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2 years ago

What is the length of the line segment between A(–5, 8) and B(7, 8)? (1 point) • 2 units • 3 units • 12 units • 13 units

Mathematics

1 answer:

tensa zangetsu [6.8K]2 years ago

3 0

Answer:

12 units

Step-by-step explanation:

B - A = (7, 8) - (-5, 8) = (7 -(-5), 8 -8) = (12, 0)

The distance is then ...

d = √(12² + 0²) = √12²

d = 12

_____

Of course, when you graph these points (actually or mentally), you can see that they are 12 units apart on the line y=8. You don't need to do any fancy calculation to discover the difference between -5 and 7.

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If the graph of the function h is defined by = h(x)+x^2+9 is translated vertically downward by 9 units, it becomes the graph of

natulia [17]

The answer is g(x) = x².
Solution:
The graph of h(x) = x²+9 translated vertically downward by 9 units means that each point (x, h(x)) is shifted onto the point (x, h(x) - 9), that is,
(x, h(x)) → (x, h(x) - 9)
The translated graph that represents the function is defined by the expression for g(x):
g(x) = h(x) - 9 = x² + 9 - 9 = x²

h(x) = x²+9 → g(x) = x² shows that the graph of the equation g(x) = x² moves the graph of h(x) = x²+9 down nine units.

6 0

3 years ago

Help me I want to understand this finnall part

allochka39001 [22]

6.7 x -2.3
You would multiply this just like any other multiplication question.
There is some work of course, but I'm not going to write it out here...
You are going to end up with -15.41, which is already rounded to the nearest hundredth.

Hope this helped :)

4 0

3 years ago

Read 2 more answers

If -3x–7 =23 then 6x=?

vovangra [49]

Answer:

6x = -60

Step-by-step explanation:

-3x - 7 = 23

-3x = 30

x = -10

6(-10) = -60

8 0

3 years ago

Learning Task 3. Find the equation of the line. Do it in your notebook.

Wewaii [24]

Answer:

1) The equation of the line in slope-intercept form is $y = 5\cdot x +9$ . The equation of the line in standard form is $-5\cdot x + y = 9$ .

2) The equation of the line in slope-intercept form is $y = \frac{2}{5}\cdot x +\frac{14}{5}$ . The equation of the line in standard form is $-2\cdot x +5\cdot y = 14$ .

3) The equation of the line in slope-intercept form is $y = 3\cdot x +4$ . The equation of the line in standard form is $-3\cdot x +y = 4$ .

4) The equation of the line in slope-intercept form is $y = 2\cdot x + 6$ . The equation of the line in standard form is $-2\cdot x +y = 6$ .

5) The equation of the line in slope-intercept form is $y = \frac{5}{6}\cdot x -\frac{7}{6}$ . The equation of the line in standard from is $-5\cdot x + 6\cdot y = -7$ .

Step-by-step explanation:

1) We begin with the slope-intercept form and substitute all known values and calculate the y-intercept: ( $m = 5$ , $x = -1$ , $y = 4$ )

$4 = (5)\cdot (-1)+b$

$4 = -5 +b$

$b = 9$

The equation of the line in slope-intercept form is $y = 5\cdot x +9$ .

Then, we obtain the standard form by algebraic handling:

$-5\cdot x + y = 9$

The equation of the line in standard form is $-5\cdot x + y = 9$ .

2) We begin with a system of linear equations based on the slope-intercept form: ( $x_{1} = 3$ , $y_{1} = 4$ , $x_{2} = -2$ , $y_{2} = 2$ )

$3\cdot m + b = 4$ (Eq. 1)

$-2\cdot m + b = 2$ (Eq. 2)

From (Eq. 1), we find that:

$b = 4-3\cdot m$

And by substituting on (Eq. 2), we conclude that slope of the equation of the line is:

$-2\cdot m +4-3\cdot m = 2$

$-5\cdot m = -2$

$m = \frac{2}{5}$

And from (Eq. 1) we find that the y-Intercept is:

$b=4-3\cdot \left(\frac{2}{5} \right)$

$b = 4-\frac{6}{5}$

$b = \frac{14}{5}$

The equation of the line in slope-intercept form is $y = \frac{2}{5}\cdot x +\frac{14}{5}$ .

Then, we obtain the standard form by algebraic handling:

$-\frac{2}{5}\cdot x +y = \frac{14}{5}$

$-2\cdot x +5\cdot y = 14$

The equation of the line in standard form is $-2\cdot x +5\cdot y = 14$ .

3) By using the slope-intercept form, we obtain the equation of the line by direct substitution: ( $m = 3$ , $b = 4$ )

$y = 3\cdot x +4$

The equation of the line in slope-intercept form is $y = 3\cdot x +4$ .

Then, we obtain the standard form by algebraic handling:

$-3\cdot x +y = 4$

The equation of the line in standard form is $-3\cdot x +y = 4$ .

4) We begin with a system of linear equations based on the slope-intercept form: ( $x_{1} = -3$ , $y_{1} = 0$ , $x_{2} = 0$ , $y_{2} = 6$ )

$-3\cdot m + b = 0$ (Eq. 3)

$b = 6$ (Eq. 4)

By applying (Eq. 4) on (Eq. 3), we find that the slope of the equation of the line is:

$-3\cdot m+6 = 0$

$3\cdot m = 6$

$m = 2$

The equation of the line in slope-intercept form is $y = 2\cdot x + 6$ .

Then, we obtain the standard form by algebraic handling:

$-2\cdot x +y = 6$

The equation of the line in standard form is $-2\cdot x +y = 6$ .

5) We begin with a system of linear equations based on the slope-intercept form: ( $x_{1} = -1$ , $y_{1} = -2$ , $x_{2} = 5$ , $y_{2} = 3$ )

$-m+b = -2$ (Eq. 5)

$5\cdot m +b = 3$ (Eq. 6)

From (Eq. 5), we find that:

$b = -2+m$

And by substituting on (Eq. 6), we conclude that slope of the equation of the line is:

$5\cdot m -2+m = 3$

$6\cdot m = 5$

$m = \frac{5}{6}$

And from (Eq. 5) we find that the y-Intercept is:

$b = -2+\frac{5}{6}$

$b = -\frac{7}{6}$

The equation of the line in slope-intercept form is $y = \frac{5}{6}\cdot x -\frac{7}{6}$ .

Then, we obtain the standard form by algebraic handling:

$-\frac{5}{6}\cdot x +y =-\frac{7}{6}$

$-5\cdot x + 6\cdot y = -7$

The equation of the line in standard from is $-5\cdot x + 6\cdot y = -7$ .

6 0

2 years ago

1. You are given the 3rd and 5th term of an arithmetic sequence. Describe in words how to determine the general term.

kozerog [31]

Step-by-step explanation:

1. In an arithmetic sequence, the general term can be written as

xₙ = y + d(a-1), where xₐ represents the ath term, y is the first value, and d is the common difference.

Given the third term and the fifth term, and knowing that the difference between each term is d, we can say that the 4th term is x₃+d and the fifth term is the fourth term plus d, or (x₃+d)+d =

x₃+2d. =x₅ Given x₃ and x₅, we can subtract x₃ from both sides to get

x₅-x₃ = 2d

divide by 2 to isolate d

(x₅-x₃)/2 = d

This lets us solve for d. Given d, we can say that

x₃ = y+d(2)

subtract 2*d from both sides to isolate the y

x₃ -2*d = y

Therefore, because we know x₃ and d at this point, we can solve for y, letting us plug y and d into our original equation of

xₙ = y + d(a-1)

Given the third and fifth term, with a common ratio of r, we can say that the fourth term is x₃ * r. Then, the fifth term is

x₃* r * r

= x₃*r² = x₅

divide both sides by x₃ to isolate the r²

x₅/x₃ = r²

square root both sides

√(x₅/x₃) = ±r

One thing that is important to note is that we don't know whether r is positive or negative. For example, if x₃ = 4 and x₅ = 16, regardless of whether r is equal to 2 or -2, 4*r² = 16. I will be assuming that r is positive for this question.

Given the common ratio, we can find x₆ as x₅ * r, x₇ as x₅*r², and all the way up to x₁₀ = x₅*r⁵. We don't know the general term, but can still find the tenth term of the sequence

8 0

3 years ago