30 points Please help

Help 80 points A concrete planter is formed from a square-based pyramid that was inverted and placed

Rufina [12.5K]

Answer:

The slant height of the pyramid refers to the height of a face of the pyramid.

Notice that each slant height forms a right triangle like the image attached shows, where the hypothenuse is the slant height. Additionally, the bottom leg is 1 yard, because is half of the base side, and the other leg is 2 yard. Using Pythagorean's Theorem, we have

$h^{2} =1^{2} +2^{2} \\h=\sqrt{1+4}=\sqrt{5}$

Therefore, the slant height is the square root of 5 yard.

The surface area of the composite figure is the sum of the surface area of both volumes.

<h3>Surface area of the cube.</h3>

$S_{cube}=5(2yd)^{2} =20 yd^{2}$

Notice that we only included 5 faces of the cube, that is because the sixth is the base of the pyramid, so if we include it here, we would have the wrong total surface area.

<h3>Surface area of the pyramid.</h3>

$S_{pyramid}=\frac{1}{2}pl$

Where $p$ is the perimeter of the base and $l$ is the slant height.

$S_{pyramid}=\frac{1}{2}(8yd)(\sqrt{5}yd )=4\sqrt{5}yd ^{2}$

Therefore, the surface od the composite figure is

$S_{total}=20yd^{2} +4\sqrt{5} yd^{2} \approx 29.94 yd^{2}$

To find the cubic yards of concrete, we need to subtract the volume of the figures, because the concrete would be inside the space between the cube and the pyramid.

$V_{concrete}=V_{cube} -V_{pyramid}=l^{3}- \frac{1}{3}l^{2}h$

Where $l=2yd$ and $h=2yd$

$V_{concrete}=2^{3}-\frac{1}{3}(2)^{2} (2)=8-\frac{8}{3} \approx 5.33 yd^{3}$

Therefore, it would be needed 5.33 cubic yards of concrete to make the planter.

8 0

3 years ago

What is the length of a line segment with endpoints 3, 2, and -3, -4, in

djyliett [7]

$\qquad \qquad \qquad\large\overbrace{\underbrace{\underline{ \boxed{ \sf \:Answer }}}}$

Let's use distance formula :

$\qquad \sf \dashrightarrow \: \sqrt{(3 - ( - 3)) {}^{2} + (2 - ( - 4) {}^{2} }$

$\qquad \sf \dashrightarrow \: \sqrt{(3 - ( - 3)) {}^{2} + (2 - ( - 4)) {}^{2} }$

$\qquad \sf \dashrightarrow \: \sqrt{ {6}^{2} + {6}^{2} }$

$\qquad \sf \dashrightarrow \: \sqrt{ {6}^{2}(1 + 1) }$

$\qquad \sf \dashrightarrow \: 6\sqrt{ 2 } \: \: units$

6 0

3 years ago

Use either the shell method or the disk/washer method to find the volume of the solid (Calculus Help!!!)

Rufina [12.5K]

Integrating with shells is the easier method.

V = 2π ∫₁³ x (√x + 3x) dx

That is, at various values of x in the interval [1, 3], we take n shells of radius x, height y = √x + 3x, and thickness ∆x so that each shell contributes a volume of 2π x (√x + 3x) ∆x. We then let n → ∞ so that ∆x → dx and sum all of the volumes by integrating.

To compute the integral, just expand the integrand:

V = 2π ∫₁³ (x ³ʹ² + 3x ²) dx

V = 2π (2/5 x ⁵ʹ² + x ³) |₁³

V = 2π ((2/5 × 3⁵ʹ² + 3³) - (2/5 × 1⁵ʹ² + 1³))

V = 4π/5 (9√3 + 64)