If it is the midpoint, then FG= FM+MG
Also, this means that they are of equal lenght, so
FM=MG
<span>5y + 13= 5-3y
5y+3y=5+13
8y=18
y=
=2.25
</span>
which means that the two halves are equal to 5y+13=5*2.25+13=24.25
and FG is double the lenght:
24.25*2=49
Answer:
(a) 7.5 seconds
(b) The horizontal distance the package travel during its descent is 1737.8 ft
Step-by-step explanation:
(a) The given function for the height of the object is s = -16·t² + v₀·t + s₀
The initial height of the object s₀ = 900 feet
The initial vertical velocity of the object v₀
= 0 m/s
The time it takes the package to strike the ground is found as follows;
0 = -16·t² + 0×t + 900
900 = 16·t²
t² = 900/16 = 62.25
t = √62.25 = 7.5 seconds
(b) Given that the horizontal velocity of the package is given as 158 miles/hour, we have
158 miles/hour = 231.7126 ft/s
The horizontal distance the package covers in the 7.5 second of vertical flight = 231.7126 ft/s × 7.5 s = 1737.8445 feet = 1737.8 ft to one decimal place
The horizontal distance the package travel during its descent = 1737.8 ft.
15+v=18
The answer is 3
V equaling vaccinations
If <em>f(x)</em> = 7/(1<em> </em>+ <em>x</em>), then
<em>f</em> (2) = 7/3
<em>f '(x)</em> = -7/(<em>x</em> + 1)² ==> <em>f '</em> (2) = -7/9
<em>f ''(x)</em> = 14/(<em>x</em> + 1)³ ==> <em>f ''</em> (2) = 14/27
<em>f '''(x)</em> = -42/(<em>x</em> + 1)⁴ ==> <em>f '''</em> (2) = -14/27
Then the Taylor series of <em>f(x)</em> about <em>a</em> = 2 is
7/3 + 1/1! (-7/9) (<em>x</em> - 2) + 1/2! (14/27) (<em>x</em> - 2)² + 1/3! (-14/27) (<em>x</em> - 2)³
= 7/3 - 7/9 (<em>x</em> - 2) + 7/27 (<em>x</em> - 2)² - 7/81 (<em>x</em> - 2)³
