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A line that passes through the point [tex]$(x, y)$[/tex], with a [tex]$y$[/tex]-intercept of [tex]$b$[/tex] and a slope of [tex]$m$[/tex], can be represented by the equation [tex]$y=mx+b$[/tex].

A line is drawn on the coordinate plane that passes through the point [tex]$(10,1)$[/tex] and has a slope of [tex]$-0.5$[/tex]. The [tex]$y$[/tex]-intercept of the line is [tex]$\square$[/tex].



Answer :

GinnyAnswer
To find the [tex]\( y \)[/tex]-intercept ([tex]\( b \)[/tex]) of the line that passes through the point [tex]\((10, 1)\)[/tex] with a slope ([tex]\( m \)[/tex]) of [tex]\(-0.5\)[/tex], we need to use the equation of the line in slope-intercept form:

[tex]\[ y = mx + b \][/tex]

Given:
- [tex]\( x = 10 \)[/tex]
- [tex]\( y = 1 \)[/tex]
- [tex]\( m = -0.5 \)[/tex]

We can substitute these values into the equation to solve for [tex]\( b \)[/tex]:

[tex]\[ 1 = (-0.5)(10) + b \][/tex]

Simplify the right side:

[tex]\[ 1 = -5 + b \][/tex]

To isolate [tex]\( b \)[/tex], add 5 to both sides:

[tex]\[ 1 + 5 = b \][/tex]

Thus:

[tex]\[ b = 6 \][/tex]

So, the [tex]\( y \)[/tex]-intercept of the line is [tex]\( \boxed{6} \)[/tex].

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