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1 year ago

Find the volume of the regular hexagonal prism with side lengths of 4 ft, height of 3ft and apothem of approximately 3.5 ft

Mathematics

1 answer:

nexus9112 [7]1 year ago

3 0

Answer:

A(p) = 84 ft³

Step-by-step explanation:

Volume of regular prism is equal to V(p):

V(p) = Area of the face * h

we know h = 3 ft

A regular hexagonal prism has 6 equal sides forming its face.

Finding the area (A₂) of a triangle formed by center of the prism, straight lines between the center and two adjacents vertex and side between these two vertex, and then multiply that area by 6 (number of equal triangles inside hexagonal prism) we get the area of the face, but we need further consideration, the triangle described above, has doble area of (A₁), the triangle formed by apothem, half side, and straight line between center of the face and the vertex, therefore.

Area of small triangle = base * height

A₁ = (1/2) * 4 * 3,5

A₁ = 2*3,5

A₁ = 7 ft²

A₂ = 2* 7

A₂ = 14 ft²

Finally volume of the hexagonal prism is:

V(p) = 6 * 14

A(p) = 84 ft³

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Determine the area (in units2) of the region between the two curves by integrating over the x-axis. y = x2 − 24 and y = 1

astra-53 [7]

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The area of the region between the two curves by integration over the x-axis is 9.9 square units.

Step-by-step explanation:

This case represents a definite integral, in which lower and upper limits are needed, which corresponds to the points where both intersect each other. That is:

$x^{2} - 24 = 1$

Given that resulting expression is a second order polynomial of the form $x^{2} - a^{2}$ , there are two real and distinct solutions. Roots of the expression are:

$x_{1} = -5$ and $x_{2} = 5$ .

Now, it is also required to determine which part of the interval $(x_{1}, x_{2})$ is equal to a number greater than zero (positive). That is:

$x^{2} - 24 > 0$

$x^{2} > 24$

$x < -4.899$ and $x > 4.899$ .

Therefore, exists two sub-intervals: $[-5, -4.899]$ and $\left[4.899,5\right]$ . Besides, $x^{2} - 24 > y = 1$ in each sub-interval. The definite integral of the region between the two curves over the x-axis is:

$A = \int\limits^{-4.899}_{-5} [{1 - (x^{2}-24)]} \, dx + \int\limits^{4.899}_{-4.899} \, dx + \int\limits^{5}_{4.899} [{1 - (x^{2}-24)]} \, dx$

$A = \int\limits^{-4.899}_{-5} {25-x^{2}} \, dx + \int\limits^{4.899}_{-4.899} \, dx + \int\limits^{5}_{4.899} {25-x^{2}} \, dx$

$A = 25\cdot x \right \left|\limits_{-5}^{-4.899} -\frac{1}{3}\cdot x^{3}\left|\limits_{-5}^{-4.899} + x\left|\limits_{-4.899}^{4.899} + 25\cdot x \right \left|\limits_{4.899}^{5} -\frac{1}{3}\cdot x^{3}\left|\limits_{4.899}^{5}$