Answer:
V = (1/3)πr²h
Step-by-step explanation:
The volume of a cone is 1/3 the volume of a cylinder with the same radius and height.
Cylinder Volume = πr²h
Cone Volume = (1/3)πr²h
where r is the radius (of the base), and h is the height perpendicular to the circular base.
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<em>Comment on area and volume in general</em>
You will note the presence of the factor πr² in these formulas. This is the area of the circular base of the object. That is, the volume is the product of the area of the base and the height. In general terms, ...
V = Bh . . . . . for an object with congruent parallel "bases"
V = (1/3)Bh . . . . . for a pointed object with base area B.
This is the case for any cylinder or prism, even if the parallel bases are not aligned with each other. (That is, it works for oblique prisms, too.)
Note that the cone, a pointed version of a cylinder, has 1/3 the volume. This is true also of any pointed objects in which the horizontal dimensions are proportional to the vertical dimensions*. (That is, this formula (1/3Bh), works for any right- or oblique pyramid-like object.)
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* in this discussion, we have assumed the base is in a horizontal plane, and the height is measured vertically from that plane. Of course, any orientation is possible.
5.3×10^5=53×10^4
53×10^4+3.8×10^4
=56.8×10^4
=5.68×10^5. Hope it help!
Answer:
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Step-by-step explanation:
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Answer:
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Step-by-step explanation:
Answer:
1. 
= 
2. 
= 
3. 
= 
4. 
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Step-by-step explanation:
1.
Taking LCM of (x+3) and (x+5) which is: (x+3)(x+5)
Prove closure: The value of x≠-3 and x≠-5 because if there values are -3 and -5 then the denominator will be zero.
2.
Factors of x^2-16 = (x)^2 -(4)^2 = (x-4)(x+4)
Factors of x^2+5x+6 = x^2+3x+2x+6 = x(x+3)+2(x+3) =(x+2)(x+3)
Putting factors
Prove closure: The value of x≠-2 and x≠4 because if there values are -2 and 4 then the denominator will be zero.
3.
Factors of x^2-9 = (x)^2-(3)^2 = (x-3)(x+3)
Factors of x^2-5x+6 = x^2-2x-3x+6 = x(x-2)+3(x-2) =(x-2)(x+3)
Putting factors
Taking LCM of (x-3)(x+3) and (x-2)(x+3) we get (x-3)(x+3)(x-2)
Prove closure: The value of x≠3 and x≠-3 and x≠2 because if there values are -3,3 and 2 then the denominator will be zero.
4.
Factors of x^2-5x+6 = x^2-3x-2x+6 = x(x-3)-2(x-3) = (x-2)(x-3)
Factors of x^2-16 = (x)^2 -(4)^2 = (x-4)(x+4)
Converting ÷ sign into multiplication we will take reciprocal of the second term
Prove Closure: The value of x≠2 and x≠4 because if there values are 2 and 4 then the denominator will be zero.
