Line s has an equation of y=-(7)/(10)x-1 Line t, which is parallel to line s includes the point (-6,5) What is the equation of line 1? __ slope-intercept form. Write the numbers in the equation as simplified proper fractions. __ __ er fractions, or integers.

To find the equation of line t, which is parallel to line s and passes through the point (-6, 5), we need to use the fact that parallel lines have the same slope.<br /><br />Given:<br />- The equation of line s is \( y = -\frac{7}{10}x - 1 \).<br />- The slope of line s is \( -\frac{7}{10} \).<br /><br />Since line t is parallel to line s, it will have the same slope, which is \( -\frac{7}{10} \).<br /><br />We need to find the y-intercept (b) of line t using the point (-6, 5). We can use the point-slope form of the equation of a line:<br /><br />\[ y - y_1 = m(x - x_1) \]<br /><br />Here, \( (x_1, y_1) = (-6, 5) \) and \( m = -\frac{7}{10} \).<br /><br />Substitute these values into the point-slope form:<br /><br />\[ y - 5 = -\frac{7}{10}(x + 6) \]<br /><br />Now, solve for y to put it in slope-intercept form (y = mx + b):<br /><br />\[ y - 5 = -\frac{7}{10}x - \frac{7}{10} \cdot 6 \]<br />\[ y - 5 = -\frac{7}{10}x - \frac{42}{10} \]<br />\[ y - 5 = -\frac{7}{10}x - 4.2 \]<br /><br />Add 5 to both sides to isolate y:<br /><br />\[ y = -\frac{7}{10}x - 4.2 + 5 \]<br />\[ y = -\frac{7}{10}x + 0.8 \]<br /><br />Therefore, the equation of line t in slope-intercept form is:<br /><br />\[ y = -\frac{7}{10}x + \frac{4}{5} \]<br /><br />So, the equation of line t is:<br /><br />\[ y = -\frac{7}{10}x + \frac{4}{5} \]