Hi there!
Let me help you out a bit here. What we need to do is subtract the area of the smaller square from the area of the larger square. To do this, we first need to find the length of one of the larger sides. Since the outer polygon is a square, we can use the 45-45-90 relationship to determine that each side of the larger square is equal to

. Next, we can do the same thing for the smaller square, determining that one outer side of the smaller square is equal to

. To figure out the area of both squares, we need to square each of the outer lengths. This gives us 162 for the bigger square and 32 for the smaller square. Now, all we need to do is subtract 32 from 162. This gives us a value of 130cm^2 for the shaded area.
Hope this helps!! :)
If there's anything else that I can help you with, please let me know!
Sllope intercept form is y=mx+b
m is the slope , b is the y itnercept
so early pay
cost=70 flat rate
x=infinity
the equation would be y=70
deposit pluss
you have to pay an initial one time payment of 15 so
y=mx+15
then 4 dollars per day
4 times number of days
y=4x+15
daily pay is 6 dollars per day so 6 times number of days
y=6x
1.
a. y=70
b. y=4x+15
c. y=4x
In triangle ACE,
we know C=93,E can be calulated by using arch angle AEC...what ever that is....,using this we get A=180-(E+93)
So, by alternate segment theorem, DCE= A.
thats all i can say.
Hello!
First you can distribute the 3
4 + 6r + 6s + 3r
Then you combine like terms
4 + 9r + 6s
The answer is 9r + 6s + 4
Hope this helps!