Question 1:
48 + 63 + 28 + 44 + 43 = 226 minutes
41 + 40 + 15 + 1 + 39 = 136 seconds
136 seconds = 2 minutes and 16 seconds
The answer is 228 minutes and 16 seconds.
Question 2: 228 minutes = 3 hours and 48 minutes
The answer is 3 hours and 48 minutes, but if you're also counting seconds, add 16 seconds to the end of it.
Answer:
B or C or A or E or D i dont know
Step-by-step explanation:
Suppose the triangle is labeled ABC. That is, angle A, angle B and angle C. The opposite sides for the respective angles are labeled a, b and c.
let a = 300, b = 250, c = 420
The smallest angle should correspond to the shortest side. So 36 degrees is angle B. Thus, we have sides a and c, with an included angle B. The area of the triangle is calculated as half of the products of the two sides with sine of the included angle.
Area = (a × c × sinB) / 2
Area = (300 ft × 420 ft × sin36) / 2
Area is 37,030 ft2.
Thus, the answer is letter C.
You make the denominators the same. The LCM of 6 and 8 is 48, so we are going to make the denominators 48.
6 x 8 = 48
8 x 6 = 48
So now we have 5/48 and 5/48. But because we changed the denominators, we need to change the nominators accordingly.
So for 5/48 (originally 5/6) we multiply 5 by 8, and for 5/48 (originally 5/8) we multiply 5 by 6.
5 x 8 = 40
5 x 6 = 30
So now we have 40/48 (originally 5/6) and 30/48 (originally 5/8).
5/6 is greater than 5/8
Hope this helps!
The area of the composite figure can be solved by separating the figure into 3 portions, which are 2 identical rectangles with one rectangle.
The image of the composite figure will be shown below
Let us sketch out the image of the two identical rectangles
The formula for the area(A) of a rectangle is,
![A=length\times width](https://tex.z-dn.net/?f=A%3Dlength%5Ctimes%20width)
where,
![\begin{gathered} l=length=5m \\ w=width=2m \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20l%3Dlength%3D5m%20%5C%5C%20w%3Dwidth%3D2m%20%5Cend%7Bgathered%7D)
Therefore, the area(A1) of the two identical rectangles are
![\begin{gathered} A_1=2\times(5\times2)=2\times5\times2=20m^2 \\ \therefore A_1=20m^2 \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20A_1%3D2%5Ctimes%285%5Ctimes2%29%3D2%5Ctimes5%5Ctimes2%3D20m%5E2%20%5C%5C%20%5Ctherefore%20A_1%3D20m%5E2%20%5Cend%7Bgathered%7D)
Let me sketch the second rectangle
Therefore, the area(A2) will be
![\begin{gathered} A_2=3\times2=6m^2 \\ \therefore A_2=6m^2 \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20A_2%3D3%5Ctimes2%3D6m%5E2%20%5C%5C%20%5Ctherefore%20A_2%3D6m%5E2%20%5Cend%7Bgathered%7D)
Hence, the area(A) of the composite figure is
![\begin{gathered} A=A_1+A_2=20m^2+6m^2=26m^2 \\ \therefore A=26m^2 \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20A%3DA_1%2BA_2%3D20m%5E2%2B6m%5E2%3D26m%5E2%20%5C%5C%20%5Ctherefore%20A%3D26m%5E2%20%5Cend%7Bgathered%7D)
Therefore, the area is
![26m^2](https://tex.z-dn.net/?f=26m%5E2)