I think 10 is the answer
HOPE THIS HELPS
Answer:
See explanation
Step-by-step explanation:
Consider triangles PTS and QTR. In these triangles,
- given;
- given;
- as vertical angles when lines PR and SQ intersect.
Thus, 
by AAS postulate.
Congruent triangles have congruent corresponding sides, so
Consider segments PR and QS:
![PR=PT+TR\ [\text{Segment addition postulate}]\\ \\QS=QT+TS\ [\text{Segment addition postulate}]\\ \\PT=QT\ [\text{Proven}]\\ \\ST=RT\ [\text{Given}]](https://tex.z-dn.net/?f=PR%3DPT%2BTR%5C%20%5B%5Ctext%7BSegment%20addition%20postulate%7D%5D%5C%5C%20%5C%5CQS%3DQT%2BTS%5C%20%5B%5Ctext%7BSegment%20addition%20postulate%7D%5D%5C%5C%20%5C%5CPT%3DQT%5C%20%5B%5Ctext%7BProven%7D%5D%5C%5C%20%5C%5CST%3DRT%5C%20%5B%5Ctext%7BGiven%7D%5D)
So,
![PR=SQ\ [\text{Substitution property}]](https://tex.z-dn.net/?f=PR%3DSQ%5C%20%5B%5Ctext%7BSubstitution%20property%7D%5D)
Answer:
the "acceleration" is -15m/s (it's negative because it is slowing down, not speeding up[it might just be a positive 15m/s if it is called deceleration and not acceleration])
Step-by-step explanation:
30m/s (starting speed)/2 (the number of seconds it took to stop from 30m/s
Answer:-1/32
Step-by-step explanation:
First term=a=1024
Common ratio=r=-512/1024
r=-1/2
Using them formula
Tn=a x r^(n-1)
T16=1024 x (-1/2)^(16-1)
T16=1024 x (-1/2)^15
T16=1024 x -1/32768
T16=-1024/32768
T16=-1/32
Answer:
20 ft by 60 ft
Step-by-step explanation:
"What should the dimensions of the garden be to maximize this area?"
If y is the length of the garden, parallel to the stream, and x is the width of the garden, then the amount of fencing is:
120 = 3x + y
And the area is:
A = xy
Use substitution:
A = x (120 − 3x)
A = -3x² + 120x
This is a downward facing parabola. The maximum is at the vertex, which we can find using x = -b/(2a).
x = -120 / (2 · -3)
x = 20
When x = 20, y = 60. So the garden should be 20 ft by 60 ft.
