If the length of a rectangle is a two-digit number with identical digits and the width is 1/10 the length and the perimeter is 2 times the area of the rectangle, what is the the length and the width
Solution:
Let the length of rectangle=x
Width of rectangle=x/10
Perimeter is 2(Length+Width)
= 2(x+x/10)
Area of Rectangle= Length* Width=x*x/10
As, Perimeter=2(Area)
So,2(x+x/10)=2(x*x/10)
Multiplying the equation with 10, we get,
2(10x+x)=2x²
Adding Like terms, 10x+x=11x
2(11x)=2x^2
22x=2x²
2x²-22x=0
2x(x-11)=0
By Zero Product property, either x=0
or, x-11=0
or, x=11
So, Width=x/10=11/10=1.1
Checking:
So, Perimeter=2(Length +Width)=2(11+1.1)=2*(12.1)=24.2
Area=Length*Width=11*1.1=12.1
Hence, Perimeter= 2 Area
As,24.2=2*12.1=24.2
So, Perimeter=2 Area
So, Answer:Length of Rectangle=11 units
Width of Rectangle=1.1 units
The first number is 9 and the second number is 10. 10+9=10. 9*2=18 and 10 is 8 less than 18. All the requirements of the numbers are met.
Answer:
A.
Step-by-step explanation:
The answer is A because well 3x would not change and 17x would not change but 7 and 5 would change because they are not variable numbers So when u do the mah you would get 3x2+17x -2
Draw a number line number it with negative number on the left and postive numbers at the right start at postive 6 now go to the left twice your answer should be 4
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