What is the product (-2d2+s)(5d2-6s)
2 answers:
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(a)
The inverse is when you swap the variables and solve for y.
g(t) = 2t - 1 (Note: g(t) represents y)
rewrite as: y = 2t - 1
swap the variables: t = 2y - 1
solve for y: t + 1 = 2y
= y
Answer for (a):
=
(b)
Same steps as part (a) above:
h(t) = 4t + 3
rewrite as: y = 4t + 3
swap the variables: t = 4y + 3
solve for y:
Answer for (b):
= 
(c)
replace all t's in the equation with
= 
=
= 
= 
Answer for (c):
= 
(d)
h(g(t)) = h(2t - 1) = 4(2t - 1) + 3 = 8t - 4 + 3 = 8t - 1
Answer for (d): h(g(t)) = 8t - 1
(e)
h(g(t)) = 8t - 1
y = 8 t - 1
t = 8y - 1
t + 1 = 8y
= y
Answer for (e): inverse of h(g(t)) =
The inverse is when you swap the variables and solve for y.
g(t) = 2t - 1 (Note: g(t) represents y)
rewrite as: y = 2t - 1
swap the variables: t = 2y - 1
solve for y: t + 1 = 2y
Answer for (a):
(b)
Same steps as part (a) above:
h(t) = 4t + 3
rewrite as: y = 4t + 3
swap the variables: t = 4y + 3
solve for y:
Answer for (b):
(c)
replace all t's in the
=
Answer for (c):
(d)
h(g(t)) = h(2t - 1) = 4(2t - 1) + 3 = 8t - 4 + 3 = 8t - 1
Answer for (d): h(g(t)) = 8t - 1
(e)
h(g(t)) = 8t - 1
y = 8 t - 1
t = 8y - 1
t + 1 = 8y
Answer for (e): inverse of h(g(t)) =