An ordered pair is written like this, ( x, y ). In this case x = 0 and y = -3. On a graph the vertical line is the y-axis and the horizontal line is the x-axis. The origin is point ( 0, 0 ). To the left of the origin on the x-axis is the negative number line and to the right is the positive number line. On the y-axis, south of the origin is the negative number line and north is the positive number line. When you plot a point on a graph you do x first, so if x equals 1, you would move one right, -1, one left. IF y were to equal 2 then from the place where you are on the x-axis, 1, you would move two up, -2, two down. In this case x = 0 so you would stay at the origin, and y = -3 so you would move 3 down. So ( 0, -3 ) would lie negative y-axis. The answer is D.
Answer:
29
Step-by-step explanation:
using Pythagorean theory.
equate it to the hypotenus.
thus- hyp sq=adjacent sq +opposite sq
Answer:
The segment GT is a median of triangle GFH
Step-by-step explanation:
we know that
A <u><em>median</em></u> of a triangle is a line segment joining a vertex to the midpoint of the opposite side.
The <u><em>angle bisector</em></u> of a triangle is a line segment that bisects one of the vertex angles of a triangle.
In this problem
SH represent a line segment that bisects the vertex angle H of triangle GFH, so represent an angle bisector.
GT represent a median of triangle GFH, because the point T is the midpoint of segment FH (FT=TH)
therefore
The segment GT is a median of triangle GFH
Answer:
0
Step-by-step explanation:
We find the determinant of a matrix by the method below. If we have a matrix:
![\left[\begin{array}{cc}a&b\\c&d\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Da%26b%5C%5Cc%26d%5Cend%7Barray%7D%5Cright%5D)
The determinant is 
Now, using cramer's rule, we find x-value by the formula:

Where D is the determinant of the original problem and
is the determinant of the x-value matrix. How do we get those?
<u><em>To get original matrix and thus D, we set up the matrix as the coefficients of x and y (s) of both the equations and to get matrix of x-value and thus
, we replace the x values of the matrix with the numbers in the right hand side of the 2 equations.</em></u> We show this below:
<em />
<em>To get D:</em>
![\left[\begin{array}{cc}3&4\\1&-6\end{array}\right] \\D=(3)(-6)-(1)(4)=-18-4=-22](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D3%264%5C%5C1%26-6%5Cend%7Barray%7D%5Cright%5D%20%5C%5CD%3D%283%29%28-6%29-%281%29%284%29%3D-18-4%3D-22)
<em>To get
:</em>
<em>
</em>
<em />
<em>Putting into the formula, we get:</em>
<em>
</em>
<em />
<em>Thus, the value of x is 0</em>
Answer:
the answer is 7 you have to multiple 5x2 then subtract that from 3