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Dimas [21]
3 years ago
12

How to find the midpoint using coordinates A(-1,5), B(2,-3)

Mathematics
2 answers:
Helen [10]3 years ago
5 0
The midpoint formula is (x1+x2)/(2),(y1+y2)/(2). The coordinates will be substituted into this formula: (-1+2)/(2), (5+-3)/(2). This simplifies to (1/2,1). This is the midpoint. Hope this helped!
Alex Ar [27]3 years ago
4 0
Count the line and the distance between them
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A solution to the compound inequality 17 less than or equal to 3 - 2x less than or equal to 21 is given below. Choose the statem
aliina [53]
<span>17<=3-2x<=21 subtract 3 14<=-2x<=18 divide by 2 7<=-x<=9 now mutiply whole inequality by -1, this will reveres the inequality -7>=x>=-9 or -9<=x<=-7 so there are mistakes in the line the corrected lines are 17 less than or equal to 3 - 2x and 3 - 2x less than or equal to 21 (line one) 14 less than or equal to -2x and -2x less than or equal to 18 (line two) -7 greater than or equal to x and x greater than or equal to -9 (line three)</span>
6 0
3 years ago
Find each value of the five-number summary for this set of data. [Note: Type your answers as numbers. Do not round.] 46, 19, 38,
frozen [14]

Answer:

Maximum : 51

Minimum: 12

Median: 38

Lower quartile: 19

Upper quartile: 46

Step-by-step explanation:

To answer this question, we first order the data from largest to smallest:

46, 19, 38, 27, 12, 38, 51

12, 19, 27, 38, 38, 46, 51

It can be seen that the maximum value is:

51

The minimum value is

12

Let's call N the amount of data.

To calculate the quartiles we use the following formula:

Q_1 = X_{\frac{1}{4}(N + 1)}

So:

Q_1 = X_{\frac{1}{4}(7 + 1)}

Q_1 = X_2

Where X_2 is the data number 2.

Q1 = 19

As the number of data is imapar, then, the median is the central data:

12, 19, 27, {38}, 38, 46, 51

Median = 38

Finally:

Q_3 = X_ {\frac{3}{4}(n + 1)}\\\\Q_3 = X_ {6}

Where X_6 represents the data number 6

Q_3 = 46

4 0
3 years ago
Given a polynomial function f(x), describe the effects on the y-intercept, regions where the graph is increasing and decreasing,
erma4kov [3.2K]
<span>The y-intercept of  is  .
Of course, it is 3 less than  , the y-intercept of  .
Subtracting 3 does not change either the regions where the graph is increasing and decreasing, or the end behavior. It just translates the graph 3 units down.
It does not matter is the function is odd or even.

 is the mirror image of  stretched along the y-direction.
The y-intercept, the value of  for  , is</span><span>which is  times the y-intercept of  .</span><span>Because of the negative factor/mirror-like graph, the intervals where  increases are the intervals where  decreases, and vice versa.
The end behavior is similarly reversed.
If  then  .
If  then  .
If  then  .
The same goes for the other end, as  tends to  .
All of the above applies equally to any function, polynomial or not, odd, even, or neither odd not even.
Of course, if polynomial functions are understood to have a non-zero degree,  never happens for a polynomial function.</span><span> </span>
4 0
4 years ago
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Mandarinka [93]
So in this problem we need to do some addition. 

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