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viva [34]
1 year ago
13

Show that b^2 is greater than equal to or less than ac, according as a, b, c are in A.P., G.P.

Mathematics
1 answer:
ValentinkaMS [17]1 year ago
5 0

Answer/Step-by-step explanation  (ac > b² or b² < ac. )

A/c to question, we have to show:-

b² >ac in A.P ........ (1)

b² = ac in G.P .....(2)

b² < ac in H.P. ..... (3)

b = a+c/2 (A.P)

b = √ac ( G.P)

b = 2ac/a+c (H.P)

In A.P :

b² > ac = b² - ac

= (a+c/2)² - ac

= (a²+2ac+c²/4) - ac  = a² + 2ac + c² - 4ac / 4

= a² - 2ac + c² / 4  = ( a - c ) ² / 4 > 0  Hence, b²>ac

In G.P:-

b = √ac

Hence, b² = ac

In H.P :-  b² < ac = ac > b²  = ac - b²  = ac - ( 2ac / a+c)

= ac(a+c) - 2ac / a+c

= a²c + ac² - 2ac / a+c

= ac(2ac - 2) / a+c > 0

Hence, ac > b² or b² < ac.

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A company manufactures aluminum mailboxes in the shape of a box with a half-cylinder top. The company will make 1863 mailboxes t
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Answer:

3433 m²

Step-by-step explanation:

From the image, we have a rectangular box without cover and half a cylinder on top.

Formula for surface area of rectangular box with top is;

S = 2(lh + wh + lw)

From the image,

l = 0.6 m

w = 0.4 m

h = 0.55 m

Thus;

S = 2((0.6 × 0.55) + (0.4 × 0.55) + (0.6 × 0.4))

S = 1.58 m²

Now, since the top is not included for this figure, then;

Surface area of this rectangular box is;

S1 = 1.58 - (lw) = 1.58 - (0.4 × 0.6) = 1.34 m²

Surface area of a cylinder is;

S = 2πr² + 2πrh

r is radius and in this case = 0.4/2 = 0.2 m

h = 0.6

S = 2π(0.2² + (0.2 × 0.6))

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Since it is half cylinder, then we have;

S2 = 1.005/2

S2 = 0.5025 m²

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S_t = 1.8425 m²

This is the surface area of one mail box.

Thus, for 1863 mailboxes, total surface area is;

S = 1863 × 1.8425 = 3432.5775 m²

Approximating to the nearest Sq.m gives;

S = 3433 m²

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