The arithmetic sequence a; is defined by the formula: a1=4 +3(i-1)
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Answer/Step-by-sep explanation:
To determine whether the lines given in each box are parallel, perpendicular, or neither, take the following simple steps:
1. Ensure the equations for both lines being compared are in the slope-intercept form, y = mx + b. Where m is the slope.
2. If both lines have the same slope value, m, then both lines are parallel.
3. If the slope of one line is the negative reciprocal of the other, then both lines are perpendicular. That is, x = -1/x.
4. If the slope of both lines are not the same, nor the negative reciprocal of each other, then they are neither parallel nor perpendicular.
1. y = 3x - 7 and y = 3x + 1.
Both have the same slope value of 3. Therefore, they are parallel.
2. ⬜ and
The slope of both lines are not the same, nor is the slope of one the negative reciprocal of the other. The slope of one is -⅖ and the slope of the other is ⅖. Therefore, they are neither parallel nor perpendicular.⬜
3. and
The slope of the first line, ¼, is the negative reciprocal of the slope of the second line, 4.
Therefore, they are perdendicular.
4. 2x + 7y = 28 and 7x - 2y = 4.
Rewrite both equations in the slope-intercept form, y = mx + b.
2x + 7y = 28
7y = -2x + 28
y = -2x/7 + 28/7
y = -²/7 + 4
And
7x - 2y = 4
-2y = -7x + 4
y = -7x/-2 + 4/-2
y = ⁷/2x - 2
The slope of the first line, -²/7, is the negative reciprocal of the slope of the second line, ⁷/2.
Therefore, they are perdendicular.
5.⬜ y = -5x + 1 and x - 5y = 30.
Rewrite the second line equation in the slope-intercept form.
x - 5y = 30
-5y = -x + 30
y = -2x/-5 + 30/-5
y = ⅖x - 6
The slope of both lines are not the same, nor is the slope of one the negative reciprocal of the other. The slope of one is -5 and the slope of the other is ⅖. therefore, they are neither parallel nor perpendicular.⬜
6.⬜ 3x + 2y = 8 and 2x + 3y = -12.
Rewrite both line equations in the slope-intercept form.
3x + 2y = 8
2y = -3x + 8
y = -3x/2 + 8/2
y = -³/2x + 4
And
2x + 3y = -12
3y = -2x -12
y = -2x/3 - 12/3
y = -⅔x - 4
The slope of both lines are not the same, nor is the slope of one the negative reciprocal of the other. The slope of one is -³/2 and the slope of the other is -⅔ therefore, they are neither parallel nor perpendicular.⬜
7. y = -4x - 1 and 8x + 2y = 14.
Rewrite the equation of the second line in the slope-intercept form.
8x + 2y = 14
2y = -8x + 14
y = -8x/2 + 14/2
y = -4x + 7
Both have the same slope value of -4. Therefore, they are parallel.
8.⬜ x + y = 7 and x - y = 9.
Rewrite the equation of both lines in the slope-intercept form.
x + y = 7
y = -x + 7
And
x - y = 9
-y = -x + 9
y = -x/-1 + 9/-1
y = x - 9
The slope of both lines are not the same, nor is the slope of one the negative reciprocal of the other. The slope of one is -1, and the slope of the other is 1, therefore, they are neither parallel nor perpendicular.⬜
9. y = ⅓x + 9 And x - 3y = 3
Rewrite the equation of the second line.
x - 3y = 3
-3y = -x + 3
y = -x/-3 + 3/-3
y = ⅓x - 1
Both have the same slope value of ⅓. Therefore, they are parallel.
10.⬜ 4x + 9y = 18 and y = 4x + 9
Rewrite the equation of the first line.
4x + 9y = 18
9y = -4x + 18
y = -4x/9 + 18/9
y = -⁴/9x + 2
The slope of both lines are not the same, nor is the slope of one the negative reciprocal of the other. The slope of one is -⁴/9, and the slope of the other is 4, therefore, they are neither parallel nor perpendicular.⬜
11.⬜ 5x - 10y = 20 and y = -2x + 6
Rewrite the equation of the first line.
5x - 10y = 20
-10y = -5x + 20
y = -5x/-10 + 20/-10
y = ²/5x - 2
The slope of both lines are not the same, nor is the slope of one the negative reciprocal of the other. The slope of one is ⅖, and the slope of the other is -2, therefore, they are neither parallel nor perpendicular.⬜
12. -9x + 12y = 24 and y = ¾x - 5
Rewrite the equation of the first line.
-9x + 12y = 24
12y = 9x + 24
y = 9x/12 + 24/12
y = ¾x + 2
Both have the same slope value of ¾. Therefore, they are parallel.
The first thing you should do when dealing with implicit derivatives is to respect the rules of derivation of both the logarithm and the exponential
Then, you must regroup the terms correctly until you get dy / dx
The answer for this case is D
I attach the solution
Then, you must regroup the terms correctly until you get dy / dx
The answer for this case is D
I attach the solution