How do variables help you model real-world situations? Give an example of a model that uses variables and explain how it helps people make predictions.

The required solution of the system is (x, y) = (9, 23). Option D is correct.

Coordinate, is represented as the values on the x-axis and y-axis of the graph. while the coordinate x is called abscissa and the coordinate of the y is called ordinate.

Here,
To solve the system of equations using the substitution method, we substitute x = 9 into the first equation:

y = 3(9) - 4

y = 27 - 4

y = 23

Therefore, the solution of the system is (x, y) = (9, 23).

So, the correct answer is option D, (9, 23).

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6) answer is B , random sampling produces a representative sample

7)  0.1 = 10% so the answer is A, he saves 10% of his income

area = PIr^2

radius = D/2 = 1KM

3.14 x 1^2 =3.14 square km

247 x 3.14 = 775.58 = 776 acres

Orio and Ted buy an equal number of comic books every month. Orio has 9 more comic books than Ted.

Linear equation is equation in which each term has at max one degree.

Linear equation in variable x and y can be written in the form y = mx + c

Linear equation with two variables, when graphed on the cartesian plane with axes of those variables, give a straight line.

Orio and Ted buy an equal number of comic books every month.

The following equation shows the number of comic books, y, that Orio has after x months:

y = 2x + 15

Using the equation y=mx+b to find Orio's number of comic books, we can plug 2 months to substitute for the variable x.

y = 2(2) + 15

y = 4+15,

y = 19.

We know Ted has 10 comics after 2 months from the graph.

19-10= 9

Therefore, Orio has 9 more comic books than Ted.

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2. The Graph Contains The Point (5, 12)

Remember that in a linear function of the form [tex]y=mx+b[/tex], [tex]m[/tex] is the slope/rate of change of the line and [tex]b[/tex] is the y-intercept.

From our equation we can infer that [tex]b=12[/tex], so the y-intercept of our graph is 12 units above the origin on the point (0, 12). Therefore, choice 2 correctly describe the graph of y = 5x + 12

We can also infer that [tex]m=5[/tex], so the slope of our line is 5, which is constant, so our line has a constant rate of change. Also, since every integer can be written as a fraction with denominator 1, we can write our slope as [tex]m=\frac{5}{1}[/tex]. Therefore, choices 1 and 2 correctly describe the graph of y = 5x + 12.

Now, remember that any point on the plane has coordinates [tex](x,y)[/tex], so, to check if a point is on a line, we just need to replace the [tex]x[/tex] and [tex]y[/tex] values in the line equation and check if the equation holds. Our point is (5, 12), so x = 5 and y = 12. Lets replace the values in our line:

[tex]y=5x+12[/tex]

[tex]12=5(5)+12[/tex]

[tex]12=25+12[/tex]

[tex]12\neq 37[/tex]

Since the equation equation doesn't hold (12 is not equal 37), we can conclude that the graph doesn't contain the point (5, 12). Therefore, choice 2 does NOT correctly describe the graph y = 5x +12.

Answer:

The graph contains the point (5,12)

I got it right on t t m

Answer:

Sample Answer: To draw a trend line with the divide-center method, first exclude any outliers. Draw a vertical line to divide the data cluster into two groups with approximately the same number of points. Find the center of each group and connect those centers to form the trend line.

Okay, so first, we find a number that is between 0 and 1.

A number that qualifies for that condition would be ½.

A property of exponents is x^(n/m) = √x m sqrt(x^n), so x^(1/2) = √x

Therefore, such expression is:

x^(1/2) = 4
Where you now solve for x.

Exponential notation is a common language that is needed in order to communicate mathematical ideas efficiently and clearly. Efficiently written repeated multiplication was the reason that this notation was developed.

An example of this is the division of cells that causes growth to occur in living organisms. A single type of cell is divided 2 times in an hour, therefore in 12 hours, the cell will divide into 2 • 2 • 2 • 2 • 2 • 2 • 2 • 2 • 2 • 2 • 2 • 2 time, or as efficiently written, 2^12.

Consider a real number in two dimensional plane that X axis and Y axis having ordered pair : P(x,y) or (3,4).

To find it's absolute value we will find it's Distance from origin.

As Coordinate of Origin = (0,0)

Absolute Value of OP = [tex]\left | OP \right |[/tex]

     =  [tex]\sqrt{(0-x)^2+(0-y)^2}=\sqrt {x^2+y^2}[/tex]

    If the point is Q(3,4)

[tex]\left | OQ \right |=\sqrt{3^2+4^2}=\sqrt{5^2}=5[/tex]

In case of a complex number :

Z = a + bi , we consider two axes Real Axis and Imaginary Axis.

We represent Z that is (a,b) in the same way that we represent point in two dimensional plane.

To find it's absolute value , we will find it's distance from origin.

[tex]\left | Z \right |= \sqrt{a^2+b^2}[/tex]

If [tex]Z_{1}[/tex] = 6 + 8 i

[tex]\left | Z_{1}\right |=\sqrt{6^2+8^2}=\sqrt{36+64}=\sqrt{100}=10[/tex]

To find out how many ants of the same length would fit end-to-end across the length of the picnic table, divide the length of the table by the length of the ant.

To find out how many ants of the same length would fit end-to-end across the length of the picnic table, we need to convert the length of the table and the length of the ant to the same unit. Since the table is 1 meter across and the ant is 1 centimeter in length, we need to convert centimeters to meters.

There are 100 centimeters in 1 meter, so the ant is 1/100 = 0.01 meters in length. Now, we can determine how many ants would fit across the length of the table by dividing the length of the table by the length of the ant: 1 meter / 0.01 meters = 100 ants.

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Answer:

probability of getting a tail = 1/2

probability of getting 3 Tails = 1/2 × 1/2 × 1/2

= 1/8

The sum of the alternating subtraction and addition sequence 300-297+294-291+...+6-3 can be calculated by pairing the numbers and determining the partial sums, which consistently equal 3. With 49 complete sets of these operations within the range given, the total sum is 147.

The sequence provided is a series of alternating subtractions and additions of integers in a specific pattern. To determine the sum of such a sequence, we can pair the numbers in such a way that each subtractive pair (e.g., 300-297) is combined with its subsequent additive pair (e.g., 294). This will result in a series of partial sums which are consistently 3 (e.g., 300-297+294 equals 297). As the pattern continues, each group of four numbers adds up to 3.

Since all groups add up to 3, and we know that each set of two operations (a subtraction followed by an addition) gives us 3, we need to find out how many such sets there are. Given that our numbering starts with 300 and ends with 3 (decreasing by 3 every time), we can see there are 99 numbers in the sequence (since (300 - 3) / 3 + 1 = 99). And because every four numbers make up one set, there would be 99 / 2 = 49.5 sets. We only consider the entire sets, which would be 49 in total, ignoring the last number (3) that does not form a complete set.

Therefore, multiplying 49 sets by 3 gives us the total sum of the pattern, which would be 147. It is important to double-check calculations to ensure accuracy when dealing with sequences and series.