Answer:
Step-by-step explanation:
Starting from the y-intercept of ![\displaystyle [0, -2],](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5B0%2C%20-2%5D%2C)
you do 
by either moving four blocks <em>south</em><em> </em>over one block <em>west</em><em> </em>or four blocks <em>north</em><em> </em>over one block<em> east</em><em> </em>[<em>west</em> and <em>south</em> are negatives]. Next, we have to determine the types of inequality symbols that are suitable for this graph, which will be <em>less</em><em> </em><em>than</em><em> </em>and <em>greater</em><em> </em><em>than</em><em> </em>since this is a <em>dashed</em><em> </em><em>line</em><em> </em>graph. We then use the zero-interval test [test point (0, 0)] to ensure whether we shade the opposite portion [portion that does not contain the origin] or the portion that DOES contain the origin. At this step, we must verify the inequalities as false or true:
<em>Greater</em><em> </em><em>than</em>
![\displaystyle 0 > 4[0] - 2 → 0 > -2](https://tex.z-dn.net/?f=%5Cdisplaystyle%200%20%3E%204%5B0%5D%20-%202%20%E2%86%92%200%20%3E%20-2)
☑
<em>Less</em><em> </em><em>than</em><em> </em>
![\displaystyle 0 < 4[0] - 2 → 0 ≮ -2](https://tex.z-dn.net/?f=%5Cdisplaystyle%200%20%3C%204%5B0%5D%20-%202%20%E2%86%92%200%20%E2%89%AE%20-2)
This graph is shaded in the portion of the origin, so you would choose the <em>greater</em><em> </em><em>than</em><em> </em>inequality symbol to get this inequality:
I am joyous to assist you anytime.
Ok...I considered it....oh..u want me to answer it...lol..ok
5/6(1/2x + 6) - 3x
5/12x + 30/6 - 3x
5/12x - 36/12x + 5
- 31/12x + 5 <===
5(x-1) + 3 = 13
5x - 5 + 3 = 13
5x -2 = 13
5x = 15
x = 15/5
x = 3
Step-by-step explanation:
Let the number with two decimal places be 1.23
Let the number with one decimal place be 4.5
Then 1.23 multiplied by 4.5
= 1.23 × 4.5
= 5.535
The result has more than two nonzero digits.
Note: When you multiply two decimal numbers, the decimal place(s) of the result is the addition of the decimal places of each of the numbers being multiplied.
