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3 years ago

Wich one is greater 7.6 or 7.41

Mathematics

2 answers:

Andre45 [30]3 years ago

8 0

7.6 is greater than 7.41. Since the 6 is greater than the 4, it makes 7.6 a bigger number :)

svetlana [45]3 years ago

4 0

7.6 is greater than 7.41

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If the cross-section shown has an approximate area of 78.54 cm2, what is the approximate volume of the cylinder if its height is

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3 years ago

Jason can travel 24 3/4 miles in one half hour what is his average speed in miles per hour

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A circus tent is cylindrical upto a height of 3.3 m and conical above it if the diameter of the base is 100m and slant height of

Rina8888 [55]

Total cost of canvas required in making the tent is rupees 79128

Solution:

Given:

Height of the cylindrical tent = h = 3.3 m

Diameter of it's base = d = 100m

$radius = \frac{diameter}{2} = \frac{100}{2} = 50$

r = 50 m

Curved surface area of cylinder:

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$C.S.A = 2 \times 3.14 \times 50 \times 3.3 = 1036.2$

Thus Curved surface area of cylinder = 1036.2 square meter

Curved surface area of cone:

$C.S.A = \pi rl$

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Thus Curved surface area of cone = 8854.8 square meter

Total curved area of tent:

⇒ curved area of cone + curved area of cylinder

$\rightarrow 8854.8 + 1036.2=9891$

Thus total curved area of tent = 9891 square meter

find total cost of canvas required in making the tent:

The cost of canvas is 8 rupees per square meter

$\rightarrow \text{total cost } = 9891 \times 8 = 79128$

Thus total cost of canvas required in making the tent is rupees 79128

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3 years ago

Which of the following points are more than 5 units from the point P(−2, −2)? Select all that apply. A A (2, 1) B B (4, −1) C C

netineya [11]

The distance between any 2 points P(a,b) and Q(c,d) in the coordinate plane, is given by the formula:

 $|PQ|= \sqrt{ (a-c)^{2} + (b-d)^{2}}$

Using this formula we calculate the distances |PA|, |PB|, |PC|, |PD| and |PE| and compare to 5.

$|PA|= \sqrt{ (-2-2)^{2} + (-2-1)^{2}}= \sqrt{16+9}= \sqrt{25}=5$

$|PB|= \sqrt{ (-2-4)^{2} + (-2+1)^{2}}= \sqrt{36+1}= \sqrt{37} \approx 6$

$|PC|= \sqrt{ (-2-2)^{2} + (-2+3)^{2}}= \sqrt{16+1}= \sqrt{17}\approx4$

$|PD|= \sqrt{ (-2+6)^{2} + (-2+6)^{2}}= \sqrt{16+16}= \sqrt{32}\ \textgreater \ \sqrt{25}=5$

$|PE|= \sqrt{ (-2+5)^{2} + (-2-1)^{2}}= \sqrt{9+9}= \sqrt{16}=4$

Answer: B and D

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3 years ago