Richelle can fold a hexagon in half and if they match up they will be congruent.
all angles would be congruent
Answer:
x=4
y=2
Step-by-step explanation:
2x+2y=12
x−y=2
In order to solve by elimination, coefficients of one of the variables must be the same in both equations so that the variable will cancel out when one equation is subtracted from the other.
2x+2y=12,x−y=2
To make 2x and x equal, multiply all terms on each side of the first equation by 1 and all terms on each side of the second by 2.
2x+2y=12,2x+2(−1)y=2×2
Simplify.
2x+2y=12,2x−2y=4
Subtract 2x−2y=4 from 2x+2y=12 by subtracting like terms on each side of the equal sign.
2x−2x+2y+2y=12−4
Add 2x to −2x. Terms 2x and −2x cancel out, leaving an equation with only one variable that can be solved.
2y+2y=12−4
Add 2y to 2y.
4y=12−4
Add 12 to −4.
4y=8
Divide both sides by 4.
y=2
Substitute 2 for y in x−y=2. Because the resulting equation contains only one variable, you can solve for x directly.
x−2=2
Add 2 to both sides of the equation.
x=4
The system is now solved.
x=4,y=2
Correct choice is B) x=4.
<u>Answer:</u>
If the spool has 10 cm of circumference or perimeter, this means that 10 cm of thread fits the whole circumference of the spool, which also means that
<h2>
1 unwind is equivalent to 10 cm of thread.
</h2>
Then, we have to make the following question:
If 10 cm of thread is equivalent to 1 unwind, to how many unwinds is equivalent 80 cm of thread?
If we want to know how many time does Alex need to unwind the spool, we can use the following relation also called the <u>Rule of Three:
</u>
<u>
</u>
The answer is:
Alex must unwind the spool 8 times in order to get 
cm of thread
A) The total number is the sum of all the frequencies
2+5+8+12+11+6=44
Answer: 44
B) The width is (upper limit - lower limit) / 2
(35-21) / 2 = 7
(50-36) / 2 = 7
(65-51) / 2 = 7
(80-66) / 2 = 7
(95-81) / 2 = 7
(110-96) / 2 = 7
Answer: the width is 7
C) The midpoint is the lower limit + the width
36+7=43
Answer: the midpoint is 43
D)The modal is the class with more frequency
In this case 66-80 with 12
Answer: class 66-80
E) We use the width formula
lower limit = 111
width = 7
upper = (width * 2) - lower
upper = (7 * 2) + 111
upper = 125
Answer: class 111-125
Hi
3n + 16 = 7n
3n - 7n = -16
-4n = -16
x = -16/-4
x= 4
