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3 years ago

Find the area of the triangle defined by the coordinates (-3, -4), (-7, 0), and (-3, 4).

1 answer:

3 0

The area of a Triangle is 1/2BH (B=Base length, H=Height)
When looking at the points, we can figure out the following:

The distance between the two points at the base is 8 (to get from -4 to 4 you must add 8) and the height is 4 (to get from -4 to 0 as well as 0 to 4, the distance is 4).

Now we plug in our values:
Area= 1/2(8)(4)
=1/2(32)
=16 Units squared

Therefore, your answer is 16 units squared.

I hope this Helps!
-Sinnamin

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Answer:

The median age is 7.

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3 years ago

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3 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

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4 years ago