The correct answer to this question is 6 in
Answer:
A non-equilateral rhombus.
Step-by-step explanation:
We can solve this graphically.
We start with square:
ABCD
with:
A = (11, - 7)
B = (9, - 4)
C = (11, - 1)
D = (13, - 4)
Only with the vertices, we can see that ABCD is equilateral, as the length of each side is:
AB = √( (11 - 9)^2 + (-7 -(-4))^2) = √( (2)^2 + (3)^2) = √(4 + 9) = √13
BC = √( (11 - 9)^2 + (-1 -(-4))^2) = √13
CD = √( (11 - 13)^2 + (-1 -(-4))^2) = √13
DA = √( (11 - 13)^2 + (-7 -(-4))^2) = √13
And we change C by C' = (11, 1)
In the image you can see the 5 points and the figure that they make:
The figure ABCD is a rhombus, and ABC'D is also a rhombus, the only difference between the figures is that ABCD is equilateral while ABC'D is not equilateral.
It can be read as "six cubed."
It can be read as "six to the power of three."
It has a base of 6.
It is the same as multiplying three factors of 6.
It has an exponent of 3.
Have a Nice Day
I think a, bc for the x it’s 2x and the y is +5
Step-by-step explanation:
there 5^(n+1) + 5^(n+2) = 5^n x 5^1 + 5^n x 5^2 breaking them as x^(a+b) = x^a x x^b
then taking common 5^n from both terms