Answer:
sec(B)=b. ab/bc
sin(B)=c. ac/ab
csc(B)=f. ab/ac
cot(B)=a. bc/ac
tan(B)=e. ac/bc
cos(B)=d. bc/ab
Step-by-step explanation:
<u><em>SOHCAHTOA</em></u>
<u><em>Sine=Opposite/Hypotenuse</em></u>
<u><em>Cosine=Adjacent/Hypotenuse</em></u>
<u><em>Tangent=Opposite/Adjacent</em></u>
<u><em>SHACHOCotAO</em></u>
<u><em>Secant=Hypotenuse/Adjacent</em></u>
<u><em>Cosecant=Hypotenuse/Opposite</em></u>
<u><em>Cotangent=Adjacent/Opposite</em></u>
<u><em /></u>
Answer:
nahhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh
Answer:
-216
I dont know if that's right
Answer:
a) t(n) = 8n -23
b) t(n) = 7n +24
c) t(n) = -30n +9
d) t(n) = 1.8n +7
Step-by-step explanation:
The general equation for the n-th term of an arithmetic sequence is ...
an = a1 +d(n -1)
Expanding this, we find ....
an = a1 +d·n -d = d·n +(a1 -d)
Comparing that to the form ...
an = m·n +b
we see that ...
m = d
b = (a1 -d)
If you're given a term other than a1, say ak, then you have ...
ak = d·k +(a1 -d)
(a1 -d) = ak -d·k
For a given term and common difference, we can write the desired equation as ...
t(n) = d·n +(ak -d·k) . . . . . . given t(k) = ak and d
__
Once again, it is convenient to use a spreadsheet to do the calculations for a number of different instances of this problem.
a) t(n) = 8n -23
b) t(n) = 7n +24
c) t(n) = -30n +9
d) t(n) = 1.8n +7
That looks very hard hood luck