I did the problem myself and my teacher always said thatif you don't get the exact answer then round to the closest one. The answer that i got was 31,808.6, the closest one to that would be letter C.

8 0

3 years ago

#2 Find the slope given the points (6,-12) and (15,-3)*

castortr0y [4]

Greetings :)

To find slope of two points we will need to use this equation: $\frac{y^2-y^1}{x^2-x^1}$

$x^1$ $x^2$ $y^1$ $y^2$

( 6 , -12 ) ( 15 , -3 )

Now let's replace the equation with the numbers. (it will be a fraction).

$\frac{-3-(-12)}{15-12}$ = $\frac{9}{9}$

The equation equals 9 over 9, which also equals 1.

The slope of the line is 1.

4 0

3 years ago

i need help asap please dont type random anwsers, that will result in it being deleted. GIVING BRAINLIEST ONLY TO CORRECT, INCOR

Veseljchak [2.6K]

Answer:

The area of the rectangle TOUR is 80.00 unit².

Step-by-step explanation:

The area of a rectangle is computed using the formula:

$Area\ of\ a\ Rectangle=length\times width$

Since the dimensions of the rectangle are not provided we can compute the dimensions using the distance formula for two points.

The distance formula using the two point is:

$distance=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$

Considering the rectangle TOUR the area formula will be:

Area of Rectangle TOUR = TO × OU

The co-ordinates of the four vertices of a triangle are:

T = (-8, 0), O = (4, 4), U = (6, -2) and R = (-6, -6)

Compute the distance between the vertices T and O as:

$TO=\sqrt{(4-(-8))^{2}+(4-0)^{2}}\\=\sqrt{12^{2}+4^{2}} \\=\sqrt{160} \\=4\sqrt{10}$

Compute the distance between the vertices O and U as:

$OU=\sqrt{(6-4)^{2}+(-2-4)^{2}}\\=\sqrt{2^{2}+6^{2}} \\=\sqrt{40} \\=2\sqrt{10}$

Compute the area of rectangle TOUR as follows:

$Area\ of\ TOUR=TO\times OU\\=4\sqrt{10}\times 2\sqrt{10}\\=80\\\approx80.00 unit^{2}$

Thus, the area of the rectangle TOUR is 80.00 unit².

6 0

3 years ago