Answer:
242.52 cubic inches
Step-by-step explanation:
Volume of the cake pan = Length × Width × Height
From the about question, we have the following dimensions for the cake pan
8 inches wide = Width
11 inches long = Length
7 cm deep = Height
We are asked to find the maximum volume in inches. Hence all the dimensions have to be in inches.
Converting Height in cm to inches
From the question,
2.54 cm = 1 inch
7cm = x inch
Cross Multiply
2.54 × x = 7 × 1
x = 7/2.54
x = 2.7559055118 inches
Volume of the cake pan =
8 × 11 × 2.7559055118
= 242.51968504 cubic inches
Approximately, the volume of the cake pan = 242.52 cubic inches
What is the maximum volume, in cubic
inches, the cake pan can hold is 242.52 cubic inches
Answer:
See below.
Step-by-step explanation:
f(x) = -|x + 9|
compared to
f(x) = |x|
has a graph that opens downward in an upside down V shape.
It has been moved 9 units left compared to the graph of f(x) = |x|.
It reflects over the x-axis.
Range is the difference between the lowest and highest values.
997- 247= 750
Answer:
i. <DCB = 
ii. Sin of <DCB = 0.8
Step-by-step explanation:
Let <DCB be represented by θ, so that;
Sin θ = 
Thus from the given diagram, we have;
Sin θ = 
= 0.8
This implies that,
θ =
0.8
= 53.1301
θ = 
Therefore, <DCB =
.
So that,
Sin of <DCB = Sin 
= 0.8
Sin of <DCB = 0.8
Answer:
0.5 m
Step-by-step explanation:
Given that the dimensions of the garden are 15 meters by 20 meters.
The area of the new garden (original garden + stone placed around the perimeter) is 336m².
Let d be the width of the border having stepping stones as shown in the figure. The shaded region in the figure is the area having stepping stones.
The area including the shaded region 

As the width of the stone border can't be a negative value, so taking the positive value.
Hence, the width of the stone border is 0.5 m.
So, the wi