.tell where are u sleeping

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see message i sent.

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5.

It is not clear if the diagram is a parallelogram.

Sum of angles = 180 – b + x + 180 – a + y = 360 deg

X + y = a + b

If it is a parallelogram, as the lines seem to be parallel,

X = y = b = a.

6.

Join OB. OA = OB = radius. So OAB is an isosceles triangle. So perpendicular OM on to AB will meet at the center of AB.

AM = AB/2 = 12cm. OA = 13 cm

Applying Pythagoras theorem, OM = √(OA² – AM²) = 5 cm

12.

Mass of a solid sphere = volume * density

= 4/3 π R³ * d = 4/3 π * 4.2³ * 10 grams

13.

side of the cube = a = 8 cm

surface area of one face of a cube = a² = 8² cm

when two cubes are joined side by side, only 5 faces of each cube are visible. So surface area will be = 5 * a² + 5 a². When a third cube is joined, one surface gets hidden on the third cube and one of the first two cubes. Hence

total surface area = 5 a² + (10 - 1) a² = 14 a² (totally 14 faces are only visible.

15.

You have to do this on a graph paper.

3 x - 2 y = 4 and x + y - 3 = 0

the point of intersection will be, x = 2 and y = 1.

drawing graph is easy. Join points (0, -2) , (2, 1), (4, 4) to get the first line.

Join points (0, 3), (2,1), (3, 0) to get the second line.

Have the scale : 1 division (square) on graph = 1 or 1/2.

16.

C = 5/9 * (F - 32)

Take F on the x- axis and C on the y axis. Keep the origin at C = 0, F = 32.

Mark the point C = 0, F = 32. Mark 1 division on x axis as 9 deg F and on y-axis scale is 1 division (square) = 5 deg C.

Now mark a point for F = 41, C = 5 deg. and F = 50 deg, C = 10 deg.

Join the marked points.. Draw a horizontal line from y axis at y = 5 divisions (ie., C = 25 deg), then from the intersection of the line, draw a vertical line on to x - axis. You should find F = 77 deg. (between 8 and 9 divisions on x axis.

17.

Please draw the diagram: marking ABCD, E , and F. Draw perpendiculars from C and D on to AB. Perpendicular from C intersects Ef at H and AB at J. Perpendicular from D intersects EF at G and AB at I.

triangles DEG and DAI are similar, as sides are all parallel in both.

As DE = 1/2 DA, so EG = 1/2 AI.

triangles CHF and CJB are similar as their sides are all parallel.

as CF = 1/2 CB, then HF = 1/2 JB.

NOW, EF = EG + GH + HF = 1/2 AI + CD + 1/2 JB

= 1/2 (AI + 2 CD + JB ) = 1/2 ( AI + IJ + JB + CD ) = 1/2 (AB+CD)

we used the fact that , CD = GH = IJ, they are parallel and cut by perpendicular lines.

19.

Please take a paper or a graph and do as follows.

1. Draw BC horizontally and of length 4.5 cm.

2. Draw a line at 45 deg at B using instruments you have. Let it be 9 cm or so (long enough).

3. mark 2.5 cm on this line from B. Call this point D.

4. Draw a perpendicular bisector of CD to intersect the inclined line. This point will be A.

5. Join ABC.

Justification : AB = AD + 2.5 cm. AC = AD, as ADC is isosceles. I hope you know to draw perpendicular bisector of CD. at mid point of CD, draw a perpendicular. Or there is a better way by drawing circular arcs from D and C.

20.

curved surface area = 2 π R H total surface area = 2 π R H + 2 π R ²

ratio = H / (H + R ) = 2/3 => 3 H = 2 H + 2 R => H/R = 2

21.

So after deducting the waste of 2 m², the lateral surface area of Conical tent = 550 m².

Let radius of base = R = 7 m and Height = H at the center, and lateral length (height) be L.

π R L = 550 m² => L = 25 m.

H² = L² - R² => H = 24 m

Volume = 1/3 π R² H = 1/2 π * 7² * 24 m³

22.

Take a graph sheet and draw vertical bars of width 5 mm each. The height of bars in millimeters will be amount of manure produced. 1 mm = 1 thousand tonne.

Draw x axis (year) and y axis (amount of manure). Have a distance of 1 cm between one bar and another on x axis.

Draw first bar of height 18 mm, next 35 mm, 45 mm, 30 mm, 40 mm, 20 mm. The highest decrease occurred during the last year 1996-97, as its 20 mm and durin g 1994-95, the decrease is 15 mm. During other years, there is an increase.

23 .

Probability (age >= 40 yrs) = number of persons with age >= 40 / total number of persons

= (95+55+15) / (50+35+95+55+15) = 165 / 250 - simplify

Probability (30 <= age <=39) = 35 / 250 - simplify

P (39 < age < 60) = (95 + 55) / 250

24.

total number of cards = 24.

numbers which are multiples of 2 and 3 are multiples of 6: 6, 12, 18, 24.

Probability = 4 / 24 , as there are four such cards, multiples of 2 and 3.

25.

(2 x - 3) / 5 + ( x + 3)/4 = (2 x + 3)/ 4

multiply by 20: 4(2 x - 3 ) + 5 ( x + 3 ) = 5 ( 2 x + 3)

8 x - 12 = 5 x => x = 4

28.

Let centers of circles be AB. Let them intersect at P and Q. Let PQ intersect AB at C. PQ is perpendicular to AB. PC = CQ. APC and BPC are right angle triangles.

PC² = AP² - AC² = 10² - AC²

PC² = BP² - BC² = 17² - (21 - AC)² = - 152 - AC² + 42 AC

So, AC = 252 / 42 = 6 cm

Then PC = √(10²-6²) = 8 cm

Chord = 2 * PC = PQ = 16 cm

29.

volume of water in cone = V = 4/3 π R₁² H₁ = 1/3 π 20² * 30 cm³

volume of water in cylinder = V = π R₂² H₂ = π 8² H₂

H₂ = π/3 * 400 * 30 / π 64 = 62.50 cm

30.

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27.

Draw a square ABCD. Draw the diagonals AC and BD, intersecting at O.

Triangles ABO and CDO are similar, as OD || OB, OA || OC, included angles COD = angle AOB, and AB || CD.

AB || CD as angle B = 90 and angle A = 90 and C = 90 and hence exterior angle at C is also 90.

AB/CD = OB/OD = OC/ OA = 1, as AB = CD in a square.

So O is the midpoint of AC and BD. Hence proved.

30.

See diagrams enclosed. histogram and frequency polygon both.

31.

Origin is O (0,0)

Line is X + Y = 2

Let x = 0, y =2, means the Line intersects y axis at A (0,2)

Let y = 0, x = 2, so line intersects x axis at B (2, 0)

Area of RIGHT angle triangle AOB, 1/2 * AB * AO

= 1/2 * 2 * 2 = 2 units square

31 part (ii)

Draw a parallelogram ABCD with base AB horizontal. See diagram enclosed.

Area of both is same = b * h

Now perimeter of parallelogram P1 = 2 ( b + L ) = 2 ( b + √(h²+CE²)

perimeter of rectangle = 2 ( b + h) = P2 < P1

proved.

Draw a square ABCD. Draw the diagonals AC and BD, intersecting at O.

Triangles ABO and CDO are similar, as OD || OB, OA || OC, included angles COD = angle AOB, and AB || CD.

AB || CD as angle B = 90 and angle A = 90 and C = 90 and hence exterior angle at C is also 90.

AB/CD = OB/OD = OC/ OA = 1, as AB = CD in a square.

So O is the midpoint of AC and BD. Hence proved.

30.

See diagrams enclosed. histogram and frequency polygon both.

31.

Origin is O (0,0)

Line is X + Y = 2

Let x = 0, y =2, means the Line intersects y axis at A (0,2)

Let y = 0, x = 2, so line intersects x axis at B (2, 0)

Area of RIGHT angle triangle AOB, 1/2 * AB * AO

= 1/2 * 2 * 2 = 2 units square

31 part (ii)

Draw a parallelogram ABCD with base AB horizontal. See diagram enclosed.

Area of both is same = b * h

Now perimeter of parallelogram P1 = 2 ( b + L ) = 2 ( b + √(h²+CE²)

perimeter of rectangle = 2 ( b + h) = P2 < P1

proved.