How do you find #h(-1)# given #h(n)=4/3n+8/5#?

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Answer 1 · 2017-03-20T00:59:10

#h(-1)= 4/15#

Steps:
Lets first begin by finding the LCD of the two fractions so we can add them.

The LCD is #15# so we must manipulate each fraction individually so that its denominator is #15#

Lets start with #4/3 n#:
To make the denominator #15# we must multiply both the numerator and denominator by #5#, thus:

#5/5*4/3 n = 20/15 n#

Similarly, we must multiply both the numerator and denominator by #3# for #8/5#

#3/3*8/5=24/15#

So now we have
#h(n)= 20/15 n+24/15#

So...
#h(-1)= 20/15 (-1)+24/15#

#h(-1)= -20/15+24/15#
#h(-1)= 4/15#

Answer 2 · 2017-03-20T01:01:04

#h(-1)=4/15#

Since it's #h(-1)# plug in -1 for n.

h(-1)=#4/3(-1)+8/5#

=#-4/3+8/5#

=#-20/15+24/15#

=#4/15#

ANSWER: #4/15#