Answer:
8^2+x + 2^ 3x = 32^ 1/2
2^3(2+x) + 2^3x = 2^5^1/2
All two's will cancel out
3(2+x) + 3x = 5^1/2
6+3x + 3x = 5^1/2
6 +6x = 2.23
6x = 2.23 - 6
6x = -3.77
x = -3.77÷ 6
x = -0.63


Consequently, t<span>he limit of

as x approaches infinity is

.
In other words,

approaches the line y=x,
</span><span>
so oblique asymptote is y=x.
I'm Japanese, if you find some mistakes in my English, please let me know.</span>
Answer:
e < 2
Step-by-step explanation:
Given
e - 2 < 0 ( ad 2 to both sides )
e < 2
(a) Compare your quadratic for h to the general quadratic ax² +bx +c. Perhaps you can see that ...
a = -16
b = 128
You use these numbers in the given formula to find the time when the ball is highest.
t = -b/(2a) = -128/(2(-16)) = 4 . . . . . . the time at which the ball is highest
(b) Evaluate the quadratic to find the height at t=4.
h = -16(4)² +128(4) +21
h = -256 +512 +21
h = 277
The maximum height of the ball is 277 ft.
Answer:
Isolate the variable by dividing each side by factors that don't contain the variable.
Exact Form:
x = 1
/9
Decimal Form:
x = 0.
1111111111111111111111111
Hope this helps! Have a great day!