The first point would be at (0,2). Then you would have to go up 10 units, and then 1 to the right. When there is no no space to graph, you extend downwards from (0,2). So you would go down 10 units, and to the left 1 unit.

4 0

3 years ago

In 2007, there were more than 8.14 million cars for sale. The number of cars then decreased by 23% per year. Write an exponentia

tatuchka [14]

Answer:

In 2013 there were approximately 2.2 million cars for sale.

Step-by-step explanation:

Since the number of cars is decreasing by 23% per year, it is a decay model. This types of decay can be expressed in the following expression:

$cars(x) = cars(2007)*(1 - 0.23)^{x - 2007}$

Where x is the year, cars(2007) is the number of cars in 2007. Applying this data we have:

$cars(x) = 8.14*(0.77)^{x - 2007}$

If we want to know the number of cars in 2013, we need to apply that value to x and solve the expression.

$cars(2013) = 8.14*(0.77)^{2013 - 2007}\\cars(2013) = 8.14*(0.77)^{5} = 2.20332$

In 2013 there were approximately 2.2 million cars for sale.

7 0

3 years ago

From a point P outside line segment AB =6cm, construct a line PQ parallel to line segment AB.

goldfiish [28.3K]

Answer:

We drawn the graph.

Step-by-step explanation:

We have a line segment AB = 6cm and a point P outside that a line segment.

We want to construct a line PQ that is parallel to line segment AB.

If PQ is parallel to longer AB, this means that a line PQ is equidistant from line segment AB everywhere.

We use the geogebra.org site to draw it.

We drawn the graph.

4 0

4 years ago