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3 years ago

Is ABC DEF? If so, identify the similarity postulate or theorem that applies.

Mathematics

2 answers:

zubka84 [21]3 years ago

8 0

Answer:

Similar AA

Step-by-step explanation:

VladimirAG [237]3 years ago

7 0

We are given traingle ABC and Triangle DEF.

Angle <A is congruent to Angle <D

Angle <B is congruent to Angle <E

Angle <C is congruent to Angle <F.

If two pairs of corresponding angles are congruent to each other, then the triangles are similar by Angle-Angle Similarity.

<em>By Angle-Angle similarity postulate triangle ABC and triangle DEF are similar.</em>

Therefore, option C is correct option.

<h3>C. Similar - AA.</h3>

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Answer:

There are 4 possible ways to place the baseballs into containers.

Step-by-step explanation:

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2 years ago

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The table below represents a linear function f(x) and the equation represents a function g(x):

Elza [17]

Answer:

(a) The slope of f(x) is greater than the slope of g(x)

(b) f(x) has a greater y intercept

Step-by-step explanation:

Given

$\begin{array}{cc}x & {f(x)} & -1&-9&0&-1&1&7 \ \end{array}$

$g(x) = 3x - 2$

Solving (a): Compare the slopes

The slope (m) of f(x) is calculated as;

$m = \frac{f(x_2) - f(x_1)}{x_2 - x_1}$

This gives:

$m = \frac{f(0) - f(1)}{0 - 1}$

$m = \frac{f(0) - f(1)}{- 1}$

Substitute values for f(0) and f(1)

$m = \frac{-1 - 7}{- 1}$

$m = \frac{-8}{- 1}$

$m = 8$

The slope of g(x) can be gotten using the following comparison

$g(x) = mx + b$

$m \to slope$

So:

$g(x) = 3x -2$

$m = 3$

$m_{f(x)} > m_{g(x)}$

Solving (b): Compare the y intercept

y intercept is when $x = 0$

From the table of f(x)

$f(0) = -1$

From the equation of g(x)

$g(x) = 3x -2$

$g(0) = 3*0-2$

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2 years ago

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The table would fit because if he rolled the table in which sideways it would fit since its smaller than the door

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2 years ago

G verify that the divergence theorem is true for the vector field f on the region

Alenkasestr [34]

$\mathbf f(x,y,z)=\langle z,y,x\rangle\implies\nabla\cdot\mathbf f=\dfrac{\partial z}{\partial x}+\dfrac{\partial y}{\partial y}+\dfrac{\partial x}{\partial z}=0+1+0=1$

Converting to spherical coordinates, we have

$\displaystyle\iiint_E\nabla\cdot\mathbf f(x,y,z)\,\mathrm dV=\int_{\varphi=0}^{\varphi=\pi}\int_{\theta=0}^{\theta=2\pi}\int_{\rho=0}^{\rho=6}\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi=288\pi$

On the other hand, we can parameterize the boundary of $E$ by

$\mathbf s(u,v)=\langle6\cos u\sin v,6\sin u\sin v,6\cos v\rangle$

with $0\le u\le2\pi$ and $0\le v\le\pi$ . Now, consider the surface element

$\mathrm d\mathbf S=\mathbf n\,\mathrm dS=\dfrac{\mathbf s_v\times\mathbf s_u}{\|\mathbf s_v\times\mathbf s_u\|}\|\mathbf s_v\times\mathbf s_u\|\,\mathrm du\,\mathrm dv$
$\mathrm d\mathbf S=\mathbf s_v\times\mathbf s_u\,\mathrm du\,\mathrm dv$
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So we have the surface integral - which the divergence theorem says the above triple integral is equal to -

$\displaystyle\iint_{\partial E}\mathbf f\cdot\mathrm d\mathbf S=36\int_{v=0}^{v=\pi}\int_{u=0}^{u=2\pi}\mathbf f(x(u,v),y(u,v),z(u,v))\cdot(\mathbf s_v\times\mathbf s_u)\,\mathrm du\,\mathrm dv$
$=\displaystyle36\int_{v=0}^{v=\pi}\int_{u=0}^{u=2\pi}(12\cos u\cos v\sin^2v+6\sin^2u\sin^3v)\,\mathrm du\,\mathrm dv=288\pi$

as required.

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Olenka [21]

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