Answer:
f(g(x)) = 15x³ - 5
Step-by-step explanation:
how confusingly described.
let me try and summarize what I understood :
f(x) = 3x² - 5
money earned when baking x cookies.
g(x) = sqrt(5x³)
the amount of cookies baked in x hours.
f(g(x)) now calculates how much money she earns when baking for x hours.
it is basically very simple : instead of f(x) we have f(g(x)), so g(x) is used as argument/variable in f instead of just plain x.
therefore,
f(g(x)) = 3×(sqrt(5x³))² - 5
with x now representing the baking hours, but f(...) calculating the overall money earned, by implicitly (!) calculating the amount of cookies baked in that time and taking that result automatically to calculate the earned money.
let's simplify this a little bit more.
f(g(x)) = 3×(sqrt(5x³))² - 5 = 3×(5x³) - 5 = 15x³ - 5
Answer:
t(g)= -4g + 20
Step-by-step explanation:
James is playing his favorite game at the arcade. After playing the game 3 times, he has 8 tokens remaining. He initially had 20 tokens, and the game costs the same number of tokens each time. The number tt of tokens James has is a function of gg, the number of games he plays
Solution
Let
g=No. of games James plays
t= No. of tokens James has.
Find the slope using
y=mx + b
Where,
m = Slope of line,
b = y-intercept.
Before James started playing the games, he has a total of 20 tokens.
That is, when g=0, t=20
After James played the games 3 times, he has 8 tokens left
That is, when g=3, t=8
(x,y)
(0,20) (3,8)
m=y2-y1 / x2-x1
=(8-20) / (3-0)
= -12 / 3
m= -4
Slope of the line, m= -4
y=mx + b
No. of tokens left depend on No. of games James plays
t is a function of g.
t(g)
t(g)= -4g + 20
Answer:
See proof below
Step-by-step explanation:
Assume that V is a vector space over the field F (take F=R,C if you prefer).
Let
. Then, we can write x as a linear combination of elements of s1, that is, there exist
and
such that
. Now,
then for all
we have that
. In particular, taking
with
we have that
. Then, x is a linear combination of vectors in S2, therefore
. We conclude that
.
If, additionally
then reversing the roles of S1 and S2 in the previous proof,
. Then
, therefore
.