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4 years ago

Notebooks cost $1.20 each. This weekend they will be $.80 What percentage is the sale?

2 answers:

5 0

The sale is 33% off notebooks. This can be found by first subtracting .8 from 1.2

1.2-.8=.4 Now divide .4 by 1.2 to get your answer.

$\frac{.4}{1.2}$ =.33

.33=33%

You can also solve this mentally because if you know your basic multiplication tables, 3*4=12 and so $\frac{2}{3}$ of 12 would equal 8.

Does you understand it now?

Taya2010 [7]4 years ago

3 0

In the question it is already given that the actual cost of each notebook is $1.20. For the weekend, the notebooks would sold at a discounted rate of $0.80
he
The amount of discount during the weekend = (1.20 - 0.80) dollars
   = 0.40 dollars
So the amount of discount given during the weekend is $0.40
Now we need to find the percentage of discount given during the weekend.
Then
Percentage of discount during the weekend = (0.40/1.20) * 100
   = 0.33 * 100
   = 33
Then the percentage of discount given is 33%.

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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

Please helppppp meeeeeee!!!!!!!!!!!!!!!!!!!!

kirill115 [55]

Answer:

parallel lines

Step-by-step explanation:

cause they go straight don't meet at a point

8 0

3 years ago

HELP FAST, I'LL GIVE BRAINLIEST, 50 POINTS!!!

liraira [26]

Answer:

Decimal = -1.375, Fraction = -11/8

Step-by-step explanation:

Decimal:

-2.25 + 0.4x = -2.8

0.4x = -0.55

x = -1.375

Fraction:

-9/4 + 2/5x = -14/5

-45/20 + 8/20x = -56/20

8/20x = -11/20

x = -11/8

Hopefully this helps!

Brainliest please?

5 0

2 years ago

Read 2 more answers

What is the answer to this solution. -7r−4≥ 4r+2

tia_tia [17]

Answer:

The solution is:

$-7r-4\ge \:\:4r+2\quad \::\quad \:\begin{bmatrix}\mathrm{Solution:}\:&\:r\le \:\:-\frac{6}{11}\:\\ \:\:\mathrm{Decimal:}&\:r\le \:\:-0.54545\dots \:\\ \:\:\mathrm{Interval\:Notation:}&\:(-\infty \:\:,\:-\frac{6}{11}]\end{bmatrix}$