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3 years ago

Rob changes his car every year and prefers a car that is always under warranty. Which option will suit him best?

Mathematics

2 answers:

Mamont248 [21]3 years ago

8 0

The Real Answer is B: Leasing a car for a certain period

larisa [96]3 years ago

6 0

The Real Answer is B: Leasing a car for a certain period.

Hope this helps!

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Which expression is equivalent to 6w + 5 - 3w + 7

love history [14]

Answer:

the answer is B

Step-by-step explanation:

6w + 5 - 3w + 7

<h3>i) collect like terms</h3>

= 6w - 3w + 5 + 7

= 3w + 5 + 7

<h3>ii) add the numbers</h3>

= 3 w + 5 + 7

= 3w + 12

4 0

3 years ago

Read 2 more answers

Simple 9th grade math

Liono4ka [1.6K]

Answer:

-1078

Step-by-step explanation:

To find the nth term in an arithmetic sequence we use the following formula

aₙ=a₁+(n-1)*d

Where a₁ is the first time, and d is the common difference (in this case -20)

So we have

2+(55-1)*-20

2+(54)*-20

2-1080

-1078

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3 years ago

HELP ME PLEASE!!!!!!!

kotegsom [21]

Answer:

1. (0,0) (3,1) (6,2)

2. proportinal

3. im so sorry but i cant see that one i really am sorry!

Step-by-step explanation:

i hope i helped brainliest?

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3 years ago

I don't understand it maybe y'all do

kap26 [50]

Right angle triangle

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3 years ago

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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

3 years ago