Question

5) Find the sum below, showing all steps and any formulas used. 1 2 22 23 214 4 4 4 (7 points) + ++ 4. 6) Solve the problem below using Cramer's Rule. Be sure to label what cach variable represents, show the equations in the system of equations, and show all matrices and determinants used. No other method will receive credit. Molly attended a coin auction and purchased some rare "Flowing Hair” fifty-cent pieces, and a number of very rare two-cent pieces from the Civil War Era. If she bought 47 coins with a face value of $10.06, how many of each type of coin did she buy? (12 points)

Answer 1

5)

14 1 92 1 + ܕ|ܠܛ 1 4 4 4

this is a geometric progression

total terms ,n=15

first term , a =(1/4)

common ratio, r =

2 44

sum of n terms of gp=$a(\frac{1-r^n}{1-r}),r\neq 1$

so

1 2. 22 + 214 4 = 1-215 -), r + 1 4 1-2 + + + 4 4

1 2 2- 32767 + 24 4 + + - 4 4 1 4

6)

let a =number of $0.50 coins

b= number of $0.02 coins

It is given she bought 47 coins

a + b =47

with face value of $10.06

0.50a + 0.02b=10.06

now the matrix form of above two equation are

$\begin{bmatrix} 1 & 1 & |47\\ 0.50&0.02 &|10.06 \end{bmatrix}$

the main matrix is

$\begin{bmatrix} 1 & 1 \\ 0.50&0.02 \end{bmatrix}$

it determinant= 0.02-0.50 =0.48

now replace the first coulumn of the main matrix with the solution vector and we will find its determinant

$\begin{bmatrix} 47& 1 \\ 10.06&0.02 \end{bmatrix}$ determinant= 47(0.02)-10.06 =-9.12

now replace the second coulumn of the main matrix with the solution vector and we will find its determinant

$\begin{bmatrix} 1& 47 \\ 0.50&10.06 \end{bmatrix}$ determinant =10.06-47(0.50) =-13.44

so we get

$a=\frac{-9.12}{-0.48}=19$

$b=\frac{-13.44}{-0.48}=28$

so she bought 19, fifty cents coins

and 28 , two cents coins

7)

$(3x-z)^9$ for the expansion we will use binomial theorem

$(a+b)^n=\sum_{i=0}^{n}(\frac{n}{i})a^{(n-i)}b^i$

here a=3x and b=-z

$(3x-z)^9=\sum_{i=0}^{9}(\frac{9}{i})(3x)^{(9-i)}(-z)^i$

expanding this

$\sum_{i=0}^{9}(\frac{9}{i})(3x)^{(9-i)}(-z)^i=\frac{9!}{0!(9-0)!}(3x)^8(-z)^0+\frac{9!}{1!(9-1)!}(3x)^7(-z)^1+\frac{9!}{2!(9-2)!}(3x)^9(-z)^2+....$

the third term of the expansion is

$\frac{9!}{2!(9-2)!}(3x)^7(-z)^2$

simplifying this,

$\frac{9!}{2!(9-2)!}(3x)^7(-z)^2=\frac{72}{2}(3x)^7(-z)^2$

$\frac{72}{2}(3x)^7(-z)^2=36*3^7x^7z^2$

$=78732x^7z^2$

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