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let the numerator be x then, denominator will be x+2.

according to given condition we getx/x+2 + x+1/x+2+1 = 19/15

solving this we get quadratic equation 11x²-5x-84=0

solving this we have x=3

original fraction = x / x+2= 3 / 3+2= 3/5

Ans: 3/5

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Circumference = 2×pi×r where pi = 22/7 a) 2×22×7/7 = 44 cm b) 2×22×1.4/7 = 8.8 mm c) 2×22×28/7 = 176 cm

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6 1/4=25/4 now 2hrs 40minutes=2*60+40=120+40=160minuteshence25/4% of 16025/400*160=10minutes

(a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)

2(ab + bc + ca) = (a + b + c)^2 - (a^2 + b^2 + c^2) = 169 - 69 = 100.

(ab + bc + ca) = 50.

Thanks for the answer

We know,(a+b+c)² = a²+b²+c² + 2(ab+ac+bc)

0² = 20 + 2(ab+ac+bc)

-20/2 = (ab+ac+bc)

(ab+ac+bc) = -10

P=4a=4·6.4= 25.6 is the answer

Weight of the heavier sphere = 11840 g and weight of the lighter sphere = 1480 g, and diameter of lighter sphere = 10 cm or radius = 5 cmLet the volume of the heavier sphere be 'V1' and the volume of the lighter sphere be 'V2'. Radius of the heavier sphere be 'r1' because the diameter of the lighter sphere is given so the radius is 5cm.Weight of an object = density× volume of that objectSo,(Weight of the heavier sphere/weight of the lighter sphere) = (Density× V1/density× V2)As both the spheres are

As both the spheres are made up of same metal, therefore, the ratio of their weights will be equal to the ratio oftheir volumes.(weight of the heavier sphere/weight of the lighter sphere) = (V1/V2)⇒ (11840/1480) = (4/3πr₁³)/(4/3πr³)⇒ (11840/1480) = (4/3×22/7×r₁³)/(4/3× 22/7× 5× 5× 5)

10y = 20 so y = 25x = 15 so x = 3.

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f) 0.850 kmg) 9.01 km



Secondary School

Science5 points

Explain the role and importance of information communication in Science and Technology

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FollowReportbySurya519516.09.2018

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bhaumikkvora12

Helping Hand

Due to communication in science and technology the communication helps to do the advertisment of the things made or the new technology so that the people take and use for there betterment or their help.Information and technology has got efficient storage systems which may be used to store information which may be of very vital impact in future. Communication is made possible faster and efficiently between individual;s who are very apart and there is provision of immediate feedback, thus ensuring easy provision of goods and services

Thecomputerdoes its primaryworkin a part of the machine we cannot see, a control center that converts data input to information output. This control center, called the central processing unit (CPU), is a highly complex, extensive set of electronic circuitry that executes stored program instructions.

The formula for thevolume of a prismis V = B h , where is the base area and is the height.Area of base√3/4a^2*height( Area of equilateral Triangle)for aperimeter is givenso 3a= 15( perimeter of equilateral Triangle a+a+a)a= 5now volume√3/4*25*5

54.12cm^3

very very thank-you

As the base of triangular prism is equilateral.side=15cm/3=5 cm∴ Area of base = √ 3 / 4 (side)2⇒ Area of base = √3 / 4 x 52⇒ ⇒ Area =( 25÷4 )x 1.73 = 10.83 cm square.Volume of prism = Area of base x heightV= 10.83x 5Volume = 54.15 cm3

L= 25;cmr=15 mh=20 cmvolume =1/3π*r*r*h4710 cubic centimetres

Radius of base = 105/2 m

Curved Surface Area of Cylinder = 2πrh

=> 2 * (π) * (105/2) * 5 = 525π

Curved Surface Area of Cone = πrl

=> (π) * 105/2 * 60 = 3150 (π)

Therefore total area of cloth required = 3150 (π) + 525 (π) = 3675 (π) = 11545.353

Therefore cost of making the tent = 11545.353/15 = ₹769.69

1 Draw a side of the square using ruler. Keep track of the length of this side so you can make all four sides the same length.

2Considering the side drawn in the previous step as one of the arms, construct a right angle on each end of it. Thus the end points of the side drawn in the previous step would also be the two vertices of these right angles

Mark a point on each of the newly drawn arms (of the two right angles), at a distance (measured from the respective vertex of the right angle) which is same as the length of the side drawn initially.Join these two points.

join oters as well

Perimeter = 4a= 160a= 40area = axa= 1600m^21600*100 cm^2

perimeter=4a=160a=40maesa=a^2=1600m^2 =160(100cm)^2 =160×100×100cm^2 1600000cm^2 =16×10^5 cm^2

Charge on one electron = -1.6×10^(-19) C

no. of electrons =2.4*10^-18/1.6*10^-19=24/1.6=15

Since the charge on the conductor is positive, the conductor is 15 electrons short.