Total amount of money invested - $11500
X +Y = $11500. ----------(1)
Total yearly interest for thetwo a/c is - $201
-0.03X + 0.04Y = $201
Form (1)
X = 11500 - Y
Substitute for X
-0.03(11500 - Y) + 0.04 Y = $201
-345 + 0.03Y + 0.04Y =201
0.07Y = 546
Y = 7800 at 4 %
Calculate X
X = 11500 - Y
= 11500 - 7800
=3700 at - 3%
Check
-0.03×3700 + 0.04 × 7800 = $201
-111 + 312 = $201
$201 = $201

6 0

3 years ago

Which number can each term of the equation be multiplied by to eliminate the fractions before solving?

DiKsa [7]

Answer:

Hope it helps

4 0

2 years ago

A line is defined by the equation y = two-thirds x minus 6. The line passes through a point whose y-coordinate is 0. What is the

DedPeter [7]

$y = \cfrac{2}{3}x~~ - ~~6~\hspace{10em} (\stackrel{x}{?}~~,~~\stackrel{y}{0}) \\\\\\ 0=\cfrac{2}{3}x~~ - ~~6\implies 6=\cfrac{2x}{3}\implies 18=2x\implies \cfrac{18}{2}=x\implies 9=x$

3 0

2 years ago

find the spectral radius of A. Is this a convergent matrix? Justify your answer. Find the limit x=lim x^(k) of vector iteration

Natasha_Volkova [10]

Answer:

The solution to this question can be defined as follows:

Step-by-step explanation:

Please find the complete question in the attached file.

$A = \left[\begin{array}{ccc} \frac{3}{4}& \frac{1}{4}& \frac{1}{2}\\ 0 & \frac{1}{2}& 0\\ -\frac{1}{4}& -\frac{1}{4} & 0\end{array}\right]$

now for given values:

$\left[\begin{array}{ccc} \frac{3}{4} - \lambda & \frac{1}{4}& \frac{1}{2}\\ 0 & \frac{1}{2} - \lambda & 0\\ -\frac{1}{4}& -\frac{1}{4} & 0 -\lambda \end{array}\right]=0 \\\\$

$\to (\frac{3}{4} - \lambda ) [-\lambda (\frac{1}{2} - \lambda ) -0] - 0 - \frac{1}{4}[0- \frac{1}{2} (\frac{1}{2} - \lambda )] =0 \\\\\to (\frac{3}{4} - \lambda ) [(\frac{\lambda}{2} + \lambda^2 )] - \frac{1}{4}[\frac{\lambda}{2} - \frac{1}{4}] =0 \\\\\to (\frac{3}{8}\lambda + \frac{3}{4} \lambda^2 - \frac{\lambda^2}{2} - \lambda^3 - \frac{\lambda}{8} + \frac{1}{16}=0 \\\\\to (\lambda - \frac{1}{2}) (\lambda -\frac{1}{4}) (\lambda - \frac{1}{2}) =0\\\\$

$\to \lambda_1=\lambda_2 =\frac{1}{2}\\\\\to \lambda_3 = \frac{1}{4} \\\\\to A = max {|\lambda_1| , |\lambda_2|, |\lambda_3|}\\\\$

$= max{\frac{1}{2}, \frac{1}{2}, \frac{1}{4}}\\\\= \frac{1}{2}\\\\(A) =\frac{1}{2}$

In point b:

Its

spectral radius is less than 1 hence matrix is convergent.

In point c:

$\to c^{(k+1)} = A x^{k}+C \\\\\to x(0) = \left(\begin{array}{c}3&1&2\end{array}\right) , c = \left(\begin{array}{c}2&2&4\end{array}\right)\\\\ \to x^{(k+1)} = \left[\begin{array}{ccc} \frac{3}{4}& \frac{1}{4}& \frac{1}{2}\\ 0 & \frac{1}{2}& 0\\ -\frac{1}{4}& -\frac{1}{4} & 0\end{array}\right] x^k + \left[\begin{array}{c}2&2&4\end{array}\right] \\\\$

after solving the value the answer is

$\lim_{k \to \infty} x^k=o = \left[\begin{array}{c}0&0&0\end{array}\right]$

4 0

2 years ago