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4 years ago

10

A new car has an MSRP of $23,450. It comes with a premium package priced

1 answer:

Gekata [30.6K]4 years ago

4 0

Answer:

its $25,850 just did this on apex

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What is equivalent to 2.43?<br><br> ~2 1/100<br> ~2 1/43<br> ~2 43/100<br> ~2 43/1000

Jet001 [13]

Answer:

2 43/100

Step-by-step explanation:

first convert into a fraction

2.43=2.43×100/100

243/100

convert into a mixed fraction

2 43/100

5 0

3 years ago

The graph shows the number of hours that Tammy spends typing for work, x, and the amount of pay that she earns, y.

san4es73 [151]

Answer:

How much Tammy is getting paid for each hour she works.

Step-by-step explanation:

If y, is the amount she gets paid then the line on the graph shows the ratio of how much she gets paid for each our of work.

8 0

3 years ago

Read 2 more answers

A car repair shop charged $80 per hour for labor and $210 for the parts of the car. If the total cost of repair for the car was

Ainat [17]

C(h)=80h+210

80h+210=490 subtract 210 from both sides

80h=280 divide both sides by 80

h=3.5 hours

6 0

4 years ago

If f(−1) = −3 and f prime of x equals the quotient of 4 times x squared and the quantity x cubed plus 3 , which of the following

lora16 [44]

<span>f(–1.1) = 4(-1.1)^2 / [(-1.1)^3 + 3]
</span>f(–1.1) = 4.8 / 1.7
f(-1.1) = 2.8

answer
<span>2.8</span>

7 0

3 years ago

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Consider the matrix A. A = 1 0 1 1 0 0 0 0 0 Find the characteristic polynomial for the matrix A. (Write your answer in terms of

dusya [7]

Answer with Step-by-step explanation:

We are given that a matrix

$A=\left[\begin{array}{ccc}1&0&1\\1&0&0\\0&0&0\end{array}\right]$

a.We have to find characteristic polynomial in terms of A

We know that characteristic equation of given matrix $\mid{A-\lambda I}\mid=0$

Where I is identity matrix of the order of given matrix

I= $\left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right]$

Substitute the values then, we get

$\begin{vmatrix}1-\lambda&0&1\\1&-\lambda&0\\0&0&-\lambda\end{vmatrix}=0$

$(1-\lambda)(\lamda^2)-0+0=0$

$\lambda^2-\lambda^3=0$

$\lambda^3-\lambda^2=0$

Hence, characteristic polynomial = $\lambda^3-\lambda^2=0$

b.We have to find the eigen value for given matrix

$\lambda^2(1-\lambda)=0$

Then , we get $\lambda=0,0,1-\lambda=0$

$\lambda=1$

Hence, real eigen values of for the matrix are 0,0 and 1.

c.Eigen space corresponding to eigen value 1 is the null space of matrix $A-I$

$E_1=N(A-I)$

$A-I=\left[\begin{array}{ccc}0&0&1\\1&-1&0\\0&0&-1\end{array}\right]$

Apply $R_1\rightarrow R_1+R_3$

$A-I=\left[\begin{array}{ccc}0&0&1\\1&-1&0\\0&0&0\end{array}\right]$

Now,(A-I)x=0[/tex]

Substitute the values then we get

$\left[\begin{array}{ccc}0&0&1\\1&-1&0\\0&0&0\end{array}\right]\left[\begin{array}{ccc}x_1\\x_2\\x_3\end{array}\right]=0$

Then , we get $x_3=0$

And $x_1-x_2=0$

$x_1=x_2$

Null space N(A-I) consist of vectors

x= $\left[\begin{array}{ccc}x_1\\x_1\\0\end{array}\right]$

For any scalar $x_1$

$x=x_1\left[\begin{array}{ccc}1\\1\\0\end{array}\right]$

$E_1=N(A-I)=Span(\left[\begin{array}{ccc}1\\1\\0\end{array}\right]$

Hence, the basis of eigen vector corresponding to eigen value 1 is given by

$\left[\begin{array}{ccc}1\\1\\0\end{array}\right]$

Eigen space corresponding to 0 eigen value

$N(A-0I)=\left[\begin{array}{ccc}1&0&1\\1&0&0\\0&0&0\end{array}\right]$

$(A-0I)x=0$

$\left[\begin{array}{ccc}1&0&1\\1&0&0\\0&0&0\end{array}\right]\left[\begin{array}{ccc}x_1\\x_2\\x_3\end{array}\right]=0$

$\left[\begin{array}{ccc}x_1+x_3\\x_1\\0\end{array}\right]=0$

Then, $x_1+x_3=0$

$x_1=0$

Substitute $x_1=0$

Then, we get $x_3=0$

Therefore, the null space consist of vectors

$x=x_2=x_2\left[\begin{array}{ccc}0\\1\\0\end{array}\right]$

Therefore, the basis of eigen space corresponding to eigen value 0 is given by

$\left[\begin{array}{ccc}0\\1\\0\end{array}\right]$

5 0

3 years ago