What is the slope of the line that passes through the given points? (2,12) and (6,11)

( – 1, – 5) 7x+2y=13, 4x+4y=14

serious [3.7K]

Answer:

{x,y} = {6/5,23/10}

Step-by-step explanation:

[1] 7x + 2y = 13

[2] 4x + 4y = 14 <---------- linear equations given

Graphic Representation of the Equations : PICTURE
2y + 7x = 13 4y + 4x = 14

Solve by Substitution :

// Solve equation [2] for the variable y

[2] 4y = -4x + 14

[2] y = -x + 7/2

// Plug this in for variable y in equation [1]

[1] 7x + 2•(-x +7/2) = 13

[1] 5x = 6

// Solve equation [1] for the variable x

[1] 5x = 6

[1] x = 6/5

// By now we know this much :

x = 6/5

y = -x+7/2

// Use the x value to solve for y

y = -(6/5)+7/2 = 23/10

// Plug this in for variable y in equation [1]

[1] 7x + 2•(-x +7/2) = 13

[1] 5x = 6

// Solve equation [1] for the variable x

[1] 5x = 6

[1] x = 6/5

// By now we know this much :

x = 6/5

y = -x+7/2

// Use the x value to solve for y

y = -(6/5)+7/2 = 23/10

4 0

2 years ago

Which equation matches the graph of the absolute value function seen here?

nignag [31]

use a calculator instead of asking

7 0

3 years ago

It's timed plzzz help!!!

Novay_Z [31]

Answer:

Last Choice / D

Step-by-step explanation:

you want to isolate v!

E = 1/2 m v^2

multiply both sides by 2

2E = mv^2

divide both sides by m

2E/m = v^2

square root both sides

v = ± $\sqrt{2E/m}$

please click heart to give thanks :)

3 0

3 years ago

Read 2 more answers

6/(x+1)-5/2=6/(3x+3)

jolli1 [7]

If you would like to calculate 6/(x+1)-5/2=6/(3x+3), you can do this using the following steps:

6/(x+1)-5/2=6/(3x+3)
6/(x+1)-5/2=6/(3(x+1)) /*(x+1)
6 - 5/2 * (x+1) = 6/3
6 - 2 = 5/2 * (x+1)
4 = 5/2 * (x+1) /*2/5
4 * 2/5 = x + 1
8/5 - 1 = x
x = 8/5 - 5/5 = 3/5

The correct result would be 3/5.

6 0

3 years ago

The focus of a parabola is (−5,−1) and the directrix is y=−3. What is an equation of the parabola? (x+5)2=8(y+5)

Anna35 [415]

Graph x2 = 4y and state the vertex, focus, axis of symmetry, and directrix.This is the same graphing that I've done in the past: y = (1/4)x2. So I'll do the graph as usual: The vertex is obviously at the origin, but I need to "show" this "algebraically" by rearranging the given equation into the conics form:x2 = 4y Copyright © Elizabeth Stapel 2010-2011 All Rights Reserved
(x – 0)2 = 4(y – 0)This rearrangement "shows" that the vertex is at (h, k) = (0, 0). The axis of symmetry is the vertical line right through the vertex: x = 0. (I can always check my graph, if I'm not sure about this.) The focus is "p" units from the vertex. Since the focus is "inside" the parabola and since this is a "right side up" graph, the focus has to be above the vertex.From the conics form of the equation, shown above, I look at what's multiplied on the unsquaredpart and see that 4p = 4, so p = 1. Then the focus is one unit above the vertex, at (0, 1), and the directrix is the horizontal line y = –1, one unit below the vertex.vertex: (0, 0); focus: (0, 1); axis of symmetry: x = 0; directrix: y = –1Graph y2 + 10y + x + 25 = 0, and state the vertex, focus, axis of symmetry, and directrix.Since the y is squared in this equation, rather than the x, then this is a "sideways" parabola. To graph, I'll do my T-chart backwards, picking y-values first and then finding the corresponding x-values for x = –y2 – 10y – 25:To convert the equation into conics form and find the exact vertex, etc, I'll need to convert the equation to perfect-square form. In this case, the squared side is already a perfect square, so:y2 + 10y + 25 = –x 
(y + 5)2 = –1(x – 0)This tells me that 4p = –1, so p = –1/4. Since the parabola opens to the left, then the focus is 1/4 units to the left of the vertex. I can see from the equation above that the vertex is at (h, k) = (0, –5), so then the focus must be at (–1/4, –5). The parabola is sideways, so the axis of symmetry is, too. The directrix, being perpendicular to the axis of symmetry, is then vertical, and is 1/4 units to the right of the vertex. Putting this all together, I get:vertex: (0, –5); focus: (–1/4, –5); axis of symmetry: y = –5; directrix: x = 1/4Find the vertex and focus of y2 + 6y + 12x – 15 = 0The y part is squared, so this is a sideways parabola. I'll get the y stuff by itself on one side of the equation, and then complete the square to convert this to conics form.y2 + 6y – 15 = –12x 
y2 + 6y + 9 – 15 = –12x + 9 
(y + 3)2 – 15 = –12x + 9 
(y + 3)2 = –12x + 9 + 15 = –12x + 24 
(y + 3)2 = –12(x – 2) 
(y – (–3))2 = 4(–3)(x – 2)Then the vertex is at (h, k) = (2, –3) and the value of p is –3. Since y is squared and p is negative, then this is a sideways parabola that opens to the left. This puts the focus 3 units to the left of the vertex.vertex: (2, –3); focus: (–1, –3)

8 0

3 years ago