Mary jogged from Point P to Point Q while Wyatt jogged from Point Q to Point P.
Point P and point Q are 7.5 kilometers apart.
Mary’s motion is represented by d = 8t and Wyatt’s motion is represented by 2d + 14t = 15, where t hours is the time and d kilometers is the distance from point P. Solve for linear equations. When did Mary and Wyatt meet? minutes later
How far from point Q did they meet? km

When did Mary and Wyatt meet? 0.5 minutes later

How far from point Q did they meet? 3.5 km

Step-by-step explanation:

Mary’s motion is represented by d = 8t----(1)

Wyatt’s motion is represented by 2d + 14t = 15----(2)

We solve for d in equation (2) to get:d=7.5-7t----(3)

We equate equation (1) and (3) to get:

We now solve for t to get:

Now put t=0.5 into equation (1) to get d=0.5*8=4

This means they met 4 km from point P.

Therefore they meet 7.5-4=3.5km from point Q

2 x^2 -3x+p

sum of roots = 3/2

let other root be x

3+ x = 3/2

x= 3/2 - 3 = -3/2

product of roots = p/2

3× -3/2 = p/2

-9 = p

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so far i think you have the right thing going for you just remember soh cah toh

step-by-step explanation:

sine over hypotenuse

cosine   equals adjacent

tangent over hypotenuse