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Zepler [3.9K]
3 years ago
6

How can you construct perpendicular lines and prove theorems about perpendicular lines

Mathematics
1 answer:
omeli [17]3 years ago
7 0

Answer:

See descriptions below.

Step-by-step explanation:

To construct a perpendicular bisector, draw a line segment. From each end of the line segment, draw arcs above and below which intersect from each side. Be sure to maintain the same radius on each. Where the arcs intersect above and below, mark points. Connect these two points. This is a perpendicular bisector.

To prove theorems about parallel lines, use angle relationships. For instance, when two parallel lines are cut by a transversal, specific angle are congruent. When these relationships are congruent, you must have parallel lines:

  • Alternate Interior
  • Alternate Exterior
  • Corresponding Angles
  • Same side interior add to 180
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A data value is considered​ _______ if its​ z-score is less than minus2 or greater than 2.
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Step-by-step explanation:

we know that

The z-score is a measure of how close the given data point is to the mean of the values given with the standard deviation

so

if its z-score is greater than or equal to -2, or less than or equal to 2., then the data value is considered ordinary

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3 years ago
Write an equation of a line, in slope-intercept form, that is perpendicular to the line y=2x+5 and passes through the point (8,-
Eva8 [605]

Answer:

22

Step-by-step explanation:

7 0
3 years ago
Please help!!<br> Will mark brainliest
Over [174]
Y would equal to -5. hope this helps
8 0
3 years ago
6 less than twice a number is at least 15. Write an inequality and solve.
Ivan

Answer:

2a-6>=15 this is because the key word at least means it can be 15 or higher but it cannot be less than 15.

Step-by-step explanation:

6 0
3 years ago
Explain how to find the relationship between two quantities, x and y, in a table. How can you use the relationship to calculate
Morgarella [4.7K]

Explanation:

In general, for arbitrary (x, y) pairs, the problem is called an "interpolation" problem. There are a variety of methods of creating interpolation polynomials, or using other functions (not polynomials) to fit a function to a set of points. Much has been written on this subject. We suspect this general case is not what you're interested in.

__

For the usual sorts of tables we see in algebra problems, the relationships are usually polynomial of low degree (linear, quadratic, cubic), or exponential. There may be scale factors and/or translation involved relative to some parent function. Often, the values of x are evenly spaced, which makes the problem simpler.

<u>Polynomial relations</u>

If the x-values are evenly-spaced. then you can determine the nature of the relationship (of those listed in the previous paragraph) by looking at the differences of y-values.

"First differences" are the differences of y-values corresponding to adjacent sequential x-values. For x = 1, 2, 3, 4 and corresponding y = 3, 6, 11, 18 the "first differences" would be 6-3=3, 11-6=5, and 18-11=7. These first differences are not constant. If they were, they would indicate the relation is linear and could be described by a polynomial of first degree.

"Second differences" are the differences of the first differences. In our example, they are 5-3=2 and 7-5=2. These second differences are constant, indicating the relation can be described by a second-degree polynomial, a quadratic.

In general, if the the N-th differences are constant, the relation can be described by a polynomial of N-th degree.

You can always find the polynomial by using the given values to find its coefficients. In our example, we know the polynomial is a quadratic, so we can write it as ...

  y = ax^2 +bx +c

and we can fill in values of x and y to get three equations in a, b, c:

  3 = a(1^2) +b(1) +c

  6 = a(2^2) +b(2) +c

  11 = a(3^2) +b(3) +c

These can be solved by any of the usual methods to find (a, b, c) = (1, 0, 2), so the relation is ...

   y = x^2 +2

__

<u>Exponential relations</u>

If the first differences have a common ratio, that is an indication the relation is exponential. Again, you can write a general form equation for the relation, then fill in x- and y-values to find the specific coefficients. A form that may work for this is ...

  y = a·b^x +c

"c" will represent the horizontal asymptote of the function. Then the initial value (for x=0) will be a+c. If the y-values have a common ratio, then c=0.

__

<u>Finding missing table values</u>

Once you have found the relation, you use it to find missing table values (or any other values of interest). You do this by filling in the information that you know, then solve for the values you don't know.

Using the above example, if we want to find the y-value that corresponds to x=6, we can put 6 where x is:

  y = x^2 +2

  y = 6^2 +2 = 36 +2 = 38 . . . . (6, 38) is the (x, y) pair

If we want to find the x-value that corresponds to y=27, we can put 27 where y is:

  27 = x^2 +2

  25 = x^2 . . . . subtract 2

  5 = x . . . . . . . take the square root*

_____

* In this example, x = -5 also corresponds to y = 27. In this example, our table uses positive values for x. In other cases, the domain of the relation may include negative values of x. You need to evaluate how the table is constructed to see if that suggests one solution or the other. In this example problem, we have the table ...

  (x, y) = (1, 3), (2, 6), (3, 11), (4, 18), (__, 27), (6, __)

so it seems likely that the first blank (x) will be between 4 and 6, and the second blank (y) will be more than 27.

6 0
3 years ago
Read 2 more answers
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