We are given with the following vertices:
A (0,0)
B (-1,1)
C(6,1)
D (2.0)

Next is to plot the vertices in the Cartesian plane so that we can identify the figure. See attached image.

The identified shape is a "Parallelogram".

4 0

2 years ago

For the problem 1/5g- 1/10- g + 1 3/10g -1/10, Tyson created an equivalent expression using the following steps. 1/5g+-1g+1 3/10

professor190 [17]

The true statements are:

Tyson's expression is not equivalent to the original expression
The equivalent expression is: $\frac{1}{2}g- \frac 1{5}$

<h3>What are equivalent expressions?</h3>

Equivalent expressions are expressions that have equal values

The original expression is given as:

$\frac 15g- \frac 1{10}- g + 1 \frac{3}{10}g -\frac{1}{10}$

Collect like terms

$\frac 15g- \frac 1{10}- g + 1 \frac{3}{10}g -\frac{1}{10} = \frac 15g - g + 1 \frac{3}{10}g- \frac 1{10} -\frac{1}{10}$

Evaluate the like terms

$\frac 15g- \frac 1{10}- g + 1 \frac{3}{10}g -\frac{1}{10} = \frac{1}{2}g- \frac 1{5}$

Tyler's equivalent expression is given as:

$\frac15g-g+ 1 \frac3{10}g-\frac 1{10}-\frac 45g+1 \frac{1}{10}$

Collect like terms

$\frac15g-g+ 1 \frac3{10}g-\frac 1{10}-\frac 45g+1 \frac{1}{10} = \frac15g-g+ 1 \frac3{10}g-\frac 45g-\frac 1{10}+1 \frac{1}{10}$

Evaluate the like terms

$\frac15g-g+ 1 \frac3{10}g-\frac 1{10}-\frac 45g+1 \frac{1}{10} = -\frac 3{10}g$

The simplified expressions of the original expression, and Tyson's equivalent expressions are not equal.

Hence, Tyson's expression is not equivalent to the original expression

Read more about equivalent expressions at:

brainly.com/question/9603710

3 0

1 year ago