Question

100 points! Help on 9 and 10 I will mark brainliest. Random answer will be reported

Answer 1

Answer:

Question 9)

It is not possible.

Question 10)

The shortest route (moving only vertically/horizontally) will take .25 miles.

From A(2, 4), we translate two units to the left.

And from A'(0, 4), we translate three units downwards to arrive at A''(0, 1).  

Step-by-step explanation:

Question 9)

We are given the three collinear points A(-2, 3), A'(x, y), and A''(3, 7).

We want to determine (if possible) the values of x and y.

Since the three points are collinear, the slope from any point to another must be equivalent to each other.

So, using the to known points A(-2, 3) and A''(3, 7), we can determine the slope to be:

$\displaystyle m=\frac{7-3}{3-(-2)}=\frac{4}{5}$

So, the slope between A and A' should be 4/5. So:

$\displaystyle m=\frac{y-3}{x+2}=\frac{4}{5}$

By cross-multiplication, we acquire:

$4x+8=5y-15$

Unfortunately, it is now impossible to determine the values of x and y. Any values of x and y that satisfy the above equation can be the values for A'.

Two examples are given:

Say that x = 0. Then by our equation, y is:

$4(0)+8=5y-15\Rightarrow y=4.6$

So, a possible point for A' is (0, 4.6).

Now, say that y = 0. Then by our equation, x is:

$4x+8=5(0)-15\Rightarrow x=-5.75$

So, another possible point for A' is (-5.75, 0).

You can substitute any number for x and solve for y or vice versa and you will get a unique point. Hence, we do not have enough information to solve for x and y.

The same result applies if you do the slope between A' and A''.

Question 10)

I'm assuming we can only move horizontally or vertically.

So, from our starting point A(2, 4), we can move two units to the left to arrive at A'(0, 4).

Now, we can move three units downwards to arrive at the post-office at A'' (0, 1).

In total, we moved 2 + 3 = 5 grids. So, we traveled a total of 5(0.05) or 0.25 miles.