Given:
The two functions are:
To find:
The statement that best compares the graph of g(x) with the graph of f(x).
Solution:
The horizontal stretch is defined as:
...(i)
If 
, the function f(x) is horizontally stretched by factor 
.
If 
, the function f(x) is horizontally compressed by factor .
We have,
Using these functions, we get
...(ii)
On comparing (i) and (ii), we get
Since , the function f(x) is horizontally stretched by factor 
.
Hence, the correct option is D.
recalling that d = rt, distance = rate * time.
we know Hector is going at 12 mph, and he has already covered 18 miles, how long has he been biking already?
so Hector has been biking for those 18 miles for 3/2 of an hour, namely and hour and a half already.
then Wanda kicks in, rolling like a lightning at 16mph.
let's say the "meet" at the same distance "d" at "t" hours after Wanda entered, so that means that Wanda has been traveling for "t" hours, but Hector has been traveling for "t + (3/2)" because he had been biking before Wanda.
the distance both have travelled is the same "d" miles, reason why they "meet", same distance.
![\bf \begin{array}{lcccl} &\stackrel{miles}{distance}&\stackrel{mph}{rate}&\stackrel{hours}{time}\\ \cline{2-4}&\\ Hector&d&12&t+\frac{3}{2}\\[1em] Wanda&d&16&t \end{array}\qquad \implies \begin{cases} \boxed{d}=(12)\left( t+\frac{3}{2} \right)\\[1em] d=(16)(t) \end{cases}](https://tex.z-dn.net/?f=%5Cbf%20%5Cbegin%7Barray%7D%7Blcccl%7D%20%26%5Cstackrel%7Bmiles%7D%7Bdistance%7D%26%5Cstackrel%7Bmph%7D%7Brate%7D%26%5Cstackrel%7Bhours%7D%7Btime%7D%5C%5C%20%5Ccline%7B2-4%7D%26%5C%5C%20Hector%26d%2612%26t%2B%5Cfrac%7B3%7D%7B2%7D%5C%5C%5B1em%5D%20Wanda%26d%2616%26t%20%5Cend%7Barray%7D%5Cqquad%20%5Cimplies%20%5Cbegin%7Bcases%7D%20%5Cboxed%7Bd%7D%3D%2812%29%5Cleft%28%20t%2B%5Cfrac%7B3%7D%7B2%7D%20%5Cright%29%5C%5C%5B1em%5D%20d%3D%2816%29%28t%29%20%5Cend%7Bcases%7D)
Answer:
3rd choice
Step-by-step explanation:
Add 6 to both sides to get 5x >= 15, or x >=3. This is the 3rd choice.
Answer:
(d-3)/2
Step-by-step explanation:
Difference is subtraction
(d-3)/2
Area of the Base = 6 * side^2 / 4 * tan (180/6)
Area of the Base = 6 * 16 / 4 * 0.57735
Area of the Base = 96 /
<span>
<span>
<span>
2.3094
</span>
</span>
</span>
Area of the Base =
<span>
<span>
<span>
41.5692387633
</span>
</span>
</span>
Area of 1 Face = 2 * 6 = 12
Area of 6 Faces = 72
<span>
<span>
Total Area = 41.5692387633
</span>
+ 72</span>
Total Area = 113<span>.5692387633</span>
