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
Answer:
p=22
Step-by-step explanation:
180-70-2p-3p=0
110=2p+3p
110=5p
p=110÷5
p=22
Answer:
y=3/5x+3
Step-by-step explanation:
4 multiplied by 3.99 will = your answer....Please mark brainliest...
Answer:
Removing pennies and dimes from the bag
Step-by-step explanation:
From the beginning, we have a 1 in 3 chance of drawing a quarter. If we were to add dimes and/or pennies to the bag, this would make us more likely to draw a dime or a penny, and less likely to draw a quarter. Proportionally, we want the most quarters in the bag to get the highest probability of drawing a quarter from the bag.
