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3 years ago

Can someone please help with this?

Mathematics

1 answer:

Mkey [24]3 years ago

6 0

Answer:

true

Step-by-step explanation:

first find the median of the lower and upper half of the data. These values are quartile 1 (Q1) and quartile 3 (Q3). The IQR is the difference between Q3 and Q1.

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A solid right pyramid has a square base. The length of the base edge is4 cm and the height of the pyramid is 3 cm period what is

Illusion [34]

Answer:

The volume of this pyramid is 16 cm³.

Step-by-step explanation:

The volume $V$ of a solid pyramid can be given as:

$\displaystyle V = \frac{1}{3} \cdot b \cdot h$ ,

where

$b$ is the area of the base of the pyramid, and
$h$ is the height of the pyramid.

Here's how to solve this problem with calculus without using the previous formula.

Imaging cutting the square-base pyramid in half, horizontally. Each horizontal cross-section will be a square. The lengths of these squares' sides range from 0 cm to 3 cm. This length will be also be proportional to the vertical distance from the vertice of the pyramid.

Refer to the sketch attached. Let the vertical distance from the vertice be $x$ cm.

At the vertice of this pyramid, $x = 0$ and the length of a side of the square is also $0$ .
At the base of this pyramid, $x = 3$ and the length of a side of the square is $4$ cm.

As a result, the length of a side of the square will be

$\displaystyle \frac{x}{3}\times 4 = \frac{4}{3}x$ .

The area of the square will be

$\displaystyle \left(\frac{4}{3}x\right)^{2} = \frac{16}{9}x^{2}$ .

Integrate the area of the horizontal cross-section with respect to $x$

from the top of the pyramid, where $x = 0$ ,
to the base, where $x = 3$ .

$\displaystyle \begin{aligned}\int_{0}^{3}{\frac{16}{9}x^{2}\cdot dx} &= \frac{16}{9}\int_{0}^{3}{x^{2}\cdot dx}\\ &= \frac{16}{9}\cdot \left(\frac{1}{3}\int_{0}^{3}{3x^{2}\cdot dx}\right) & \text{Set up the integrand for power rule}\\ &= \left.\frac{16}{9}\times \frac{1}{3}\cdot x^{3}\right|^{3}_{0}\\ &= \frac{16}{27}\times 3^{3} \\ &= 16\end{aligned}$ .

In other words, the volume of this pyramid is 16 cubic centimeters.

5 0

2 years ago

Consider the force field and circle defined below.

grin007 [14]

By Green's theorem,

$\displaystyle\int_{x^2+y^2=9}\vec F(x,y)\cdot\mathrm d\vec r=\iint_D\left(\frac{\partial(xy)}{\partial x}-\frac{\partial(x^2)}{\partial y}\right)\,\mathrm dx\,\mathrm dy=\iint_Dy\,\mathrm dx\,\mathrm dy$

where $C$ is the circle $x^2+y^2=9$ and $D$ is the interior of $C$ , or the disk $x^2+y^2\le1$ .

Convert to polar coordinates, taking

$\begin{cases}x=r\cos\theta\\y=r\sin\theta\end{cases}\implies\mathrm dx\,\mathrm dy=r\,\mathrm dr\,\mathrm d\theta$

Then the work done by $\vec F$ on the particle is

$\displaystyle\iint_Dy\,\mathrm dx\,\mathrm dy=\int_0^{2\pi}\int_0^3(r\sin\theta)r\,\mathrm dr\,\mathrm d\theta=\left(\int_0^{2\pi}\sin\theta\,\mathrm d\theta\right)\left(\int_0^3r^2\,\mathrm dr\right)=\boxed0$