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gizmo_the_mogwai [7]
3 years ago
8

Explain how to find the relationship between two quantities, x and y, in a table. How can you use the relationship to calculate

the missing values in the table?
Mathematics
2 answers:
motikmotik3 years ago
8 0

Here is your answer     |  To determine the relationship between quantities, you must determine what to do to the x-values to make them into y-values. The correct operation must turn every x-value into the corresponding y- value in the table. Once you know the relationship, you can use the same operation on all of the x-values that have unknown y-values.

Morgarella [4.7K]3 years ago
6 0

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.

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Pani-rosa [81]

9514 1404 393

Answer:

  a) GH or HG

  b) KL

  c) YZ or ZY

  d) EF

Step-by-step explanation:

A segment or line can be named by naming two points on it in any order. The points naming a line segment are its end points.

A ray is named by first naming its end point, then any other point on the ray.

a) GH or HG

b) KL

c) YZ or ZY

d) EF

4 0
3 years ago
I need it done asapp✔️
Alexxx [7]

Answer:

It is at 0 that mean that is the answer right?

Step-by-step explanation:

6 0
2 years ago
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julsineya [31]
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we do a sum of areas [S] 2pi [f(x)]^2 dx
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3 0
3 years ago
My boss has told me that I will need one gallon of paint for every 300 square feet of wall I must paint. Unfortunately, the stor
lubasha [3.4K]

Answer: 14 cans

Step-by-step explanation:

Given, Total area to paint = 400 square meters

approximately 3.28 feet in a meter.

So, 400 square meters = 400 x (3.28)² square feet

i.e. Total area to paint  = 4303.36 square feet

One gallon of paint requires for every 300 square feet.

One liter contains approximately 0.264 gallons

Then, One gallon = \dfrac{1}{0.264}\approx3.78\text{ liters}

So, 3.78 liters paint requires for every 300 square feet.

Paint requires for each square feet = (3.78)÷(300) liters

Total paint required  = (Total area to paint ) x (Paint requires for each square feet)

= (4303.36)x (3.78)÷(300)

≈54.22 liters

Each can contains 4 liters of paint.

Smallest number of cans required = (Total paint required )÷ 4

=(54.22 ) ÷ 4

= 13.55≈ 14

Hence, 14 cans are required .

4 0
3 years ago
How to solve? 2x - 5 = 11
shutvik [7]

Answer:

Add

5

5

5

to both sides of the equation

2

−

5

=

1

1

2

−

5

+

5

=

1

1

+

5

2

Simplify

Add the numbers

Add the numbers

2

=

1

6

3

Divide both sides of the equation by the same term

2

=

1

6

2

2

=

1

6

2

4

Simplify

Cancel terms that are in both the numerator and denominator

Divide the numbers

=8

Solution

=8

Step-by-step explanation:

6 0
3 years ago
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