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Maru [420]
3 years ago
9

Find x so that q parallel r​

Mathematics
1 answer:
kondor19780726 [428]3 years ago
3 0

Answer: x = 7

Step-by-step explanation:

Corresponding angles have congruent measures. Thus, 11x - 2 = 75

11x - 2 = 75

11x = 77

x = 7

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Multiply and simplify: 3a(4a2 – 5a + 12)
DochEvi [55]
3a(4a²-5a+12)

12a³-15a²+36a

~Hope this helped!~
3 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
A rectangular swimming pool that is 10 ft wide by 16 ft long is surrounded by a cement sidewalk of uniform width. If the area of
Alchen [17]
We are asked to solve for the width "x" in the given problem. To visualize the problem, see attached drawing.
We have the area of the swimming pool such as:
Area SP = l x w
Area SP = 10 * 16
Area SP = 160 feet2
Area of the swimming pool plus the sidewalk with uniform width:
Area SP + SW = (10 + x) * (16 + x)
160 + 155 = 160 + 10x + 16x + x2
160 -160 + 155 = 26x + x2
155 = 26 x + x2
x2 + 26x -155= 0
Solving for x, we need to use quadratic formula and the answer is 5 feet.

The value of x is <span>5 feet. </span>

5 0
3 years ago
How many square units are in an office that is 13 units by 9 units? 22 117 121 130?
julsineya [31]
Square units are the units of measuring area. Therefore, this question asks for the area of the room.

The room has the shape of a rectangle with:
length = 13 units and width = 9 units

The are of the rectangle can be calculated using the following rule:
area of rectangle = length * width
area of the room = 13 * 9 = 117 square units
8 0
3 years ago
GIVING BRAINIEST!!!!!!!!
Sav [38]

Answer:

look at the horizontal line in the picture. the degree measure of any line is 180° given there's a perpendicular ray through that horizontal line it's therfore split into two sides both with angle measure of 90°.

given f is 71° then g can be found knowing that both g and f must add to 90°. 71+g=90. g=19°

now look at f again. f and d are what's known as vertical angles and that means that they're angle measures are congruent. therfore the measure of d is 71° d=71°

Finally to find e we notice that angle d and e form a straight line which means both angles measures must add to 180°. therefore e can be found by computing d+e=180

aunaitituitmg our information we know 71+e=180 then e must equal 109° e=109°

3 0
3 years ago
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