Answer:
A cell wall is a structural layer surrounding some types of cells, just outside the cell membrane. It can be tough, flexible, and sometimes rigid. It provides the cell with both structural support and protection, and also acts as a filtering mechanism.
Answer:
Step-by-step explanation to be honest probably 12 if not i’m so sorry
Answer:
x= -4
Step-by-step explanation:
2x+4=6x+20
so 2x-6x=20-4
divide both sides by -4
-4x/-4 = 16/4
=-4
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
Let h represent the height of the trapezoid, the perpendicular distance between AB and DC. Then the area of the trapezoid is
Area = (1/2)(AB + DC)·h
We are given a relationship between AB and DC, so we can write
Area = (1/2)(AB + AB/4)·h = (5/8)AB·h
The given dimensions let us determine the area of ∆BCE to be
Area ∆BCE = (1/2)(5 cm)(12 cm) = 30 cm²
The total area of the trapezoid is also the sum of the areas ...
Area = Area ∆BCE + Area ∆ABE + Area ∆DCE
Since AE = 1/3(AD), the perpendicular distance from E to AB will be h/3. The areas of the two smaller triangles can be computed as
Area ∆ABE = (1/2)(AB)·h/3 = (1/6)AB·h
Area ∆DCE = (1/2)(DC)·(2/3)h = (1/2)(AB/4)·(2/3)h = (1/12)AB·h
Putting all of the above into the equation for the total area of the trapezoid, we have
Area = (5/8)AB·h = 30 cm² + (1/6)AB·h + (1/12)AB·h
(5/8 -1/6 -1/12)AB·h = 30 cm²
AB·h = (30 cm²)/(3/8) = 80 cm²
Then the area of the trapezoid is
Area = (5/8)AB·h = (5/8)·80 cm² = 50 cm²
