A skydiver's speed during a free fall
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Answer:
(C) reflection of f(x) = 5x² across the x-axis and translated down 2 units
<h3>Given</h3>
trapezoid PSTK with ∠P=90°, KS = 13, KP = 12, ST = 8
<h3>Find</h3>the area of PSTK
<h3>Solution</h3>It helps to draw a diagram.
∆ KPS is a right triangle with hypotenuse 13 and leg 12. Then the other leg (PS) is given by the Pythagorean theorem as
... KS² = PS² + KP²
... 13² = PS² + 12²
... PS = √(169 -144) = 5
This is the height of the trapezoid, which has bases 12 and 8. Then the area of the trapezoid is
... A = (1/2)(b1 +b2)h
... A = (1/2)(12 +8)·5
... A = 50
The area of trapezoid PSTK is 50 square units.
Answer: x" = 5.69
Step-by-step explanation:
The graphic solution is attached.
Verifying the solution:
Existence condition: x > 0
2x - 4 = √x + 5
√x =2x - 4 - 5
√x =2x - 9 (²)
x = (2x - 9)²
x = 4x² - 36x + 81
4x² - 36x - x + 81 = 0
4x² - 37x + 81 = 0
Δ = -37² - 4.4.81 = 1369 - 1296 = 73
x = 37 ±√73/8
x' = 3.55
x" = 5.69
checking:
2*3.55 - 4 = 3.1
√3.55 + 5 = 6.88 Its not the same ∴ 3.55 is not a solution
2*5.69 - 4 = 7.39
√5.69 + 5 = 7.39 ∴ it's the only solution
