)

From equation (1) and (2) AB² - AD² = BE² - DE² = (BC/2)² - (BE - BD)² = BC²/4 - {(BC/2) - (BC/3)}² = BC²/4 - (BC/6)² = BC²/4 - BC²/36 = 8BC²/36 = 2BC²/9 ∵AB = BC = CA So, AB² = AD² + 2AB²/9 9AB² - 2AB² = 9AD²Hence, 9AD² = 7AB²

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8a^3 + 27b^3 + c^3 - 18abc

Using formulaa^3+b^3+c^3–3abc = (a+b+c)(a^2 + b^2 + c^2–ab–bc–ca)

= (2a)^3 + (3b)^3 + c^3 - 3(2a)*(3b)*c

= 4a^2 + 9b^2 + c^2 - 6ab - 3bc - 2ac