The answer would be 4/85
1. 2/5 ÷ 8 * 2 + 1 over 2
2. 2/5 ÷ 16 + 1 over 2
3. 2/5 ÷ 17/2
4. 2/5 * 2/17
5. 2 * 2 over 5 * 17
6. 4 over 5 * 17
7. finally 4/85
The trial and error method is used to find an initial factor:
If we let f(x) = x³ - x² - 24x - 36 and all we have to do is sub' in values of x until
f(x) = 0, we can use this to find an initial factor by the factor theorem:
f(1) = (1)³ - (1)² - 24(1) - 36 = -60
f(2) = (2)³ - (2)² - 24(2) - 36 = -80
f(5) = (5)³ - (5)² - 24(5) - 36 = -56
*** f(6) = (6)³ - (6)² - 24(6) - 36 = 0 ***
f(6) = 0 so (x - 6) is a factor of f(x).
This means that: f(x) = x³ - x² - 24x - 36 = (x - 6)(ax² + bx + c).
To find a,b and c, use long division (or inspection) to divide x³ - x² - 24x - 36 by x - 6.
The other 2 factors of f(x) can then be found by factorizing the
ax² + bx + c quadratic the way you would with any other quadratic (i.e. by quadratic formula, CTS or inspection).
- Midpoint formula is
.
<h3>19.</h3>
So starting with this one, we will be solving for the coordinates of the unknown endpoint separately. Starting with the x-coordinate, since we know that the midpoint x-coordinate is 5 and the x-coordinate of N is 2, our equation is set up as such:
From here we can solve for the x-coordinate of Q.
Firstly, multiply both sides by 2: 
Next, subtract both sides by 2 and your x-coordinate is 
With finding the y-coordinate, it's a similar process as with the x-coordinate except that we are using the y-coordinates of the midpoint and endpoint N.

<u>Putting it together, the missing endpoint is (8,4).</u>
<em>(The process is pretty much the same with the other problems, so I'll go through them real quickly.)</em>
<h3>20.</h3>


<u>The missing endpoint is (7,2).</u>
<h3>21.</h3>


<u>The missing endpoint is (-5,1).</u>
Answer:
74
Step-by-step explanation:
You just have to substitute r with 5.
Now, the equation looks like this:

Next, solve
first according to PEMDAS
And
equals 75
![\frac{\left[\begin{array}{ccc}1&5\\5\\\end{array}\right] }{75}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%265%5C%5C5%5C%5C%5Cend%7Barray%7D%5Cright%5D%20%7D%7B75%7D)
Lastly, do
which equals 74
![\frac{\left[\begin{array}{ccc}7&5\\-1\\\end{array}\right] }{74}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D7%265%5C%5C-1%5C%5C%5Cend%7Barray%7D%5Cright%5D%20%7D%7B74%7D)
Therefore, 74 is your answer.