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: GCF of 13 and 17 is 1. 2.
Step-by-step explanation: Greatest common factor (GCF) of 13 and 17 is 1. We will now calculate the prime factors of 13 and 17, than find the greatest common factor (greatest common divisor (gcd)) of the numbers by matching the biggest common factor of 13 and 17.
I'm not sure but I think I may have the answer to number 9
It says she uses 3 draws for shirts and wants to find out the percentage. Percentage is normally out of 100% so we do 100÷4 (we divide by 4 because there are 4 drawers)
100÷4=25. So now it says she uses three drawers for her shirts so we need to do 25x3=75 and all you do with that is put it in percent. The answer is 75%
P.S I am not entirely sure that this is correct but I wanted to help ❤
Answer:
D. none
Step-by-step explanation:
The point in geometry is defined as the location.
Point A is described as the location of A on any given plane.
Point is represented as a dot.
A dot has no physical dimension like depth, height, thickness or size.
Hence, any two points can not be compared on the basis of any physical parameter.
Answer:
494
Step-by-step explanation:
You can use Symbolab to solve equations of all types with explanations
