Rob scored 34 points. 88-54=34
Solving the inequality given algebraically, the solution of the inequality for the value of m is:
<u>m < -4 OR m > 3</u>
<em><u>Given the following </u></em><em><u>inequality</u></em><em><u>,</u></em>
<em><u /></em>
or
4m + 3 > 15
Let's solve algebraically for the value of m in both inequality statements given.



Or
4m + 3 > 15
- Subtract 3 from each side
4m + 3 - 3 > 15 - 3
4m > 12

m > 3
Therefore, solving the inequality given algebraically, the solution of the inequality for the value of m is:
<u>m < -4 OR m > 3</u>
<u></u>
<u></u>
Learn more here:
brainly.com/question/24434501
<h3>
<u>Answer:</u></h3>

<h3>
<u>Step-by-step explanation:</u></h3>
A inequality is given to us and we need to convert it into standard form and see whether if it has a solution . So let's solve the inequality.
The inequality given to us is :-

Let's plot a graph to see its interval . Graph attached in attachment .
Now we can see that the Interval notation of would be ,
![\boxed{\boxed{\orange \tt \purple{\leadsto}y \in [-2,-1] }}](https://tex.z-dn.net/?f=%5Cboxed%7B%5Cboxed%7B%5Corange%20%5Ctt%20%5Cpurple%7B%5Cleadsto%7Dy%20%5Cin%20%5B-2%2C-1%5D%20%7D%7D)
<h3>
<u>Hence</u><u> the</u><u> </u><u>standa</u><u>rd</u><u> </u><u>form</u><u> </u><u>of</u><u> </u><u>inequa</u><u>lity</u><u> </u><u>is</u><u> </u><u>y²</u><u>+</u><u>3y</u><u> </u><u>+</u><u>2</u><u> </u><u>≤</u><u> </u><u>0</u><u> </u><u>and</u><u> </u><u>the </u><u>Solution</u><u> </u><u>set</u><u> </u><u>of</u><u> </u><u>the</u><u> </u><u>ineq</u><u>uality</u><u> </u><u>is</u><u> </u><u>[</u><u> </u><u>-</u><u>2</u><u> </u><u>,</u><u> </u><u>-</u><u>1</u><u> </u><u>]</u><u> </u><u>.</u></h3>
Answer:
the slope is positive
Step-by-step explanation:
Notice that as x is increasing, g(x) is also increasing. Therefore, the slope is positive. If you were to plot those points and draw the line thru them, you would see that the line goes up as x increases. A line that does that has a positive slope. Another way to look at it is to see that the change in g(x) is positive as the change in x is also positive.
The slope is change in g(x)/change in x.
So, you have positive/positive = positive.
There are several ways to solve this problem. I just gave you 3 of them.