- Triangle Inequality Theorem: States that the sum of any two sides of a triangle is greater than the length of the third side;

So for this, we are applying the triangle inequality theorem. If any of the inequalities are not true, then this cannot be a triangle. (Let A = 7.7, B = 4.0, and C = 1.7)

<u>Since the second inequality is false, these lengths cannot form a triangle.</u>
Histograms are useful when we have data which can be divided into several classes or groups. The histogram shows the trend of each class and the trend among the different classes. For example when we have about 50 different values ranging from 1 to 20, it will be a better approach to draw a histogram in this case by dividing the data into small ranges e.g 1 to 4, 5 to 9 and so on and counting the frequency for each class.
Dot plot is useful when we have a small number of individual values. In this case we can visualize how many times each individual value occurred in the data. This is useful when the number of values in the data is less.
In the given scenario, we have 12 values in total ranging from 1 to 5. So making a dot plot would be the best choice. A histogram would not be useful in this case.
Therefore, the correct answer is option D. Dot plot, because a small number of scores are reported individually
Answer:
<h2>The answer is option A</h2>
Step-by-step explanation:
<h3>

</h3>
<u>First of all cross multiply</u>
That's
g( wd + f ) = a + bc
<u>Multiply the terms at the left side of the equation</u>
We have
gwd + fg = a + bc
<u>Send fg to the right side of the equation</u>
That's
gwd = a + bc - fg
<u>Divide both sides by dg to make W stand alone</u>
<h3>

</h3>
We have the final answer as
<h2 /><h2>

</h2>
Hope this helps you
Given:

and,
K = 1
We would require to do the synthetic division of the above problem in order to know whether K=1 is a lower bound or not. Let's do it!
Steps of Synthetic Division:
1. Write the coefficients of the equation, and before that write 1. Like,
1 ║ 4 -2 2 4
As you can see, the coefficients of the equation are seperated by ║.
2. Drop down the first value after ║as it is. Like:
1 ║ 4 -2 2 4
:
----------------------
4
3. Multiply 4 with 1 and then place the resultant value under -2 and then add that resultant value with -2. Like
1 ║ 4 -2 2 4
:
4----------------------
4
24. Repeat Step 3 until you're done.
1 ║ 4 -2 2 4
: 4
2 ----------------------
4 2
4
1 ║ 4 -2 2 4
: 4 2
4----------------------
4
2 4
8
Now every value, +4, +2, +4, +8, under the bar(---------) is positive; therefore, it means that
K=1 is the upper bound, NOT lower bound.
Ans: K=1 is NOT a lower bound.
-i
Explanation:
When the inequality symbol is replaced by an equal sign, the resulting linear equation is the boundary of the solution space of the inequality. Whether that boundary is included in the solution region or not depends on the inequality symbol.
The boundary line is included if the symbol includes the "or equal to" condition (≤ or ≥). An included boundary line is graphed as a solid line.
When the inequality symbol does not include the "or equal to" condition (< or >), the boundary line is not included in the solution space, and it is graphed as a dashed line.
Once the boundary line is graphed, the half-plane that makes up the solution space is shaded. The shaded half-plane will be to the right or above the boundary line if the inequality can be structured to be of one of these forms:
- x > ... or x ≥ ... ⇒ shading is to the right of the boundary
- y > ... or y ≥ ... ⇒ shading is above the boundary
Otherwise, the shaded solution space will be below or to the left of the boundary line.
_____
Just as a system of linear equations may have no solution, so that may be the case for inequalities. If the boundary lines are parallel and the solution spaces do not overlap, then there is no solution.
_____
The attached graph shows an example of graphed inequalities. The solutions for this system are in the doubly-shaded area to the left of the point where the lines intersect. We have purposely shown both kinds of inequalities (one "or equal to" and one not) with shading both above and below the boundary lines.