Answer:
Substitute the x and y values from the ordered pairs into the equation, one by one, until both sides of the equation are equal.
Step-by-step explanation:
A. (-2,-3)
3 * (-2) - 4 * (-3) = 21
-6 + 12 = 21
6 = 21 <u>not equal</u>
B. (0,7)
3 * 0 - 4 * 7 = 21
0 - 28 = 21
- 28 = 21 <u>n</u><u>o</u><u>t</u><u> </u><u>e</u><u>q</u><u>u</u><u>a</u><u>l</u>
C. (7,0)
3 * 7 - 4 * 0 = 21
21 - 0 = 21
21 = 21 <u>equal</u>
If the slope of AB = CD and BC = AD it's a parallelogram:
Slope of AB = 6+1 / -9+5 = -7/4
CD = -2-5 / 3+1 = -74
These are equal.
BC = 5-6 / -1 +9 = -1/8
AD = -2 +1 / 3+5 = -1/8
These are also equal so it is a parallelogram.
Now to find if the diagonals are perpendicular find the slope of the perpensicular points:
AC = 5 +1 / -1 +5 = 6/4 = 3/2
BD = 6+2 / -9 -3 = 8/-12 = -2/3
Because BD is the reciprocal of AC, this means they are perpendicular.
And because AB is not perpendicular to AD ( AB and AD are not reciprocals) it is a rhombus.
<u>Answer</u>:
equation: y = x/2
<u>Explanation</u>:
let the points be (-6, -3) , ( 0, 0)
slope: 
: 
: 
Equation using:
y - y1 = m ( x - x1 )
y - - 3 = 1/2( x - - 6 )
y + 3 = x/2 + 3
= 
Volume = Length x Width x Height
V= 7.5 ft x 4.2 ft x w fr
The width is 2 as listed.
V= 7.5ft x 4.2ft x 2ft
V=63
Step-by-step explanation:
The interior angle of a polygon is given by

The exterior angle of a polygon is given by

where n is the number of sides of the polygon
The statement
The interior of a regular polygon is 5 times the exterior angle is written as

Solve the equation
That's

Since the denominators are the same we can equate the numerators
That's
180n - 360 = 1800
180n = 1800 + 360
180n = 2160
Divide both sides by 180
<h3>n = 12</h3>
<h2>I).</h2>
The interior angle of the polygon is

The answer is
<h2>150°</h2>
<h2>II.</h2>
Interior angle + exterior angle = 180
From the question
Interior angle = 150°
So the exterior angle is
Exterior angle = 180 - 150
We have the answer as
<h2>30°</h2>
<h2>III.</h2>
The polygon has 12 sides
<h2>IV.</h2>
The name of the polygon is
<h2>Dodecagon</h2>
Hope this helps you.