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Ask QuestionPosted by Rekha My Mom 2 years, 11 months ago
- 1 answers
Preeti Dabral 2 years, 11 months ago
Let the radius of the base of the cone be r cm.
h = 15 cm, Volume = 1570 cm3
⇒{tex}{1\over3}\pi r^2h{/tex}= 1570
⇒{tex}1\over3{/tex} × 3.14 × r2 × 15 = 1570
⇒ {tex}r^2={{1570\times3}\over3.14\times15}{/tex} ⇒ {tex}r^2=100{/tex}
⇒ {tex}r=\sqrt{100}{/tex} ⇒ r = 10 cm.
∴ the radius of the base of the cone is 10 cm.
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Preeti Dabral 3 years ago
On the number line we first move 9 steps to the right from 0 reaching 9 and then we move 6 steps to the left of 9 and reach 3.
Thus, 9 + (-6) = 3
Posted by Vinod Kumar 3 years ago
- 5 answers
Preeti Dabral 3 years ago
Given,
Area of square = 100 cm2
Area of square = Side x Side
Let's substitute the options in the formula
15 cm:
=15 x 15 = 225
10 cm:
=10 x 10 = 100
12 cm:
=12 x 12 = 144
20 cm:
=20 x 20 = 400
∴The side is 10 cm
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Preeti Dabral 3 years ago
52 rupees 90 paise
52 rupees + {tex}\frac {90}{100}{/tex} rupees
=52.9 rupees
Snehal Sinha 3 years ago
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Preeti Dabral 2 years, 11 months ago
Length of rectangular hall = 14 m
Breadth of rectangular hall = 12 m
Area of hall ={tex}\text { Length } \times \text { Breadth }=14 \times 12=168 \mathrm{sq} . \mathrm{m} .{/tex}
Side of square tile = 10 cm = 0.1 m
Area of square tile ={tex}0.1^2=0.01 s q . m{/tex}
No. of tiles are required = {tex}\frac{168}{0.01}=16800{/tex}
Hence 16800 tiles are required .
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