Explore topic-wise InterviewSolutions in .

volume of cube= 20 x 20 x 20 = 8000 cm^3

volume of cuboidal log of wood = 800 x 500 x 80 = 32000000 cm^3

now divide, 32000000 / 8000 = 4000So 4000 cubes

Please like the solution 👍 ✔️

2x-3y=-8-----(1)4x+3y=2-----(2)multiply with 2 in equation (1)4x-6y=-16----(3)4x+3y=2subtract(2) from (3)-9y= -18y= 2 and x is equal to -1

cosec (90° + A) +xcos A cot (90° + A) = sin (90° + A)

1/sin(90+A)+xcosAcos(90+A)/sin(90+A)=sin(90+A)⇒ 1 +xcos A cos (90° + A) = sin^2(90° + A)

⇒ 1 –xcos A sin A = cos^2A[cos (90° + A) = – sin A]

⇒ 1 – cos^2A =xcos A sin A

⇒ sin^2A =xcos A sin A

⇒ sin A (xcos A – sin A) = 0

sinA=0 X=sinA/cosA

⇒x= tan A (If sin A ≠ 0)

120

Explanation:A regular hexagon has 6 equal exterior and 6 equal interior angles. The sum of the exterior angles is 360 deg, hence each exterior angle is 360/6 = 60. The interior angle being supplementary of the exterior, its value will be 180–60 = 120 deg.

Given outer length = 10 cmOuter breadth = 8 cmOuter height = 7 cmOuter volume = length × breadth × height= 10 × 8 × 7 = 560 cm3Thickness of wooden box, w = 1 cmInner length =l– 2w = 10 – 2 = 8 cmInner breadth =b– 2w = 8 – 2 = 6 cmInner height =h– 2w = 7 – 2 = 5 cmOuter volume = 8 × 6 × 5 = 240 cm3Volume of wood used = outer volume – inner volume= 560 – 240 = 320 cm3Given cost of 1 cm3of wood = Rs 2Hence the total cost of wood required = Rs 2 × 32= Rs 640

<BOD = <AOC(vertically opposite angles)Hence, <BOD = 60°

We know sum of all angles around a point is 360°

So,<BOD + <AOC + <AOD + <BOC = 360°

Let,<AOD = <BOC = x(vertically opposite angles)

Hence, 60 + 60 + x + x = 3602x = 360 - 120 = 240x = 240/2 = 120

Therefore, <AOD = <BOC = 120°

sinQ = 7/25tanQ =7/24sin²Q=49/625cos²Q=576/625

answers of both questions are - 49/625 and 576/625

sinQ= 7/25tanQ= 7/24sin2(sin square)Q= 49/625cos2(cos square)Q= 576/625

579/625 is the correct answer

in right angled Lmn

sinQ=7/25tanQ=7/24sin2Q=49/625cos2Q=576/625

sin q = 7/25tan q =7/24sin²q=49/625cos²q=576/625

sinQ=7/25 and cosQ=24/25and tan Q=7/24

sin square equal to answer hoga 47 /625

the question answer is sinQ=7/25 tanQ7/24sin2Q=49/625cos2Q=576/625

given that cosQ= 24/25and we know that sinQ=7/25tanQ= 7/24

sinQ=7/25tanQ=7/24sin=49/625cos=576/625

NaN 45 2.7

Volume of all the cubes = 9×9×9 + 6×6×6 + 4×4×4 + 2×2×2= 729 + 216 + 64 + 8= 1017 cm^3

wastage = 17cm^31017 - 17 cm^3=1000 cmm^3

edge of the new cube formed = cube root of 1000=10 cm

therefore surface area of cube = 6a^26 × 10^26 × 100600 cm^2

therefore Ans is 600cm^2

In ∆ AOB, OD is the bisector of angle AOB

OA/OB =AD/DB---------------eq(1)Theorem used here

[The internal bisector of an angle of a triangle divides the opposite side internally in the ratio of the sides containing angle]

In ∆BOC .OE is the bisector of angle BOC

OB/OC = BE/EC---------eq(2)

In ∆COA, OF is the bisector of angle COA

OC/OA =CF/FA-----------eq(3)

Multiplying eq 1, 2, 3

(OA/OB) * (OB/OC) * (OC/OA) = (AD/DB) * (BE/EC) * (CF/FA)

1= (AD/DB) * (BE/EC) * (CF/FA)

DB*EC*FA = AD*BE*CF

AD*BE*CF = DB*EC*FA

Area =3.14*6*6=113.04 cm^2

Letthe radius of the circle bercm.Area of the circle (A) = 154 cm2 ⇒154=227×(r)2cm2⇒r2=(154×722)⇒r2=(107822)⇒r2=(49)⇒r=7cm.⇒154=227×r2cm2⇒r2=154×722⇒r2=107822⇒r2=49⇒r=7cm.

Hence, the radius of the circle is 7 cm.

86400 is the correct answer of the given question

86400seconds are there in the 2010

Since the two circles are congruent,then their areas are also equal.

THEREFORE AREA OF BOTH THE CIRCLES ARE πr²

Like my answer if you find it useful!

Let us write a=b-1;c=b+1log(1+(b-1)(b+1))=2logblog(1+b^2-1)=2logblogb^2=2logb2logb=2logb::hence proved

Thanks

198 is the least number that we add in 14931 then it will become a perfect square.so 14931+198=15129=123*123

Given- PRQ=90°M is mid point of QR so, RM=MQ=1/2RQ

----> To find- PQ^2 =4PM^2-3PR^2

SOLUTION---> In ∆PRM ( By Pythagoras Theorem)

◆ PM^2= PR^2+ RM^2

◆PM^2=PR^2+ (1/2RQ)^2 ( As M is mid point)

◆PM^2=PR^2+RQ^2/4

◆RQ^2=4(PM^2-PR^2)

◆RQ^2=4PM^2-4PR^2------(1)

◆ In ∆PRQ ( By Pythagoras Theorem)

◆PQ^2=PR^2 +RQ^2

◆PQ^2=PR^2+(4PM^2-4PR^2). { from equation 1}

◆PQ^2= PR^2+4PM^2-4PR^2

◆PQ^2=4PM^2-3PR^2

______H.PROVED__________

We know one thing about parallelogram,

diagonals of a parallelogram bisects each other

e.g., midpoint of diagonal PR = midpoint of diagonal QS

midpoint of diagonal PR = {(2 + 11)/2, (-2 -1)/2 } = (13/2,-3/2)[ From midpoint section formula, we know that if two points (x₁,y₁) and (x₂,y₂) are given then midpoint of line joining of giebn pouts is {(x₁ + x₂)/2, (y₁ + y₂)/2}]

Similarly, midpoint of QS = {(7 + 6)/2 , (3 -6)/2} = (13/2 , -3/2)Here we see midpoint of PR = midpoint of QS so, PQRS is parallelogram

hit like if you find it useful

thanks

Labour goes in 1 sec = 2 stepsthen,he get down in next 2 sec = 1 step .so, finally he goes up in 1 step in = 3sec .hence , he goes 10 steps in =3×10 = 30 sec .

1

2