What is i to the 97th power minus i

MAVERICK [17]

The first graph.

The equation is x - 1 $\leq$ 3
In order to solve it, you need to get x by itself.
Add 1 to both sides. 3+1=4
x $\leq$ 4

Since this is your sign $\leq$ the arrow has to point towards the left, and the circle needs to be colored in.

3 0

3 years ago

Write an equation in slope-intercept form of the line.

alexira [117]

Answer:

y=x-3

Step-by-step explanation:

To find slope, use m=y2-y1/x2-x1, using values of x and y from two different points. Once you have found slope, take another point and use the point-slope equation y-y1=m(x-x1) and substitute in slope for m and the point for x1 and y1. Isolate y by subtracting y1 from both sides.

3 0

3 years ago

What are the solutions to the equation x - (7/x) = 6

BlackZzzverrR [31]

Answer: c. x=-1 and x=7

Step-by-step explanation:

7 0

3 years ago

Plz help<br>urgent !!!!<br>will give the brainliest !!

Mkey [24]

Answer:

$X = \begin{bmatrix}1&3\\ 2&4\end{bmatrix}$

Step-by-step explanation:

The question we have at hand is, in other words,

$\begin{bmatrix}4&2\\ \:1&1\end{bmatrix}\left(X\right)=\begin{bmatrix}8&20\\ \:3&7\end{bmatrix}$ - where we have to solve for the value of $X$

If we have to isolate $X$ here, then we would have to take the inverse of the following matrix ...

$\begin{bmatrix}4&2\\ \:1&1\end{bmatrix}$ ... so that it should be as follows ... $\begin{bmatrix}4&2\\ \:1&1\end{bmatrix}^{-1}$

Therefore, we can conclude that the equation as to solve for " $X$ " will be the following,

$X=\begin{bmatrix}4&2\\ 1&1\end{bmatrix}^{-1}\begin{bmatrix}8&20\\ 3&7\end{bmatrix}$ - First find the 2 x 2 matrix inverse of the first portion,

$\begin{bmatrix}4&2\\ 1&1\end{bmatrix}^{-1}$ = $\frac{1}{\det \begin{pmatrix}4&2\\ 1&1\end{pmatrix}}\begin{pmatrix}1&-2\\ -1&4\end{pmatrix}$ = $\frac{1}{2}\begin{bmatrix}1&-2\\ -1&4\end{bmatrix}$ = $\begin{bmatrix}\frac{1}{2}&-1\\ -\frac{1}{2}&2\end{bmatrix}$

At this point we have to multiply the rows of the first matrix by the rows of the second matrix,

$X = \begin{bmatrix}\frac{1}{2}&-1\\ -\frac{1}{2}&2\end{bmatrix}\begin{bmatrix}8&20\\ 3&7\end{bmatrix}$ ,

$X = \begin{pmatrix}\frac{1}{2}\cdot \:8+\left(-1\right)\cdot \:3&\frac{1}{2}\cdot \:20+\left(-1\right)\cdot \:7\\ \left(-\frac{1}{2}\right)\cdot \:8+2\cdot \:3&\left(-\frac{1}{2}\right)\cdot \:20+2\cdot \:7\end{pmatrix}$ - Simplifying this, we should get ...

$\begin{bmatrix}1&3\\ 2&4\end{bmatrix}$ ... which is our solution.

7 0

3 years ago

Let X = the time (in 10−1 weeks) from shipment of a defective product until the customer returns the product. Suppose that the m

Aleks [24]

Answer:

Step-by-step explanation: see attachment for solution

5 0

3 years ago