To place a picture in his class newsletter, Joaquin must reduce the picture by a scale factor of 0.3 . Find the dimensions of the reduced picture if the origin is 15 centimeters wide and 10 centimeters high.

CMIIh
2020-11-30
Answered

Find the dimensions of the reduction.

To place a picture in his class newsletter, Joaquin must reduce the picture by a scale factor of 0.3 . Find the dimensions of the reduced picture if the origin is 15 centimeters wide and 10 centimeters high.

To place a picture in his class newsletter, Joaquin must reduce the picture by a scale factor of 0.3 . Find the dimensions of the reduced picture if the origin is 15 centimeters wide and 10 centimeters high.

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Obiajulu

Answered 2020-12-01
Author has **98** answers

A scale factor is the ratio of the dilated images

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Find the vectors T, N, and B at the given point.

$r(t)=<{t}^{2},\frac{2}{3}{t}^{3},t>$ and point $<4,-\frac{16}{3},-2>$

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Find the volume of the parallelepiped with one vertex at the origin and adjacent vertices at

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Find a nonzero vector orthogonal to the plane through the points P, Q, and R. and area of the triangle PQR

Consider the points below

P(1,0,1) , Q(-2,1,4) , R(7,2,7)

a) Find a nonzero vector orthogonal to the plane through the points P,Q and R

b) Find the area of the triangle PQR

Consider the points below

P(1,0,1) , Q(-2,1,4) , R(7,2,7)

a) Find a nonzero vector orthogonal to the plane through the points P,Q and R

b) Find the area of the triangle PQR

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Show that

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Find the basis for kernel of a matrix transformation

Let$\psi \text{}:{\left\{Mat\right\}}_{2\times 2}\left(\mathbb{R}\right)\text{}\to \text{}{\left\{Mat\right\}}_{2\times 2}\left(\mathbb{R}\right)$

$\psi :\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)\to \left(\begin{array}{cc}a+b& a-c\\ a+c& b-c\end{array}\right)$

Find basis for ker$\psi$

Let

Find basis for ker

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How can you find a reflection matrix about a given line, using matrix multiplication and the idea of composition of transformations?

The line of:$y=-\frac{2x}{3}$ , all in $\mathbb{R}}^{2$

The line of:

asked 2022-06-20

I am trying to fit data to an equation of this form using non-linear regression:

$y=1-\frac{a}{\mathrm{exp}\left(n\left(x-{x}_{0}\right)\right)}$

I would like to select three sets of $(x,y)$ data and solve the above equation for $a$, $n$ and ${x}_{0}$ in order to improve the initial estimates of these parameters. I eliminated $a$ using the following expression:

$a=\left(1-{y}_{1}\right)\mathrm{exp}\left(n\left({x}_{1}-{x}_{0}\right)\right)$

where $({x}_{1},{y}_{1})$ is one of the selected data points and then back-substituted it to give:

${y}_{2}=1-\frac{\left(1-{y}_{1}\right)\mathrm{exp}\left(n\left({x}_{1}-{x}_{0}\right)\right)}{\mathrm{exp}\left(n\left({x}_{2}-{x}_{0}\right)\right)}$

where $({x}_{2},{y}_{2})$ in another of the selected data points. This becomes

${y}_{2}=1-\left(1-{y}_{1}\right)\mathrm{exp}\left(n\left({x}_{1}-{x}_{2}\right)\right)$

and thus

$n=\frac{1}{{x}_{1}-{x}_{2}}\mathrm{ln}\left(\frac{1-{y}_{2}}{1-{y}_{1}}\right)$

If I use point $({x}_{3},{y}_{3})$ instead I get:

$n=\frac{1}{{x}_{1}-{x}_{3}}\mathrm{ln}\left(\frac{1-{y}_{3}}{1-{y}_{1}}\right)$

The problem is that I have two different expressions for n but none for ${x}_{0}$ because it was eliminated along the way. I get a similar result if I eliminate ${x}_{0}$ first.

Is it possible to solve for $a$, $n$ and ${x}_{0}$ in this manner?

$y=1-\frac{a}{\mathrm{exp}\left(n\left(x-{x}_{0}\right)\right)}$

I would like to select three sets of $(x,y)$ data and solve the above equation for $a$, $n$ and ${x}_{0}$ in order to improve the initial estimates of these parameters. I eliminated $a$ using the following expression:

$a=\left(1-{y}_{1}\right)\mathrm{exp}\left(n\left({x}_{1}-{x}_{0}\right)\right)$

where $({x}_{1},{y}_{1})$ is one of the selected data points and then back-substituted it to give:

${y}_{2}=1-\frac{\left(1-{y}_{1}\right)\mathrm{exp}\left(n\left({x}_{1}-{x}_{0}\right)\right)}{\mathrm{exp}\left(n\left({x}_{2}-{x}_{0}\right)\right)}$

where $({x}_{2},{y}_{2})$ in another of the selected data points. This becomes

${y}_{2}=1-\left(1-{y}_{1}\right)\mathrm{exp}\left(n\left({x}_{1}-{x}_{2}\right)\right)$

and thus

$n=\frac{1}{{x}_{1}-{x}_{2}}\mathrm{ln}\left(\frac{1-{y}_{2}}{1-{y}_{1}}\right)$

If I use point $({x}_{3},{y}_{3})$ instead I get:

$n=\frac{1}{{x}_{1}-{x}_{3}}\mathrm{ln}\left(\frac{1-{y}_{3}}{1-{y}_{1}}\right)$

The problem is that I have two different expressions for n but none for ${x}_{0}$ because it was eliminated along the way. I get a similar result if I eliminate ${x}_{0}$ first.

Is it possible to solve for $a$, $n$ and ${x}_{0}$ in this manner?