Since y = 9 is a horizontal line, the parallel line must also be horizontal. All horizontal lines can be written as y = (a number). That number can be found as the y-coordinate in the ordered pair, which is -8.

y = -8

8 0

3 years ago

PLEASE HELP ME!!!! I NEED THIS QUICKLY!!

zloy xaker [14]

Part A: $y=-\frac{5}{4} x+\frac{35}{2}$ is the equation of the line.

Part B: Anika worked 17.5 hours on day 0.

Part C: The setup time decreases with 1 hour and 15 min per day.

Explanation:

Part A: Let us find the equation of the line using the points (2,15) and (6,10)

The equation of the line is given by,

$y-y_{1}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}} \cdot\left(x-x_{1}\right)$

Substituting, we get,

$y-15=\frac{10-15}{6-2} (x-2)$

Simplifying, we have,

$y-15=-\frac{5}{4} (x-2)$

$y-15=-\frac{5}{4} x+\frac{5}{2}$

Subtracting both sides by 15, we get,

$y=-\frac{5}{4} x+\frac{35}{2}$

Hence, the equation of the line is $y=-\frac{5}{4} x+\frac{35}{2}$

Part B: To determine the number of hours Anika worked on day 0, let us substitute x = 0 in the equation of the line.

Thus, we get,

$y=\frac{35}{2}\\y=17.5$

Hence, Anika worked 17.5 hours on day 0.

Part C: The setup time decrease per day can be determined using the slope.

The formula for slope is given by

$m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

Substituting , we have,

$m=\frac{10-15}{6-2}=-\frac{5}{4}$

Converting the fraction into hours and minutes, we get,

$-\frac{5}{4}\times60=-75 min$

Since, 1 hour = 60 min

Subtracting 60 and 75 = 15 min

Thus, the setup time decreases with 1 hour and 15 min per day.

6 0

3 years ago

Solve x(−2) 2 =4(−2)

Simora [160]

Hi there,

X x (-2) x 2 = 4 x (-2)
- x = - 2

Hence, X = 2

Hope this helps :)

7 0

4 years ago