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3 years ago

Donna gets 3 problems incorrect on a math quiz. Her score is 85%. How many questions are on the quiz.

Mathematics

1 answer:

nata0808 [166]3 years ago

5 0

Answer:

Step-by-step explanation: 5 times 20 = 100

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Q1.How many significant figures does 39018 have

beks73 [17]

Answer:

The answer is:

Q1: 5

Q2: 5

Q3: 2

7 0

3 years ago

Please help me with this problem

Sliva [168]

Answer:

First of all the Dodge Viper wouldn't cost the price it has for listed; maybe if it was totaled you could probably get a finder for that much.

3 0

3 years ago

Pls help I struggling a lot pls pls pls I need pro math expert

Soloha48 [4]

Answer:

Option: $h\leq 16$

Step-by-step explanation:

Looking for key words in the problem:

"At Most" represents important key words, meaning the maximum amount of hotdogs that are at the barbecue.

Inequality:

$h\leq 16$

h $\leq$ 16 represents the maximum amount of hotdogs that has to be at the barbecue, also represents less than 16 can be cooked.

3 0

2 years ago

Variable y varies directly with x2, and y = 96 when x = 4.

lakkis [162]

Direct variation is y = kx, where k is the constant of variation.
But now it says y varies directly with x2 (or 2x), so now the x in the equation is 2x.

The equation is y = k(2x)
Now you find k.
y = 96 when x = 4.
(96) = k(2*4)
96 = k(8)
k = 12

The equation is now y = 12(2x)
To find the value of y when x=2, plug 2 into the equation you made. y = 12(2*2) y = 48
_________________
Now it's with a "quadratic variation," which is the same thing except x is squared.
The equation is y = kx^2

But y varies directly with x2 (same thing as 2x), so now it's y = k(2x)^2.

Now you find k by substituting y and x values that were given.
y = 180 when x = 6
(180) = k(2*6)^2
180 = k(12)^2
180 = k(144)
k = 1.25
k, 1.25, is the constant of variation.

4 0

3 years ago

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