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3 years ago

Which of the following points is a vertex for the image produced by a dilation about the origin with a scale factor of 1/2?

Mathematics

1 answer:

MA_775_DIABLO [31]3 years ago

8 0

Answer:

A (0,3)

Step-by-step explanation:

The given trapezoid has vertices:

(0,6), (7,12), (7,9) and (0,12).

We want to choose from the given options, a point that is a vertex for the image produced by a dilation about the origin with a scale factor of 1/2.

Note that the mapping for such a dilation is:

$(x,y)\to( \frac{1}{2} x, \frac{1}{2} y)$

This implies that:

$(0,6)\to(0,3)$

$(7,12)\to(3.5,6)$

$(7,9)\to(3.5,4.5)$

$(0,12)\to(0,6)$

Therefore correct choice is (0,3)

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4x + 10 = 2x - 14<br> Find the value of x.

Gre4nikov [31]

Answer:

x= -12

Step-by-step explanation:

Simplifying

4x + 10 = 2x + -14

Reorder the terms:

10 + 4x = 2x + -14

Reorder the terms:

10 + 4x = -14 + 2x

Solving

10 + 4x = -14 + 2x

Solving for variable 'x'.

Move all terms containing x to the left, all other terms to the right.

Add '-2x' to each side of the equation.

10 + 4x + -2x = -14 + 2x + -2x

Combine like terms: 4x + -2x = 2x

10 + 2x = -14 + 2x + -2x

Combine like terms: 2x + -2x = 0

10 + 2x = -14 + 0

10 + 2x = -14

Add '-10' to each side of the equation.

10 + -10 + 2x = -14 + -10

Combine like terms: 10 + -10 = 0

0 + 2x = -14 + -10

2x = -14 + -10

Combine like terms: -14 + -10 = -24

2x = -24

Divide each side by '2'.

x = -12

Simplifying

x = -12

3 0

2 years ago

Read 2 more answers

Write a phrase as a algebraic expression.4 times the sum of a number and 20.

Deffense [45]

4(20 + x)

PLEASE RATE AS THE BRAINLIEST ANSWER! THANK YOU! :)

4 0

3 years ago

See question in attached photo.

statuscvo [17]

9514 1404 393

Answer:

(a, b) = (-2, -1)

Step-by-step explanation:

The transpose of the given matrix is ...

$A^T=\left[\begin{array}{ccc}1&2&a\\2&1&2\\2&-2&b\end{array}\right]$

Then the [3,1] term of the product is ...

$(A\cdot A^T)_{31}=\left[\begin{array}{ccc}a&2&b\end{array}\right]\cdot\left[\begin{array}{ccc}1&2&2\end{array}\right]=a+2b+4$

and the [3,2] term is ...

$(A\cdot A^T)_{32}=\left[\begin{array}{ccc}a&2&b\end{array}\right]\cdot\left[\begin{array}{ccc}2&1&-2\end{array}\right]=2a-2b+2$

Both of these terms in the product matrix are 0. We can solve the system of equations by adding these two terms:

(a +2b +4) +(2a -2b +2) = (0) +(0)

3a +6 = 0

a = -2

Substituting for 'a' in term [3,1] gives ...

-2 +2b +4 = 0

b = -1

The ordered pair (a, b) is (-2, -1).

8 0

2 years ago

Given below are the graphs of two lines, y=-0.5 + 5 and y=-1.25x + 8 and several regions and points are shown. Note that C is th

zalisa [80]

We have the following equations:

$(1) \ y=-0.5x+5 \\ (2) \ y=-1.25x+8$

So we are asked to write a system of equations or inequalities for each region and each point.

Part a)

Region Example A

$y \leq -0.5x+5 \\ y \leq -1.25x+8$

Region B.

Let's take a point that is in this region, that is:

$P(0,6)$

So let's find out the signs of each inequality by substituting this point in them:

$y \ (?)-0.5x+5 \\ 6 \ (?) -0.5(0)+5 \\ 6 \ (?) \ 5 \\ 6\ \textgreater \ 5 \\ \\ y \ (?) \ -1.25x+8 \\ 6 \ (?) -1.25(0)+8 \\ 6 \ (?) \ 8 \\ 6\ \textless \ 8$

So the inequalities are:

$(1) \ y \geq -0.5x+5 \\ (2) \ y \leq -1.25x+8$

Region C.

A point in this region is:

$P(0,10)$

So let's find out the signs of each inequality by substituting this point in them:

$y \ (?)-0.5x+5 \\ 10 \ (?) -0.5(0)+5 \\ 10 \ (?) \ 5 \\ 10\ \textgreater \ 5 \\ \\ y \ (?) \ -1.25x+8 \\ 10 \ (?) -1.25(0)+8 \\ 10 \ (?) \ 8 \\ 10 \ \ \textgreater \ \ 8$

So the inequalities are:

$(1) \ y \geq -0.5x+5 \\ (2) \ y \geq -1.25x+8$

Region D.

A point in this region is:

$P(8,0)$

So let's find out the signs of each inequality by substituting this point in them:

$y \ (?)-0.5x+5 \\ 0 \ (?) -0.5(8)+5 \\ 0 \ (?) \ 1 \\ 0 \ \ \textless \ \ 1 \\ \\ y \ (?) \ -1.25x+8 \\ 0 \ (?) -1.25(8)+8 \\ 0 \ (?) \ -2 \\ 0 \ \ \textgreater \ \ -2$

So the inequalities are:

$(1) \ y \leq -0.5x+5 \\ (2) \ y \geq -1.25x+8$

Point P:

This point is the intersection of the two lines. So let's solve the system of equations:

$(1) \ y=-0.5x+5 \\ (2) \ y=-1.25x+8 \\ \\ Subtracting \ these \ equations: \\ 0=0.75x-3 \\ \\ Solving \ for \ x: \\ x=4 \\ \\ Solving \ for \ y: \\ y=-0.5(4)+5=3$

Accordingly, the point is:

$\boxed{p(4,3)}$

Point q:

This point is the $x-intercept$ of the line:

$y=-0.5x+5$

So let:

$y=0$

Then

$x=\frac{5}{0.5}=10$

Therefore, the point is:

$\boxed{q(10,0)}$

Part b)

The coordinate of a point within a region must satisfy the corresponding system of inequalities. For each region we have taken a point to build up our inequalities. Now we will take other points and prove that these are the correct regions.

Region Example A

The origin is part of this region, therefore let's take the point:

$O(0,0)$

Substituting in the inequalities:

$y \leq -0.5x+5 \\ 0 \leq -0.5(0)+5 \\ \boxed{0 \leq 5} \\ \\ y \leq -1.25x+8 \\ 0 \leq -1.25(0)+8 \\ \boxed{0 \leq 8}$

It is true.

Region B.

Let's take a point that is in this region, that is:

$P(0,7)$

Substituting in the inequalities:

$y \geq -0.5x+5 \\ 7 \geq -0.5(0)+5 \\ \boxed{7 \geq \ 5} \\ \\ y \leq \ -1.25x+8 \\ 7 \ \leq -1.25(0)+8 \\ \boxed{7 \leq \ 8}$

It is true

Region C.

Let's take a point that is in this region, that is:

$P(0,11)$

Substituting in the inequalities:

$y \geq -0.5x+5 \\ 11 \geq -0.5(0)+5 \\ \boxed{11 \geq \ 5} \\ \\ y \geq \ -1.25x+8 \\ 11 \ \geq -1.25(0)+8 \\ \boxed{11 \geq \ 8}$

It is true

Region D.

Let's take a point that is in this region, that is:

$P(9,0)$

Substituting in the inequalities:

$y \leq -0.5x+5 \\ 0 \leq -0.5(9)+5 \\ \boxed{0 \leq \ 0.5} \\ \\ y \geq \ -1.25x+8 \\ 0 \geq -1.25(9)+8 \\ \boxed{0 \geq \ -3.25}$

It is true

7 0

3 years ago

Can anyone help??? will give best answer brainliest!!

tia_tia [17]

Answer:

5xy = 2

Step-by-step explanation:

This is the only one that has two (x & y) letters/integers on one single number.

3 0

2 years ago