Area of square = s^2
Area of Rectangle = lw
l = 2s
w+3 = s
solve for w, w=s-3
Now we have l and w.
Plug both into area of rectangle formula so: (2s)(s-3)
Since both areas are equal set both equations equal to each other:
(2s)(s-3)=s^2
Now simplify
2s^2-6s=s^2
s^2-6s=0, Solve for s.
Factor polynomial. s(s-6)=0 , s can be equal to 0 or 6. HOWEVER, you cannot have a side length of 0 therefore the side length has to be 6.
Now plug in s for the length formula for the rectangle:
l = 2s so... l = 2(6) so length of rectangle = 12.
Now plug in s for the width formula for the rectangle:
w+3=s so... w+3=6 so width of rectangle = 3.
Now the dimensions of the rectangle are 12 by 3. 12 being length and 3 width.
To CHECK:
Find area of rectangle:
A=lw so A=3 times 12 so A=36
Find area of square:
We know the side is equal to 6 so
A=s^2 so 6^2 = 36
The areas are equal that verifies the answer of 12 by 3.
Answer:
C)
region C
Step-by-step explanation:
We have to use what is called the zero-interval test [test point] in order to figure out which portion of the graph these inequalities share:
0 ≤ 2 ☑ [We shade the portion of the graph that CONTAIN THE ORIGIN, which is the bottom portion.]
0 ≥ −3 ☑ [We shade the portion of the graph that CONTAINS THE ORIGIN, which is the left side.]
So, now that we got that all cleared up, we can tell that both graphs share a region where the ORIGIN IS VISIBLE. Therefore region C matches the above inequalities.
I am joyous to assist you anytime.
Which one question? there is multiple in there.
Yes you are correct B is the right answer
2 Dewey you can still go to get
