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Sure, let's go through each question step-by-step.
1. Find the slope of a line passing through the points [tex]\((1,2)\)[/tex] and [tex]\((3,4)\)[/tex]:
The formula for the slope [tex]\( m \)[/tex] between two points [tex]\((x_1, y_1)\)[/tex] and [tex]\((x_2, y_2)\)[/tex] is given by:
[tex]\[ m = \frac{y_2 - y_1}{x_2 - x_1} \][/tex]
For the points [tex]\((1, 2)\)[/tex] and [tex]\((3, 4)\)[/tex]:
[tex]\[ m = \frac{4 - 2}{3 - 1} = \frac{2}{2} = 1 \][/tex]
Therefore, the slope is [tex]\( 1 \)[/tex].
2. Find the slope of the line passing through the points [tex]\((4,-2)\)[/tex] and [tex]\((-1,7)\)[/tex]:
Using the same slope formula:
[tex]\[ m = \frac{7 - (-2)}{-1 - 4} = \frac{7 + 2}{-1 - 4} = \frac{9}{-5} = -1.8 \][/tex]
Therefore, the slope is [tex]\( -1.8 \)[/tex].
3. Write the slope-intercept equation of the line that passes through the point [tex]\((0,0)\)[/tex] and is perpendicular to the line shown at right:
Assuming the given line has a slope of [tex]\( 1 \)[/tex], since perpendicular slopes are negative reciprocals:
[tex]\[ \text{Slope of perpendicular line} = -\frac{1}{1} = -1 \][/tex]
Since the line passes through [tex]\((0,0)\)[/tex], the y-intercept [tex]\( b \)[/tex] is [tex]\( 0 \)[/tex]. Therefore, the equation is:
[tex]\[ y = -1x + 0 \quad \text{or simply} \quad y = -1x \][/tex]
4. Write the slope-intercept equation of the line that passes through the point [tex]\((1,-2)\)[/tex] and is parallel to the line [tex]\(y = \frac{4}{3} x + 11\)[/tex]:
Since parallel lines have the same slope, the slope [tex]\( m \)[/tex] is [tex]\( \frac{4}{3} \)[/tex]. Using the point-slope form [tex]\( y = mx + b \)[/tex]:
[tex]\[ -2 = \frac{4}{3}(1) + b \][/tex]
Solving for [tex]\( b \)[/tex]:
[tex]\[ -2 = \frac{4}{3} + b \implies b = -2 - \frac{4}{3} = -\frac{6}{3} - \frac{4}{3} = -\frac{10}{3} \][/tex]
Therefore, the equation is:
[tex]\[ y = \frac{4}{3}x - \frac{10}{3} \][/tex]
5. Solve the system of equations: [tex]\(\left\{\begin{array}{c} 3 x+4 y=-6 \\ 5 x+3 y=1 \end{array}\right.\)[/tex]
Using the system:
[tex]\[ \begin{cases} 3x + 4y = -6 \\ 5x + 3y = 1 \end{cases} \][/tex]
Solving this system, we get:
[tex]\[ x = 2, \quad y = -3 \][/tex]
Therefore, the solution is [tex]\((2, -3)\)[/tex].
6. A system of equations in the [tex]\(xy\)[/tex]-plane has a single solution [tex]\((x, y)\)[/tex]. If the system is given by the equations [tex]\(\left\{\begin{array}{c} 2 x+y=k \\ x-y=3-k \end{array}\right.\)[/tex] for some unknown value [tex]\(k\)[/tex], what is the value of the [tex]\(x\)[/tex]-coordinate of the solution of the system?
Solving this system of equations:
[tex]\[ x = k = 1 \][/tex]
Therefore, the [tex]\( x \)[/tex]-coordinate of the solution is [tex]\( 1 \)[/tex].
7. Solve: [tex]\(\frac{3}{5}x + 4 \leq 2\)[/tex]
Isolate [tex]\( x \)[/tex]:
[tex]\[ \frac{3}{5}x + 4 = 2 \implies \frac{3}{5}x = -2 \implies x = -\frac{10}{3} \approx -3.333 \][/tex]
Therefore, the solution set is:
[tex]\[ \{-3.333\} \][/tex]
8. Solve: [tex]\(3(x+3)>5(x-1)\)[/tex]
Expand and solve:
[tex]\[ 3x + 9 > 5x - 5 \implies 3x + 9 - 5x > -5 \implies -2x > -14 \implies x < 7 \][/tex]
Therefore, the solution set is:
[tex]\[ (-\infty, 7) \][/tex]
9. Solve: [tex]\(2 \leq 3x - 10 \leq 5\)[/tex]
Split it into two inequalities:
[tex]\[ 2 \leq 3x - 10 \implies 12 \leq 3x \implies 4 \leq x \][/tex]
[tex]\[ 3x - 10 \leq 5 \implies 3x \leq 15 \implies x \leq 5 \][/tex]
Therefore, combining both:
[tex]\[ 4 \leq x \leq 5 \][/tex]
Interval solution is:
[tex]\[ [4, 5] \][/tex]
However, since we can't find [tex]\( x \)[/tex] that satisfies both inequalities, we get:
[tex]\[ \text{Empty Set} \][/tex]
1. Find the slope of a line passing through the points [tex]\((1,2)\)[/tex] and [tex]\((3,4)\)[/tex]:
The formula for the slope [tex]\( m \)[/tex] between two points [tex]\((x_1, y_1)\)[/tex] and [tex]\((x_2, y_2)\)[/tex] is given by:
[tex]\[ m = \frac{y_2 - y_1}{x_2 - x_1} \][/tex]
For the points [tex]\((1, 2)\)[/tex] and [tex]\((3, 4)\)[/tex]:
[tex]\[ m = \frac{4 - 2}{3 - 1} = \frac{2}{2} = 1 \][/tex]
Therefore, the slope is [tex]\( 1 \)[/tex].
2. Find the slope of the line passing through the points [tex]\((4,-2)\)[/tex] and [tex]\((-1,7)\)[/tex]:
Using the same slope formula:
[tex]\[ m = \frac{7 - (-2)}{-1 - 4} = \frac{7 + 2}{-1 - 4} = \frac{9}{-5} = -1.8 \][/tex]
Therefore, the slope is [tex]\( -1.8 \)[/tex].
3. Write the slope-intercept equation of the line that passes through the point [tex]\((0,0)\)[/tex] and is perpendicular to the line shown at right:
Assuming the given line has a slope of [tex]\( 1 \)[/tex], since perpendicular slopes are negative reciprocals:
[tex]\[ \text{Slope of perpendicular line} = -\frac{1}{1} = -1 \][/tex]
Since the line passes through [tex]\((0,0)\)[/tex], the y-intercept [tex]\( b \)[/tex] is [tex]\( 0 \)[/tex]. Therefore, the equation is:
[tex]\[ y = -1x + 0 \quad \text{or simply} \quad y = -1x \][/tex]
4. Write the slope-intercept equation of the line that passes through the point [tex]\((1,-2)\)[/tex] and is parallel to the line [tex]\(y = \frac{4}{3} x + 11\)[/tex]:
Since parallel lines have the same slope, the slope [tex]\( m \)[/tex] is [tex]\( \frac{4}{3} \)[/tex]. Using the point-slope form [tex]\( y = mx + b \)[/tex]:
[tex]\[ -2 = \frac{4}{3}(1) + b \][/tex]
Solving for [tex]\( b \)[/tex]:
[tex]\[ -2 = \frac{4}{3} + b \implies b = -2 - \frac{4}{3} = -\frac{6}{3} - \frac{4}{3} = -\frac{10}{3} \][/tex]
Therefore, the equation is:
[tex]\[ y = \frac{4}{3}x - \frac{10}{3} \][/tex]
5. Solve the system of equations: [tex]\(\left\{\begin{array}{c} 3 x+4 y=-6 \\ 5 x+3 y=1 \end{array}\right.\)[/tex]
Using the system:
[tex]\[ \begin{cases} 3x + 4y = -6 \\ 5x + 3y = 1 \end{cases} \][/tex]
Solving this system, we get:
[tex]\[ x = 2, \quad y = -3 \][/tex]
Therefore, the solution is [tex]\((2, -3)\)[/tex].
6. A system of equations in the [tex]\(xy\)[/tex]-plane has a single solution [tex]\((x, y)\)[/tex]. If the system is given by the equations [tex]\(\left\{\begin{array}{c} 2 x+y=k \\ x-y=3-k \end{array}\right.\)[/tex] for some unknown value [tex]\(k\)[/tex], what is the value of the [tex]\(x\)[/tex]-coordinate of the solution of the system?
Solving this system of equations:
[tex]\[ x = k = 1 \][/tex]
Therefore, the [tex]\( x \)[/tex]-coordinate of the solution is [tex]\( 1 \)[/tex].
7. Solve: [tex]\(\frac{3}{5}x + 4 \leq 2\)[/tex]
Isolate [tex]\( x \)[/tex]:
[tex]\[ \frac{3}{5}x + 4 = 2 \implies \frac{3}{5}x = -2 \implies x = -\frac{10}{3} \approx -3.333 \][/tex]
Therefore, the solution set is:
[tex]\[ \{-3.333\} \][/tex]
8. Solve: [tex]\(3(x+3)>5(x-1)\)[/tex]
Expand and solve:
[tex]\[ 3x + 9 > 5x - 5 \implies 3x + 9 - 5x > -5 \implies -2x > -14 \implies x < 7 \][/tex]
Therefore, the solution set is:
[tex]\[ (-\infty, 7) \][/tex]
9. Solve: [tex]\(2 \leq 3x - 10 \leq 5\)[/tex]
Split it into two inequalities:
[tex]\[ 2 \leq 3x - 10 \implies 12 \leq 3x \implies 4 \leq x \][/tex]
[tex]\[ 3x - 10 \leq 5 \implies 3x \leq 15 \implies x \leq 5 \][/tex]
Therefore, combining both:
[tex]\[ 4 \leq x \leq 5 \][/tex]
Interval solution is:
[tex]\[ [4, 5] \][/tex]
However, since we can't find [tex]\( x \)[/tex] that satisfies both inequalities, we get:
[tex]\[ \text{Empty Set} \][/tex]
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