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Let breadth be x cm.A/Q L×B=144 sq.cm. or, 18×x=144 or,x=144/18 x= 8 cm.Reqd.breadth=8 cm.

area skure is18cm+144sumito

breadth=8cm is the correct answer

8cm is the correct answer of the given question is

Let the breadth be x cm.A/Q(l×b)=144 sq.cm18×b=144 b=144/18 b=8cm.

speed in downstream=9+3 =12kmphspeed in upstream=9-3=6kmph

let the distance b/w a to be xso, time in going b to a (upstream)=x/6 hrtime in going a to b (downstream)=x/12 hr

as , x/6 + x/12=3so, 2x+x/12=3so, 3x=12×3=36so, x=36/3=12 km

speed in downstream=9+3 =12kmphspeed in upstream=9-3=6kmph

let the distance b/w a to be xso, time in going b to a (upstream)=x/6 hrtime in going a to b (downstream)=x/12 hr

as , x/6 + x/12=3so, 2x+x/12=3so, 3x=12×3=36so, x=36/3=12 km.

speed in downstream=9+3=12kmph, spread upstream=9-3=6kmph. let the distance b/w to be x, time in going b to a(up)=x/6 hr.going a to b ( down stream)=x/12 hr. x/6 + x/12 = 3; so, 2x+x/12=3; 3x/12=3, 3x=12×3=36; x=36/3=12 km

12km is the correct answer of this question

Let the distance traveled by bus = xTime taken traveling by bus = x/60Distance traveled by plane = (1900-x)Time taken traveling by plane = (1900-x)/700Total time = x/60 + (1900-x)/700 = 5x/60 – x/700 = 16/78/525x = 16/7x = 150Therefore, the distance traveled by bus = 150km

Let the distance traveled by bus = x

Time taken traveling by bus = x/60Distance traveled by plane = (1900-x)

Time taken traveling by plane = (1900-x)/700

Total time = x/60 + (1900-x)/700 = 5x/60 – x/700 = 16/7

8/525x = 16/7x = 150

Therefore, the distance traveled by bus = 150km

HCF *LCM = products if 2 numbersHere HCF= 5, LCM=200Now products of 2 number = 1000Therefore ab=1000Like if you find it useful

Thanks for the answer

distance travelled by ricksaw = 45 km 600 m - 43km 500m =2 km 100 m

distance travel by rickshaw = total distance - distance travel by bus = 13 km 285 m - 9 km 470 m = 3 km 815 m

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No they will not cross each other

As we know that when a1/a2=b1/b2 is not= to c1/c2 which is the situation of parallel lines.

And same is here.

a1/a2=b1/b2 is not= to c1/c2

1/-2=-3/6 is not =to 2/5

So they will not cross each other.

The perimeter of a square is 20then the side is 20/4so the side is 5mand the area is 5^2so the area is 25m^2

the number of hen is 200and goat is 290

Let’s assume Number of chicken as X and goat as Y

2X + 4Y = 490nd X + Y = 185

Since chicken has 2 legs and goat has 4 legs.

X = 185 - Y. sub value of X in equation 1.

2(185- Y) + 4Y = 490

370-2Y + 4Y = 490

370+ 2Y = 490

2Y = 120

Y = 60, Number of goat is 60. So chicken will 125

book ka answer and question bhejo please nhi smjhe h

Let’s assume Number of chicken as X and goat as Y

2X + 4Y = 490nd X + Y = 185

Since chicken has 2 legs and goat has 4 legs.

X = 185 - Y. sub value of X in equation 1.

2(185- Y) + 4Y = 490

370-2Y + 4Y = 490

370+ 2Y = 490

2Y = 120

Y = 60, Number of goat is 60. So chicken will 125

A bird sitting on a wire, flies, circles around and comes back to its perch .That means the bird covers circular path and reaches its perch.

Distance travelled : The total path traversed by an object .Here circumferenceof circle =2πR

Displacement is the shortest path travelled between source and destination with a direction.

As it reaches its original position, displacement is zero.

d is correct answer.

The given determinant will be k^3|A|.

D) is the correct answer

Since, A and B are square matrices of the order 3 × 3.∴ (AB)2= AB.AB= A(BA)B= A(AB)B [If AB = BA]= AABB= A2B2So, (AB)2= A2B2is true only if AB = BA.

Given,The diagonal of square=8 cm

We have to find the area of region lying outside the circle and inside the square.We know that diagonal of square =side*√2

Thus side*√2=8 cm

Therefore side = 8/ root 2 = 5.65 cm

The radius of circle = 5.65/2 = 2.825 cm

Thus required area = side² -radius² = 31.9225-25.0718 = 6.85 cm

what is the meaning of region lying outside the circle and inside the square

the coordinate is (4,0) as it lying in x axis y coordinate is 0

y coordinate point are4,0 where y=0

make a çircle then make a 60° angle from center

1 line can be drown.

(tan20/cosec70)^2 + (cot20/sec70)^2 + 2tan15tan45tan75

={tan(90-20)/cosec70}^2 + {cot90-20)/sec70}^2 + 2(tan90-15)tan45 tan 75

=(cot70/cosec70)^2 + (tan70/sec70) ^2+ 2(cot75 tan 45 tan75 )

=(cos 70/sin70/1/sin70)^2 +(sin70/cos70/1/cos70) ^2 + 2(1×1)

=(cos70+sin70)^2 +2

=1+2

=3

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We know |kA| = kⁿ|A| , where n is order of A.

So, |2A| = 2³ |A| = 8 × 4 = 32

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भिन्न = x x/(6/7) = 6x/7 + 13/70 7x/6 - 6x/7 = 13/70 (49x-36x)/(6×7) = 13/70 13x/6 = 13/10 x = 6/10 = 3/5

good

The domain is the set of the first number in every pair (those are the x-coordinates): {-1, 1, 2, 3, 5}. The range is the set of the second number of all the pairs (those are the y-coordinates): {3, 4, 5, 7, 9}. This table describes y as a function of x.

Therefore, for this problem:

The domain is{-1,1,2,3,5}

The range is{3,4,5,7,9}

option c is correct

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