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ValentinkaMS [17]
3 years ago
15

How do we create a table of values to help us graph linear equations?​

Mathematics
1 answer:
VashaNatasha [74]3 years ago
6 0

Let's say for example you wanted to graph the linear equation y = 2x+5

At minimum, we'll need 2 points to form this line.

To get any point of the form (x,y), we plug in a number for x to find a corresponding number for y.

For example, replace x with 0 to find that...

y = 2x+5\\\\y = 2(0)+5\\\\y = 0+5\\\\y = 5\\\\

Therefore, x = 0 leads to y = 5. So the point (0,5) is on the line.

Then we repeat this process for other x values. Let's try x = 1

y = 2x+5\\\\y = 2(1)+5\\\\y = 2+5\\\\y = 7\\\\

Telling us that (1,7) is also on this line. We have enough to graph this equation. Plot the two points (0,5) and (1,7). Then draw a straight line through them. Extend this line as far as you can to the left and right. The graph is shown below.

We can keep going to get other points like (2,9) and (3,11) and (4,13) and so on. A table is a way to keep track of the xy values.

\begin{array}{|c|c|}\cline{1-2}\text{x} & \text{y}\\ \cline{1-2}0 & 5\\ \cline{1-2}1 & 7\\ \cline{1-2}2 & 9\\ \cline{1-2}3 & 11\\ \cline{1-2}4 & 13\\ \cline{1-2}\end{array}

Each time x goes up by 1, y goes up by 2. This directly ties to the slope of 2/1 = 2.

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Answer:

Samia Suluhu HassanStep-by-step explanation:

8 0
3 years ago
What is the solution set to the inequality (4x-3)(2x-1)20?
Fynjy0 [20]

The given question is wrong.

Question:

What is the solution set to the inequality (4x – 3) (2x – 1) ≥ 0?

(A) \{x| x\leq 3\ \text {or} \ x\geq 1

(B) \{x| x\leq 2\ \text {or} \ x\geq \frac{4}{3}

(C) \{x| x\leq \frac{1}{2}\ \text {or} \ x\geq \frac{3}{4}

(D) \{x| x\leq \frac{-1}{2}\ \text {or} \ x\geq \frac{-3}{4}

Answer:

The solution set to the given inequality is \{x| x\leq \frac{1}{2}\ \text {or} \ x\geq \frac{3}{4}.

Solution:

Given expression is (4x – 3) (2x – 1) ≥ 0.

Let us take the expression is equal to zero.

(4x – 3) (2x – 1) = 0

By quadratic factor, If AB = 0, then A = 0 or B = 0.

(4x – 3) = 0 or (2x – 1) = 0

Let us take the first factor equal to zero.

⇒ 4x – 3 = 0

⇒ 4x = 3

$x=\frac{3}{4}

Now, take the second factor equal to zero.

⇒ 2x – 1 = 0

⇒ 2x = 1

$x=\frac{1}{2}

So, $x=\frac{1}{2},x=\frac{3}{4}.

Now, write it in the inequality to make the statement true.

$x\geq \frac{1}{2}\ \text{(or)}\ x\leq \frac{3}{4}

Option C is the correct answer.

The solution set to the given inequality is \{x| x\leq \frac{1}{2}\ \text {or} \ x\geq \frac{3}{4}.

6 0
4 years ago
How do i do this???????
Fittoniya [83]

Answer:

2y-4-6y-12=4

-4y=20

y= -5

Step-by-step explanation:

because (y+2)(y-2)=y^2-4

so we have the step to let the every part have the same denominator y^2-4

2(y-2)/(y+2)(y-2)  -6(y+2)/(y-2)(y+2)=4/(y-2)(y+2)

the denominators are the same so the equality .we just need the part over be the same 2y-4-6y-12=4

y= -5

3 0
3 years ago
Find the distance between the points (-1,-2) and (5,2). Simplified radical form.
frozen [14]

Answer:

2\sqrt{13}

Step-by-step explanation:

Using the distance formula,

\sqrt{( x_{2}-x_{1} )^2+(y_{2} -y_{1})^2 }

We just need to plug in the numbers.

So \sqrt{(-1-5)^2+(-2-2)^2}

Then we simplify it to

\sqrt{(-6)^2+(-4)^2} , to \sqrt{36+16}

Finally, we get

\sqrt{52}, which is 2\sqrt{13} when simplified.

3 0
3 years ago
van budgets 12 a day for groceries for weekdays and 15 a day for weekends. write an expression in simplest form to show his groc
Ksenya-84 [330]
The budget of Van for groceries for 1 week is:

12 $ (per a weekday) * 5 (weekdays) + 15 $(per a weekend day) *2 (weekend days) = 60$ +30$ = 90$

The budget of Van for w weeks is:

90 $ (per week) * w (weeks) = 90w dollars.


Answer: 90w $
6 0
3 years ago
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