Respuesta :
Question 1:
The generic equation of the line is given by:
[tex] y = mx + b [/tex]
Where,
m: slope of the line
b: cutting point with the y axis
Substituting values we have:
[tex]y = \frac{2}3}x + 9 [/tex]
Answer:
an equation for the line with slope 2/3 and y-intercept is:
[tex]y = \frac{2}3}x + 9 [/tex]
Question 2:
The standard equation of the line is given by:
[tex]y-yo = m (x-xo) [/tex]
Where,
m: slope of the line
(xo, yo): ordered pair that belongs to the line
The slope of the line is:
[tex]m = \frac{y2-y1}{x2-x1} [/tex]
Substituting values:
[tex]m = \frac{1-(-3)}{3-1} [/tex]
[tex]m = \frac{1+3}{3-1} [/tex]
[tex]m = \frac{4}{2} [/tex]
[tex]m = 2 [/tex]
We choose an ordered pair:
[tex](xo, yo) = (3, 1) [/tex]
Substituting values:
[tex]y-1 = 2 (x-3) [/tex]
Rewriting:
[tex]y = 2x - 6 + 1 y = 2x - 5[/tex]
Answer:
An equation in slope-intercept form for the line that passes through the points (1, -3) and (3,1) is:
[tex] y = 2x - 5 [/tex]
Question 3:
For this case, since the variation is direct, then we have an equation of the form:
[tex]y = kx [/tex] Where,
k: constant of variation.
We then have the following equation:
[tex]-4y = 8x [/tex] Rewriting we have:
[tex]y = \frac{8}{-4}x [/tex]
[tex] y = -2x [/tex] Therefore, the constate of variation is given by:
[tex]k = -2 [/tex]
Answer:
the constant of variation is:
[tex]k = -2 [/tex]
Question 4:
For this case, since the variation is direct, then we have an equation of the form:
[tex]y = kx [/tex] Where,
k: constant of variation.
We must find the constant k, for this we use the following data:
y = 24 when x = 8
Substituting values:
[tex] 24 = k8 [/tex] Clearing k we have:
[tex]k = \frac{24}{8} [/tex]
[tex]k = 3 [/tex]
Therefore, the equation is:
[tex]y = 3x [/tex] Thus, substituting x = 10 we have that the value of y is given by:
[tex]y = 3 (10) y = 30[/tex] Answer:
the value of y when x = 10 is:
[tex] y = 30[/tex]
The generic equation of the line is given by:
[tex] y = mx + b [/tex]
Where,
m: slope of the line
b: cutting point with the y axis
Substituting values we have:
[tex]y = \frac{2}3}x + 9 [/tex]
Answer:
an equation for the line with slope 2/3 and y-intercept is:
[tex]y = \frac{2}3}x + 9 [/tex]
Question 2:
The standard equation of the line is given by:
[tex]y-yo = m (x-xo) [/tex]
Where,
m: slope of the line
(xo, yo): ordered pair that belongs to the line
The slope of the line is:
[tex]m = \frac{y2-y1}{x2-x1} [/tex]
Substituting values:
[tex]m = \frac{1-(-3)}{3-1} [/tex]
[tex]m = \frac{1+3}{3-1} [/tex]
[tex]m = \frac{4}{2} [/tex]
[tex]m = 2 [/tex]
We choose an ordered pair:
[tex](xo, yo) = (3, 1) [/tex]
Substituting values:
[tex]y-1 = 2 (x-3) [/tex]
Rewriting:
[tex]y = 2x - 6 + 1 y = 2x - 5[/tex]
Answer:
An equation in slope-intercept form for the line that passes through the points (1, -3) and (3,1) is:
[tex] y = 2x - 5 [/tex]
Question 3:
For this case, since the variation is direct, then we have an equation of the form:
[tex]y = kx [/tex] Where,
k: constant of variation.
We then have the following equation:
[tex]-4y = 8x [/tex] Rewriting we have:
[tex]y = \frac{8}{-4}x [/tex]
[tex] y = -2x [/tex] Therefore, the constate of variation is given by:
[tex]k = -2 [/tex]
Answer:
the constant of variation is:
[tex]k = -2 [/tex]
Question 4:
For this case, since the variation is direct, then we have an equation of the form:
[tex]y = kx [/tex] Where,
k: constant of variation.
We must find the constant k, for this we use the following data:
y = 24 when x = 8
Substituting values:
[tex] 24 = k8 [/tex] Clearing k we have:
[tex]k = \frac{24}{8} [/tex]
[tex]k = 3 [/tex]
Therefore, the equation is:
[tex]y = 3x [/tex] Thus, substituting x = 10 we have that the value of y is given by:
[tex]y = 3 (10) y = 30[/tex] Answer:
the value of y when x = 10 is:
[tex] y = 30[/tex]
(1) The equation of straight line for slope [tex]\frac{2}{3}[/tex] and y-intercept [tex]9[/tex] is [tex]\boxed{y=\dfrac{2}{3}x+9}[/tex].
(2) The equation of line that passes through the points [tex](1,-3)[/tex] and [tex](3,1)[/tex] is [tex]\boxed{y=2x-5}[/tex].
(3) The constant of variation for line [tex]-4y=8x[/tex] is [tex]\boxed{k=-2}[/tex].
(4) The value of [tex]y[/tex] is [tex]\boxed{20}[/tex].
Further explanation:
Part (1)
Concept used:
The equation of straight line for slope [tex]m[/tex] and intercept [tex]c[/tex] can be expressed as,
[tex]\boxed{y=mx+c}[/tex] ....(1)
Here, [tex]c[/tex] is the [tex]y[/tex]-intercept of the straight line.
Given:
The slope of the line is [tex]\frac{2}{3}[/tex] and [tex]y[/tex]-intercept is [tex]9[/tex].
Calculation:
Substitute [tex]m=\frac{2}{3}[/tex] and [tex]c=9[/tex] in the equation (1) to obtain the equation of straight line as follows,
[tex]y=\dfrac{2}{3}x+9[/tex]
Therefore, the equation of straight line for slope [tex]\frac{2}{3}[/tex] and [tex]y[/tex]-intercept [tex]9[/tex] is [tex]y=\frac{2}{3}x+9[/tex].
Part (2)
Concept used:
The equation of straight line that passes through the point [tex](x_{1},y_{1})[/tex] and [tex](x_{2},y_{2})[/tex] is expressed as,
[tex]\boxed{y-y_{1}=\dfrac{y_{2}-y_{1}}{x_{2}-x_{2}}(x-x_{1})}[/tex] ....(2)
Here, [tex]\frac{y_{2}-y_{1}}{x_{2}-x_{1}}[/tex] is the slope [tex](m)[/tex] of the line.
Now, the equation of line in the point slope form is,
[tex]\boxed{y-y_{1}=m(x-x_{1})}[/tex]
Calculation:
The equation passes through the point [tex](1,-3)[/tex] and [tex](3,1)[/tex].
Now, substitute [tex]1[/tex] for [tex]x_{1}[/tex], [tex]-3[/tex] for [tex]y_{1}[/tex],[tex]3[/tex] for [tex]x_{2}[/tex] and [tex]1[/tex] for [tex]y_{2}[/tex] in the equation (2) to obtain the equation of line.
[tex]\begin{aligned}y-(-3)&=\dfrac{1-(-3)}{3-1}(x-1)\\y+3&=\dfrac{4}{2}(x-1)\\y+3&=2(x-1)\\y&=2x-2-3\\y&=2x-5\end{aligned}[/tex]
Therefore, the equation in slope intercepts form for line that passes through the points [tex](1,-3)[/tex] and [tex](3,1)[/tex] is [tex]y=2x-5[/tex].
Part (3)
Given:
The equation of line is [tex]-4y=8x[/tex].
Concept used:
If the value of [tex]y[/tex] varies directly with the value of [tex]x[/tex] that means as the value of [tex]x[/tex] increases, [tex]y[/tex] increases in the same ratio.
[tex]\boxed{y=kx}[/tex]
Here, [tex]k[/tex] is the constant of variation.
Calculation:
The given equation [tex]-4y=8x[/tex] can be simplified as,
[tex]\begin{aligned}y&=-\dfrac{8x}{4}\\y&=-2x\end{aligned}[/tex]
Compare the equation [tex]y=-2x[/tex] with the equation (3) to obtain the value of [tex]k[/tex].
[tex]\boxed{k=-2}[/tex]
Therefore, the constant of variation for the given equation is [tex]k=-2[/tex].
Part (4)
Calculation:
The value of [tex]y[/tex] is [tex]24[/tex] when [tex]x[/tex] is [tex]8[/tex]. The value of [tex]y[/tex] varies directly with [tex]x[/tex].
The constant of variation [tex]k[/tex] is calculated as,
[tex]\begin{aligned}k&=\dfrac{24}{8}\\&=3\end{aligned}[/tex]
The constant of variation [tex]k[/tex] is [tex]3[/tex].
The relationship between [tex]x[/tex] and [tex]y[/tex] can be written as,
[tex]\boxed{y=kx}[/tex] ......(4)
Substitute [tex]10[/tex] for [tex]x[/tex] and [tex]3[/tex] for [tex]k[/tex] in the equation (4) to obtain the value of [tex]y[/tex].
[tex]\begin{aligned}y&=3\times10\\&=30\end{aligned}[/tex]
Therefore, the value of [tex]y[/tex] is [tex]30[/tex].
Learn more:
1. A problem on line https://brainly.com/question/1473992
2. A problem on simplification https://brainly.com/question/3658196
3. A problem on permutation https://brainly.com/question/5199020
Answer details:
Grade: Middle school
Subject: Mathematics
Chapter: Coordinate geometry
Keywords: Slope, equation of line, constant of variation, y=2/3x+9, points, intercepts, straight line, slope-intercept form.
