If there was a chart with this that would be helpful but at this time with the information you have given you can not solve this problem
The answer is 11. Hope this helped.
Answer:
The width must be greater than 3 meters.
Step-by-step explanation:
Let w represent the width. Then 5w will represent the length, which is 5 times the width. The perimeter is the total of the side lengths of the rectangle, so is ...
P = 2w + 2(5w) = 12w
We want this to be greater than 36 m, so ...
P > 36 m
12w > 36 m . . . . . . . substitute our expression for P
w > 3 m . . . . . . . . . . divide by 12
The possible values for width are those values that are more than 3 meters.
tl;dr Answer is C
Here we will have to calculate 3 different areas separately.
When calculating the area of the triangle we will use the formula
A = (h*b)/2
A = Area
h = height
b = base
To find the height we do X - Z
23 - 15 = 8 ft
To find the base we do Y - W
19 - 13 = 6 ft
Using the formula above we can now solve for A
A = (8*6)/2
A = (48)/2
A = 24 sq ft
Now we solve the two rectangles using the formula
A = wl
w = width
l = length
We will calculate the area of the left most rectangle first.
We know the length of the rectangle because it's Y - W and we are given the width of the triangle.
w = 15 ft
l = 6 ft
A = 15*6
A = 90 sq ft
Second Rectangle has the width of X and length of W
w = 23 ft
l = 13 ft
A = 23 * 13
A = 299 sq ft
Now we add all the areas to give us the total area of the warehouse.
24 + 90 + 299 = 413 sq ft
Therefore, the answer is C
Answer:
20 292 02 202 202
Step-by-step explanation:
