A. Let us assume the width of the pool = x
Then
Length of the pool = 3x + 3
Depth of the pool = 2(3x + 3) - 7
= 6x + 6 - 7
= 6x - 1
B. Area of the land space = Width * length
= x * (3x + 3)
= 3x^2 + 3x
C. Volume of the pool = (6x - 1) * (3x^2 + 3x)
= 18x^3 + 18x^2 - 3x^2 - 3x
= 18x^3 + 15x^2 - 3x
D. The highest degree is actually 3 and the number of terms that we get is also 3.
Answer:
height = 4 meters
h = 2A / ( a + b )
Step-by-step explanation:
Let A represent the area of the trapezoid, h, the height a, the length of one base and b, the length of the other base
Area of a trapezoid, A = ( a + b)* h / 2
Now let's slot in the given values
A= 18 sq meters
a = 3 meters
b = 6 meters
we have;
18 = ( 3 + 6 ) * h /2
18 = 9 * h /2
Cross multiply
18*2 = 9h
36 = 9h
divide both sides by the coefficient of h (9)
4 = h
therefore, h = 4 meters
To solve for h without substituting
From A = ( a + b )* h /2
cross multiply
2*A = ( a + b )* h
divide both sides by ( a + b )
2A / ( a + b ) = h
therefore,
h = 2A / ( a + b )
5:7, 15:21. Hope this helps!!!
Y inter (0, -3) x inter (2,0)
1s -0
5s- 1
25s- 0
125s- 0
625s- 2
3125s- 0
15625s- 2
2020010. I think this is right but would like so confirmation, just taught myself this!
