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3 years ago

10

Which is the graph of 3x-2y=6

2 answers:

alexdok [17]3 years ago

5 0

Answer:

c

Step-by-step explanation:

steposvetlana [31]3 years ago

4 0

Answer:

Graph 1

Step-by-step explanation:

The y intercept is 3 (because 6 divided by -2 equals 3)

The only graph with the y intercept of 3 is Graph 1

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A circuit is supplied with 24 volts and its load offers a total resistance of 400 ohm. What’s the total circuit power?

USPshnik [31]

Answer:

P = 1.44 Watts

Step-by-step explanation:

We know that the relation between voltage and current in a simple circuit with a resistor is

V = I.R

Where

V = the voltage of the source

I = current of the circuit

R = resistance

The power is defined as

P = V.I

Which is the same as

P = V.(V/R)

P = (24 v)^2 / (400 ohm)

P = 1.44 Watts

8 0

4 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

Can someone figure out the problem in this image?

erma4kov [3.2K]

Right side:
(1/27)^(2x+10)
=(3^-3)^(2x+10)
= 3^(-6x - 30)

Re-write both sides:
3^(4x-5) = 3^(-6x - 30)

from here, you can solve x

4x - 5 = -6x - 30
4x + 6x = -30 + 5
10x = -25
x = -2.5

7 0

4 years ago

The interior angles formed by the sides of a quadrilateral have measures that sum to 360°

Ronch [10]

Answer:

Is it x = 85

196 +2x+(3x-6)

subtract 196 on both sides

simplify

add 6 on both sides

divide by 2

5 0

3 years ago

A hotel manager recorded the percentage of rooms that were occupied each day over a period of 25 days. The data she collected is

larisa86 [58]

Answer:

D. 15

Step-by-step explanation:

I just took the test lol

3 0

3 years ago

Read 2 more answers