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.
There are two methods to solve this problem. One of them is the typical method and the other one is the short-cut. I will explain the shortcut
Short-cut:
This method is quick and easy, and works with ALL triangles, which is why I suggest you use this. You make proportions. Since the triangles are similar, we can use CPCTC. Thereafter,
Now, you just need to identify the terms, substitute, and solve for the unknown, which would be k.
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So we found k, but it is asking for BE. Therefore, plug in the value of k, which is 11, into the expression for BE, which is k-7.
BE=K-7
k=11
BE=11-7
BE=4
Your answer is 4.
The roots of the entire <em>polynomic</em> expression, that is, the product of p(x) = x^2 + 8x + 12 and q(x) = x^3 + 5x^2 - 6x, are <em>x₁ =</em> 0, <em>x₂ =</em> -2, <em>x₃ =</em> -3 and <em>x₄ =</em> -6.
<h3>How to solve a product of two polynomials </h3>
A value of <em>x</em> is said to be a root of the polynomial if and only if <em>r(x) =</em> 0. Let be <em>r(x) = p(x) · q(x)</em>, then we need to find the roots both for <em>p(x)</em> and <em>q(x)</em> by factoring each polynomial, the factoring is based on algebraic properties:
<em>r(x) =</em> (x + 6) · (x + 2) · x · (x² + 5 · x - 6)
<em>r(x) =</em> (x + 6) · (x + 2) · x · (x + 3) · (x + 2)
r(x) = x · (x + 2)² · (x + 3) · (x + 6)
By direct inspection, we conclude that the roots of the entire <em>polynomic</em> expression are <em>x₁ =</em> 0, <em>x₂ =</em> -2, <em>x₃ =</em> -3 and <em>x₄ =</em> -6.
To learn more on polynomials, we kindly invite to check this verified question: brainly.com/question/11536910