Answer:
Third option: 
Step-by-step explanation:
<h3><em> The correct form of the exercise is: "The point-slope form of the equation of a line that passes through points (8, 4) and (0, 2) is
. What is the slope-intercept form of the equation for this line?"</em></h3><h3><em /></h3>
<em> </em>The equation of the line in Slope-Intercept form is:
Where "m" is the slope and "b" is the y-intercept.
Given the equation of the line in Point-Slope form:
You need to solve for "y" in order to write the given equation of the line in Slope Intercept form.
Then, this is:
You can identify that the slope "m" is:
And the y-interecept "b" is:
Answer:
because a positive is over zero and a negative is below zero so from that you can tell that the opposite of any positive will be a negative.
Step-by-step explanation:
Answer:
Let the vectors be
a = [0, 1, 2] and
b = [1, -2, 3]
( 1 ) The cross product of a and b (a x b) is the vector that is perpendicular (orthogonal) to a and b.
Let the cross product be another vector c.
To find the cross product (c) of a and b, we have
![\left[\begin{array}{ccc}i&j&k\\0&1&2\\1&-2&3\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Di%26j%26k%5C%5C0%261%262%5C%5C1%26-2%263%5Cend%7Barray%7D%5Cright%5D)
c = i(3 + 4) - j(0 - 2) + k(0 - 1)
c = 7i + 2j - k
c = [7, 2, -1]
( 2 ) Convert the orthogonal vector (c) to a unit vector using the formula:
c / | c |
Where | c | = √ (7)² + (2)² + (-1)² = 3√6
Therefore, the unit vector is
![\frac{[7,2,-1]}{3\sqrt{6} }](https://tex.z-dn.net/?f=%5Cfrac%7B%5B7%2C2%2C-1%5D%7D%7B3%5Csqrt%7B6%7D%20%7D)
or
[ 
, 
, 
]
The other unit vector which is also orthogonal to a and b is calculated by multiplying the first unit vector by -1. The result is as follows:
[ 
, 
, 
]
In conclusion, the two unit vectors are;
[ , , ]
and
[ , , ]
<em>Hope this helps!</em>
Answer:
a/c=2/7 u are welcome thanks for ur help
Answer:
C. To make sure all temporary accounts have a zero balance.
Step-by-step explanation:
The closing process is an accounting method applied at the end of the year with the purpose of ensuring that all accounts are ready for the preparation of financial statements, as well as the start of a new accounting year.
Temporary accounts are made permanent. Also, the accounts for the present year are zeroed. By so doing, only the current balance will be reflected in the new year accounts.
