Express cos5mcos(-3m) as a sum or difference.
-1/2cos2m + 1/2cos2m
1/2cos2m - 1/2cos8m
1/2cos8m - 1/2cos2m
1/2cos8m + 1/2cos2m

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Step-by-step explanation:

Using the product to sum formula

• 2cosAcosB = cos(A+B) + cos(A - B)

note that cos(- 3m) = cos 3m, hence

cos5m cos(- 3m)

= cos 5m cos3m ← A = 5m and B = 3m

= ( 5m + 3m) + cos(5m - 3m) ]

= cos 8m + cos 2m

problem:

f(x) = - 2/3x + 5; f(5/2)

substitute 5/2 in for x:

f(5/2) = -2/3(5/2) + 5

simplify the right side:

f(5/2) = -5/3 + 5

f(5/2) = 10/3

f(5/2) = 10/3

i hope this !

yes because there is exactly one y value for every x value

for function there can't be more than one value for every x value