Answer:
The ferris wheel travelled 434.717 ft.
Step-by-step explanation:
Assuming that the ferris wheel is a perfect circle, then it's height is the same as the diameter of the circle, therefore it's radius is:
radius = height/2 = 246/2 = 123 ft
The distance travelled by the ferris wheel is the same as the length of the arc created by a angle of (9pi/8). The length of an arc is given by:
length = r*(angle in radians)
length = 123*(9pi/8)
length = 434.717 ft
The ferris wheel travelled 434.717 ft.
Answer:
396 in.^2
Step-by-step explanation:
The cushion is shaped like a cylinder with 14 inch diameter and 2 inch height.
We need to find the surface area of the cylinder.
diameter = 14 in.
radius = diameter/2 = 7 in.
height = 2 in.
A = 2(pi)r^2 + 2(pi)rh
A = 2(3.14159)(7 in.)^2 + 2(3.14159)(7 in.)(2 in.)
A = 395.84 in.^2
A = 396 in.^2
Answer:
20 students
Step-by-step explanation:
If the class decreased by 15%, the students that she has now (17) represents a percentaje of:
100% - 15% = 85%
so<u> the 17 students are 85% of what she had</u>:
Students Percentage
17 ⇒ 85%
and we are looking for how many students she had 2 years ago, thus we are looking for the <u>100%</u> of students (the original number of studens). If we represent this number by x:
Students Percentage
17 ⇒ 85%
x ⇒ 100%
and we solve this problem using the <u>rule of three</u>: multiply the cross quantities on the table( 17 and 100) and then divide by the remaining amount (85):
x = 17*100/85
x = 1700/85
x=20
2 years ago she had 20 students
Answer:
1 squared and 1 x squared
Step-by-step explanation:
Answer:
We have the equation
![c_1\left[\begin{array}{c}0\\0\\0\\1\end{array}\right] +c_2\left[\begin{array}{c}0\\0\\3\\1\end{array}\right] +c_3\left[\begin{array}{c}0\\4\\3\\1\end{array}\right] +c_4\left[\begin{array}{c}8\\4\\3\\1\end{array}\right] =\left[\begin{array}{c}0\\0\\0\\0\end{array}\right]](https://tex.z-dn.net/?f=c_1%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D0%5C%5C0%5C%5C0%5C%5C1%5Cend%7Barray%7D%5Cright%5D%20%2Bc_2%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D0%5C%5C0%5C%5C3%5C%5C1%5Cend%7Barray%7D%5Cright%5D%20%2Bc_3%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D0%5C%5C4%5C%5C3%5C%5C1%5Cend%7Barray%7D%5Cright%5D%20%2Bc_4%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D8%5C%5C4%5C%5C3%5C%5C1%5Cend%7Barray%7D%5Cright%5D%20%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D0%5C%5C0%5C%5C0%5C%5C0%5Cend%7Barray%7D%5Cright%5D)
Then, the augmented matrix of the system is
![\left[\begin{array}{cccc}0&0&0&8\\0&0&4&4\\0&3&3&3\\1&1&1&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D0%260%260%268%5C%5C0%260%264%264%5C%5C0%263%263%263%5C%5C1%261%261%261%5Cend%7Barray%7D%5Cright%5D)
We exchange rows 1 and 4 and rows 2 and 3 and obtain the matrix:
![\left[\begin{array}{cccc}1&1&1&1\\0&3&3&3\\0&0&4&4\\0&0&0&8\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%261%261%261%5C%5C0%263%263%263%5C%5C0%260%264%264%5C%5C0%260%260%268%5Cend%7Barray%7D%5Cright%5D)
This matrix is in echelon form. Then, now we apply backward substitution:
1.
2.
3.
4.
Then the system has unique solution that is 
and this imply that the vectors 
are linear independent.
