Write the expanded form of 5764321 by using exponents.​

  • matematika

    Subject:

    Math

  • Author:

    cirino

  • Created:

    1 year ago

Answers 1

Answer:he given numbers are 279404, 3006194, 2806196, 120719 and 20068.

The expansion of number 279404 is,

279404=2×105+7×104+9×103+4×102+0×101+4×100

The expansion of number 3006194 is,

3006194=3×106+0×105+0×104+6×103+1×102+9×101+4×100

The expansion of number 2806196 is,

2806196=2×106+8×105+0×104+6×103+1×102+9×101+6×100

The expansion of number 120719 is,

120719=1×105+2×104+0×103+7×102+1×101+9×100

The expansion of number 20068 is,

Step-by-step explanation:

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A farmer dug a flower bed of radius 7 m at the centre of a field. He needs to purchase fertilizer. If 1 kg of fertilizer is required for 1 square metre area. The cost of fertilizer is Rs10/kg. Answer the following questions. Q12. What is the area of flower bed ?

Answer:

Area = Pi * (r)^2

or, A = πr^2

thus, A = 3.14159 (7)^2

which is A = 3.14159 * 49

which is 150, using appropriate significant figures (I rounded 153.97 down).

Thus, the farmer will need 150Kg.

According the question,:-

Radius of the flower bed = 7 m.

∴ Area of the flower bed will be =  π r2  

                                                       = [tex]\frac{22}{7}[/tex] × (7m)²

                                                     =154m²

Hence total area of the flower bed will be 154m².

If 1 kg of fertilizer is required for 1 square meter area.

then the farmer will need 154kg of fertilizer.

cost of 1 kg of fertilizer =Rs 10

∴ cost of 154 kg will be = 154 ×Rs10 = Rs1540.

Ans :-  Total area of the flower bed will be 154m². And the farmer will need 154kg of fertilizer, which costs Rs1540.

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3. Solve for x: when 243x3*= (27)2X-5​

Answer:

243×3 = 54x-135

54x = 729+135

54x = 864

x = 864/54

x=16

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