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3 years ago

When you convert 100 feet per second to inches per second will there be more or less than 100 inches

1 answer:

7 0

Feet are bigger.
Inches are smaller.
You need more inches to make the same distance.

Just like:
Gallons are bigger.
Ounces are smaller.
You need more ounces to make the same volume.

You might be interested in

How do you work out 3^2+2^3 indices

Bingel [31]

3^2 +2^3

9+2^3

9+8

Answer : 17

4 0

3 years ago

Visual Description for Figure 1

just olya [345]

The slope in question seven is 1. Below are the solutions to the remaining questions

<h3>In what means can Transformation occur ?</h3>

Transformation of an object can occur in different ways. Such as translation, rotation, enlargement of different scale factors, reflection and dilation.

Given that Blue points, blue line segments, red points, and red line segments arranged on a Cartesian coordinate plane. Here are the blue points and their coordinates. Point A: (negative nine, 3). Point B: (negative 11, 3). Point C: (negative 10, 3). Point D: (negative 10, 5). Point F: (negative 10, 4). Point E: (negative 11, 4). Here are the red points and their coordinates. Point A-1: (negative 1, 6). Point A-2: (negative 3, 4). Point B-1: (negative 2, 4). Point B-2: (negative 5, 4). Point C-2: (negative 3, 3). Point D-1: (negative 5, 5). Point D-2: (negative 5, 3). Point E-1: (negative 6, 6). Point F-1: (negative 5, 6). A semicircle that lies below its line of symmetry AB. A semicircle that lies above its line of symmetry B-2 A-2. A triangle DEF. A triangle D-1 E-1 F-1. Line segments are drawn from C to D, from A-1 to B-1, from A-2 to B-2, and from C-2 to D-2. Triangle DEF, segment CD, and the semicircle with line of symmetry BA are arranged so that they look like a boat.

1. The transformations to use on the blue segment CD to get it to match with the red segment C2D2 is ROTATION of 180° Clockwise

2.The transformations to use on the blue triangle to get it to match with the red triangle is ROTATION of 360° Clockwise Explain your movement using the coordinates of the vertices.

3.The line segments on the boat that is parallel is line B2 to A2 because it is parallel to line D2 to C2

4.The line segments on the boat is perpendicular is line BCA because they are perpendicular to line CFD and AB.

5.The line segments on the boat have a slope of 0 is line BCA because change in y axis is 0

6.The line segments on the boat have an undefined slope is line DFC because change in x axis is 0. When denominator is zero, the fraction is undefined.

7.Given that E is at (−11,4) and D is at (−10,5), the slope of ED will be

Slope = Δy ÷ Δx

Slope = (5 - 4) ÷ (-10 + 11)

Slope = 1 ÷ 1

Slope = 1

Learn more about Transformation here: brainly.com/question/1462871

#SPJ1

8 0

2 years ago

Evaluate the surface integral S F · dS for the given vector field F and the oriented surface S. In other words, find the flux of

tresset_1 [31]

Because I've gone ahead with trying to parameterize $S$ directly and learned the hard way that the resulting integral is large and annoying to work with, I'll propose a less direct approach.

Rather than compute the surface integral over $S$ straight away, let's close off the hemisphere with the disk $D$ of radius 9 centered at the origin and coincident with the plane $y=0$ . Then by the divergence theorem, since the region $S\cup D$ is closed, we have

$\displaystyle\iint_{S\cup D}\vec F\cdot\mathrm d\vec S=\iiint_R(\nabla\cdot\vec F)\,\mathrm dV$

where $R$ is the interior of $S\cup D$ . $\vec F$ has divergence

$\nabla\cdot\vec F(x,y,z)=\dfrac{\partial(xz)}{\partial x}+\dfrac{\partial(x)}{\partial y}+\dfrac{\partial(y)}{\partial z}=z$

so the flux over the closed region is

$\displaystyle\iiint_Rz\,\mathrm dV=\int_0^\pi\int_0^\pi\int_0^9\rho^3\cos\varphi\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi=0$

The total flux over the closed surface is equal to the flux over its component surfaces, so we have

$\displaystyle\iint_{S\cup D}\vec F\cdot\mathrm d\vec S=\iint_S\vec F\cdot\mathrm d\vec S+\iint_D\vec F\cdot\mathrm d\vec S=0$

$\implies\boxed{\displaystyle\iint_S\vec F\cdot\mathrm d\vec S=-\iint_D\vec F\cdot\mathrm d\vec S}$

Parameterize $D$ by

$\vec s(u,v)=u\cos v\,\vec\imath+u\sin v\,\vec k$

with $0\le u\le9$ and $0\le v\le2\pi$ . Take the normal vector to $D$ to be

$\vec s_u\times\vec s_v=-u\,\vec\jmath$

Then the flux of $\vec F$ across $S$ is

$\displaystyle\iint_D\vec F\cdot\mathrm d\vec S=\int_0^{2\pi}\int_0^9\vec F(x(u,v),y(u,v),z(u,v))\cdot(\vec s_u\times\vec s_v)\,\mathrm du\,\mathrm dv$