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Misha Larkins [42]
3 years ago
14

What set would no longer be a A function if the ordered pair(5,3) was added

Mathematics
2 answers:
Delicious77 [7]3 years ago
4 0
The answer is D because a function means for every input  you get exactly one output so
for every x you get one y
in D there is a coordinate (5,2) meaning that if you put in 5 for x you get 2 for y

but the new ordered pair also  has 5 for x but has a difarent y value.  this violates the definition of a funciton
Pavlova-9 [17]3 years ago
3 0
I think it's A because everything else has even numbers and letter A has negative numbers... 
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if you prder two servings of pancakes and fruit cup the cost of the fruit cup is $1.50 and you leave a 15% tip your total bill i
Sindrei [870]

Both servings of pancakes cost $8.50

One serving of pancakes cost $4.35

Equation:   1.15 (2x + 1.50) = 11.50

                  2.3x + 1.725 + 11.50

                   2.3x = 9.725

                   x = 4.25 for one serving of pancakes

                   2x = 8.50 for both servings of pancakes

3 0
3 years ago
Determine the value of x in each figure.
velikii [3]

Answer:

x=15is your answer

Step-by-step explanation:

8x=120 vertically opposite angle

x=120/8=15

3 0
2 years ago
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Which of the following graphs shows the solution set for the inequality below? 3|x + 1| < 9
Bas_tet [7]

Step-by-step explanation:

The absolute value function is a well known piecewise function (a function defined by multiple subfunctions) that is described mathematically as

                                 f(x) \ = \ |x| \ = \ \left\{\left\begin{array}{ccc}x, \ \text{if} \ x \ \geq \ 0 \\ \\ -x, \ \text{if} \ x \ < \ 0\end{array}\right\}.

This definition of the absolute function can be explained geometrically to be similar to the straight line   \textbf{\textit{y}} \ = \ \textbf{\textit{x}}  , however, when the value of x is negative, the range of the function remains positive. In other words, the segment of the line  \textbf{\textit{y}} \ = \ \textbf{\textit{x}}  where \textbf{\textit{x}} \ < \ 0 (shown as the orange dotted line), the segment of the line is reflected across the <em>x</em>-axis.

First, we simplify the expression.

                                             3\left|x \ + \ 1 \right| \ < \ 9 \\ \\ \\\-\hspace{0.2cm} \left|x \ + \ 1 \right| \ < \ 3.

We, now, can simply visualise the straight line,  y \ = \ x \ + \ 1 , as a line having its y-intercept at the point  (0, \ 1) and its <em>x</em>-intercept at the point (-1, \ 0). Then, imagine that the segment of the line where x \ < \ 0 to be reflected along the <em>x</em>-axis, and you get the graph of the absolute function y \ = \ \left|x \ + \ 1 \right|.

Consider the inequality

                                                    \left|x \ + \ 1 \right| \ < \ 3,

this statement can actually be conceptualise as the question

            ``\text{For what \textbf{values of \textit{x}} will the absolute function \textbf{be less than 3}}".

Algebraically, we can solve this inequality by breaking the function into two different subfunctions (according to the definition above).

  • Case 1 (when x \ \geq \ 0)

                                                x \ + \ 1 \ < \ 3 \\ \\ \\ \-\hspace{0.9cm} x \ < \ 3 \ - \ 1 \\ \\ \\ \-\hspace{0.9cm} x \ < \ 2

  • Case 2 (when x \ < \ 0)

                                            -(x \ + \ 1) \ < \ 3 \\ \\ \\ \-\hspace{0.15cm} -x \ - \ 1 \ < \ 3 \\ \\ \\ \-\hspace{1cm} -x \ < \ 3 \ + \ 1 \\ \\ \\ \-\hspace{1cm} -x \ < \ 4 \\ \\ \\ \-\hspace{1.5cm} x \ > \ -4

           *remember to flip the inequality sign when multiplying or dividing by

            negative numbers on both sides of the statement.

Therefore, the values of <em>x</em> that satisfy this inequality lie within the interval

                                                     -4 \ < \ x \ < \ 2.

Similarly, on the real number line, the interval is shown below.

The use of open circles (as in the graph) indicates that the interval highlighted on the number line does not include its boundary value (-4 and 2) since the inequality is expressed as "less than", but not "less than or equal to". Contrastingly, close circles (circles that are coloured) show the inclusivity of the boundary values of the inequality.

3 0
2 years ago
Billy makes $5 a week in allowance plus $4 for each lawn that he mows.
Ad libitum [116K]

Answer:

$15 < $4n + $5

Step-by-step explanation:

We know that Billy needs to make more than $15 between his allowance and the lawns that he mows. This means our inequality should include $15<. Also, since Billy will make $4 per lawn, that means we need to multiply $4 by the number of lawns he needs to mow, n: $4n. So far we have the following: $15<$4n. Next, we know that he makes $5 each week, on top of what he makes mowing each law. This means we need to add the $5 to the $4n. When we put all of these pieces together, we will get the following inequality: $15<$4n+$5

7 0
3 years ago
What is the greatest common factor of 24, 40 and 32? <br> A. 1 <br> B. 2 <br> C. 4 <br> D. 8
trasher [3.6K]
It’s 8 because 24/8 is 3 and 40/8 is 5 and 32/8 is 4 so the answer is 8 YOUR WELCOME
6 0
3 years ago
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