Replace x with 10 and simplify the fraction.
40/10 = 4/1 = 4
The answer is 4
Answer:4
Step-by-step explanation:
A zero-coupon bond doesn’t make any payments. Instead, investors purchase the zero-coupon bond for less than its face value, and when the bond matures, they receive the face value.
To figure the price you should pay for a zero-coupon bond, you'll follow these steps:
Divide your required rate of return by 100 to convert it to a decimal.
Add 1 to the required rate of return as a decimal.
Raise the result to the power of the number of years until the bond matures.
Divide the face value of the bond to calculate the price to pay for the zero-coupon bond to achieve your desired rate of return.
First, divide 4 percent by 100 to get 0.04. Second, add 1 to 0.04 to get 1.04. Third, raise 1.04 to the sixth power to get 1.2653. Lastly, divide the face value of $1,000 by 1.2653 to find that the price to pay for the zero-coupon bond is $790,32.
Answer: 4=-2x -1 and y=-1 2x
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.
The answer is 1506 it is this because if you add th number you will get it
