$=(a^2-10a)-(b^2+6b) +16$
$=[(a^2-2(5)a+25)-25]-[(b^2+2(3)b+9)-9]+16$
$=(a-5)^2-25-(b+3)^2+9+16$
$=(a-5)^2-(b+3)^2$
Help? with complete solution please help me please
1 answer:
9514 1404 393
Answer:
- 13 ft
- (a) 1 second; (b) t = 0, t = 1/2
Step-by-step explanation:
<h3>1. </h3>Let w represent the length of the wire. Then the height of attachment is (w-1). The Pythagorean theorem tells us a relevant relation is ...
5² +(w -1)² = w²
w² -2w +26 = w² . . . . . . . eliminate parentheses, collect terms
26 = 2w . . . . . . . . . . . . add 2w
13 = w . . . . . . . . . . . . divide by 2
The length of the wire is 13 feet.
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<h3>2. </h3>(a) When h = 0, the equation is ...
0 = -16t^2 +8t +8
Dividing by -8 puts this into standard form:
2t^2 -t -1 = 0
Factoring, we get ...
(2t +1)(t -1) = 0
The positive value of t that makes a factor zero is t = 1.
It will take 1 second for the gymnast to reach the ground.
__
(b) When h = 8, the equation is ...
8 = -16t^2 +8t +8
Subtract 8 and divide by 8 to get ...
0 = -2t^2 +t
0 = t(1 -2t) . . . . factor out t
Values of t that make the factors zero are ...
t = 0
t = 1/2
The gymnast will be 8 feet above the ground at the start of the dismount, and 1/2 second later.
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