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3 years ago

The rectangle ABCD is translated to another rectangle A'B'C'D' using the transformation T: (x → x - 1, y → y - 2).

Mathematics

2 answers:

Vika [28.1K]3 years ago

7 0

1,1 is your answer. just apply the rule to the set points

blondinia [14]3 years ago

4 0

(-3,0) (1,1) (0,3) (-2,-2) would be the where the new rectangle is, so out of your options you would only choose
(1,1) and (-3,0)

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brilliants [131]

Answer:

(a) x=2, y=-1

(b) x=2, y=2

(c) $\displaystyle x=\frac{5}{2}, y=\frac{5}{4}$

(d) x=-2, y=-7

Step-by-step explanation:

Cramer's Rule

It's a predetermined sequence of steps to solve a system of equations. It's a preferred technique to be implemented in automatic digital solutions because it's easy to structure and generalize.

It uses the concept of determinants, as explained below. Suppose we have a 2x2 system of equations like:

$\displaystyle \left \{ {{ax+by=p} \atop {cx+dy=q}} \right.$

We call the determinant of the system

$\Delta=\begin{vmatrix}a &b \\c &d \end{vmatrix}$

We also define:

$\Delta_x=\begin{vmatrix}p &b \\q &d \end{vmatrix}$

And

$\Delta_y=\begin{vmatrix}a &p \\c &q \end{vmatrix}$

The solution for x and y is

$\displaystyle x=\frac{\Delta_x}{\Delta}$

$\displaystyle y=\frac{\Delta_y}{\Delta}$

(a) The system to solve is

$\displaystyle \left \{ {{x+y=1} \atop {x-2y=4}} \right.$

Calculating:

$\Delta=\begin{vmatrix}1 &1 \\1 &-2 \end{vmatrix}=-2-1=-3$

$\Delta_x=\begin{vmatrix}1 &1 \\4 &-2 \end{vmatrix}=-2-4=-6$

$\Delta_y=\begin{vmatrix}1 &1 \\1 &4 \end{vmatrix}=4-3=3$

$\displaystyle x=\frac{\Delta_x}{\Delta}=\frac{-6}{-3}=2$

$\displaystyle y=\frac{\Delta_y}{\Delta}=\frac{3}{-3}=-1$

The solution is x=2, y=-1

(b) The system to solve is

$\displaystyle \left \{ {{4x-y=6} \atop {x-y=0}} \right.$

Calculating:

$\Delta=\begin{vmatrix}4 &-1 \\1 &-1 \end{vmatrix}=-4+1=-3$

$\Delta_x=\begin{vmatrix}6 &-1 \\0 &-1 \end{vmatrix}=-6-0=-6$

$\Delta_y=\begin{vmatrix}4 &6 \\1 &0 \end{vmatrix}=0-6=-6$

$\displaystyle x=\frac{\Delta_x}{\Delta}=\frac{-6}{-3}=2$

$\displaystyle y=\frac{\Delta_y}{\Delta}=\frac{-6}{-3}=2$

The solution is x=2, y=2

$\displaystyle \left \{ {{-x+2y=0} \atop {x+2y=5}} \right.$

Calculating:

$\Delta=\begin{vmatrix}-1 &2 \\1 &2 \end{vmatrix}=-2-2=-4$

$\Delta_x=\begin{vmatrix}0 &2 \\5 &2 \end{vmatrix}=0-10=-10$

$\Delta_y=\begin{vmatrix}-1 &0 \\1 &5 \end{vmatrix}=-5-0=-5$

$\displaystyle x=\frac{\Delta_x}{\Delta}=\frac{-10}{-4}=\frac{5}{2}$

$\displaystyle y=\frac{\Delta_y}{\Delta}=\frac{-5}{-4}=\frac{5}{4}$

The solution is

$\displaystyle x=\frac{5}{2}, y=\frac{5}{4}$

(d) The system to solve is

$\displaystyle \left \{ {{6x-y=-5} \atop {4x-2y=6}} \right.$

Calculating:

$\Delta=\begin{vmatrix}6 &-1 \\4 &-2 \end{vmatrix}=-12+4=-8$

$\Delta_x=\begin{vmatrix}-5 &-1 \\6 &-2 \end{vmatrix}=10+6=16$

$\Delta_y=\begin{vmatrix}6 &-5 \\4 &6 \end{vmatrix}=36+20=56$

$\displaystyle x=\frac{\Delta_x}{\Delta}=\frac{16}{-8}=-2$

$\displaystyle y=\frac{\Delta_y}{\Delta}=\frac{56}{-8}=-7$

The solution is x=-2, y=-7

4 0

3 years ago

A vehicle uses 1 7/8 gallons of gasoline to travel 56 1/4 miles. At this rate, how many miles can the vehicle travel per gallon

sweet [91]

Answer:

30 miles

Step-by-step explanation:

In this context, "per" means "divided by", so to find miles per gallon, divide miles by gallons:

(56 1/4 mi)/(1 7/8 gal) = 30 mi/gal

The vehicle can travel 30 miles per gallon of gas.

_____

My calculator divides mixed numbers directly, as many graphing calculators do. If you're working this out by hand, you can convert to improper fractions, then multiply the numerator by the inverse of the denominator.

(56 1/4)/(1 7/8) = (225/4)/(15/8)

= (225/4)·(8/15) = (225/15)·(8/4)

= 15·2 = 30

4 0

3 years ago

Rewrite the equation in slope-intercept form: 8x - 3y - 5 = 0

algol [13]

Slope-Intercept Form: y = 8/3x + −5/3

6 0

3 years ago

Help me in number one please

morpeh [17]