Is the expression x3•x3•x3 equivalent to x3•3•3? Why or why not? Explain your reasoning.

1 answer:

3 0

If the first expression reads x(cube) • x(cube) • x(cube) and x(cube • cube • cube), then the answer is no. They are not equal.

x(cube) • x(cube) • x(cube) will be equivalent to x(to the 9th power) while x(cube • cube • cube) will be equivalent to x( to the 27th power).

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Two triangles are similar with a scale factor of 3. The preimage triangle has a base of 7 inches and a height of 10 inches. What

vaieri [72.5K]

Answer:

The area of the larger triangle is 315 inches²

Step-by-step explanation:

1. Let's review all the information provided for answering the questions properly:

Pre-image triangle has a base of 7 inches and a height of 10 inches

Scale used : Factor of 3

2. What is the area of the larger triangle??

For calculating the area of the larger triangle, we use the scale this way:

Base = 7 inches * 3

21 inches

Height = 10 inches * 3

30 inches

Area of the larger triangle = 1/2 (Base * Height)

Area of the larger triangle = 1/2 (21 * 30)

Area of the larger triangle = 630/2 = 315 inches²

6 0

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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

7nadin3 [17]

The equation of the cone should be $z=\sqrt{x^2+y^2}$ . Parameterize $S$ by

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

with $1\le u\le2$ and $0\le v\le2\pi$ . Take the normal vector to $S$ to be

$\vec s_v\times\vec s_u=u\cos v\,\vec\imath+u\sin v\,\vec\jmath-u\,\vec k$

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

$\displaystyle\iint_S\vec F\cdot\mathrm d\vec S=\int_0^{2\pi}\int_1^2(-u\cos v\,\vec\imath-u\sin v\,\vec\jmath+u^3\,\vec k)\cdot(u\cos v\,\vec\imath+u\sin v\,\vec\jmath-u\,\vec k)\,\mathrm du\,\mathrm dv$