I am answering with the best knowledge that i have based on what I already learned last year, so the f(5)* g(5) is the one expression that is equivalent to(fg) (5)
A recursive sequence is a sequence of numbers whose values are determined by the numbers that come before them in the sequence.
We’re given a sequence whose (n + 1)-th term f(n + 1) depends on the value of the n-th term f(n), specified by the recursive rule
f(n + 1) = -4 f(n) + 3
We’re also given the 1st term in the sequence, f(1) = 1. Using this value and the recursive rule, we can find the next term f(2). (Just replace n with 1.)
f(1 + 1) = -4 f(1) + 3
f(2) = -4 • 1 + 3
f(2) = -1
We do the same thing to find the next term f(3) :
f(2 + 1) = -4 f(2) + 3
f(3) = -4 • (-1) + 3
f(3) = 7
One more time to find the next term f(4) :
f(3 + 1) = -4 f(3) + 3
f(4) = -4 • 7 + 3
f(4) = -25
wheat = $0.96 lb rye = $1.89 lb
12w + 15r = 3987 15w + 10r = 3330 r = (3330 - 15w)/10 12w + 15r = 3987 12w + 15((3330 - 15w)/10) = 3987 12w + (15/10)(3330 - 15w) = 3987 12w + 4995 - 22.5w = 3987 10.5w = 1008 w = 96 cents r = 189 cents.
Hope this helps mate
Answer:
the answer is 14.14
Step-by-step explanation:
Answer:
See explanation
Step-by-step explanation:
In ΔABC, m∠B = m∠C.
BH is angle B bisector, then by definition of angle bisector
∠CBH ≅ ∠HBK
m∠CBH = m∠HBK = 1/2m∠B
CK is angle C bisector, then by definition of angle bisector
∠BCK ≅ ∠KCH
m∠BCK = m∠KCH = 1/2m∠C
Since m∠B = m∠C, then
m∠CBH = m∠HBK = 1/2m∠B = 1/2m∠C = m∠BCK = m∠KCH (*)
Consider triangles CBH and BCK. In these triangles,
- ∠CBH ≅ ∠BCK (from equality (*));
- ∠HCB ≅ ∠KBC, because m∠B = m∠C;
- BC ≅CB by reflexive property.
So, triangles CBH and BCK are congruent by ASA postulate.
Congruent triangles have congruent corresponding sides, hence
BH ≅ CK.
