Answer:
The value of the side PS is 26 approx.
Step-by-step explanation:
In this question we have two right triangles. Triangle PQR and Triangle PQS.
Where S is some point on the line segment QR.
Given:
PR = 20
SR = 11
QS = 5
We know that QR = QS + SR
QR = 11 + 5
QR = 16
Now triangle PQR has one unknown side PQ which in its base.
Finding PQ:
Using Pythagoras theorem for the right angled triangle PQR.
PR² = PQ² + QR²
PQ = √(PR² - QR²)
PQ = √(20²+16²)
PQ = √656
PQ = 4√41
Now for right angled triangle PQS, PS is unknown which is actually the hypotenuse of the right angled triangle.
Finding PS:
Using Pythagoras theorem, we have:
PS² = PQ² + QS²
PS² = 656 + 25
PS² = 681
PS = 26.09
PS = 26
<span>68=-16t2+64t+8
16t2-64t+60=0
4t2-16t+15=0
(2t-5)(2t-3)=0
t=5/2 or 3/2
The object is at 68 feet after 3/2 secs, and returns to 68 feet after another second..</span>
0.00067=67X10^5 so
67X10^5-2.3X10^5
64.7X10^5 which can be expressed as:
6.47X10^6
(not sure if it matters to you or not, but technically we only had two significant figures so it should be rounded to 6.5X10^6)
Answer:
Hello, The correct answer is A. $7,708. Good luck!!
