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Mathematics 8 Online
OpenStudy (anonymous):

I need help how to do this...? What exactly is this process of math problem called? h(x) = 7 -x 1) Evaluate h(-1/8)

OpenStudy (anonymous):

hey

OpenStudy (anonymous):

Evaluation of functions.

OpenStudy (anonymous):

thats what its called and how would u possibly do that if u arn't given any number for h(x)?

OpenStudy (anonymous):

i think it is called "evaluating a function" but i am not sure in any case \[h(\frac{1}{8})=7-\left(-\frac{1}{8}\right)\]

OpenStudy (anonymous):

so h(x) is the function it does not mean h * x ... h is the function and x is the variable for the input of the function.

OpenStudy (anonymous):

h(-1\8) inside ?

OpenStudy (anonymous):

so the input of x is (1/8)

OpenStudy (anonymous):

so "h(x) = 7 - x" is the function definition so if I put something like h(3) where ever I saw an x I put a 2 in its place so h(2) = 7 - 2

OpenStudy (anonymous):

\[h(\diamondsuit )=7-\diamondsuit\] put a \(-\frac{1}{8}\) where you see \(\diamondsuit\)

OpenStudy (anonymous):

so h(2) = 5

OpenStudy (anonymous):

good explanation @satellite73

OpenStudy (anonymous):

so basically are u trying to get the h alone when all is said and done ?

OpenStudy (anonymous):

good explanation @satellite73

OpenStudy (anonymous):

so far i have h (-1/8) = 7 - (-1/8)

OpenStudy (anonymous):

lol thanks, just showing off the \(\diamondsuit\) you could also write \[h(\clubsuit)=7-\clubsuit\]

OpenStudy (anonymous):

I plug in the - into the (-1/8) so I get now is 7 + 1/8

OpenStudy (anonymous):

is that right ?

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

so now I can plus the 7 into 1/8 and get h(-1/8) = 57/8

OpenStudy (anonymous):

but what do I do now?

OpenStudy (anonymous):

yeah, although being rather simpleminded i would probably have written \(7+\frac{1}{8}=7\tfrac{1}{8}\)

OpenStudy (anonymous):

go have a cup of coffee i guess, because you are done

OpenStudy (anonymous):

ah yes thanks yess!

OpenStudy (anonymous):

so now thats my answer because we can't do anything where does the h go?

OpenStudy (anonymous):

h(-1/8) = 7(1/8)

OpenStudy (anonymous):

it doesn't "go" anywhere \(h\) is a function, in this case a rather simple linear function it is a rule that assigns to each number \(x\) the number \(7-x\)

OpenStudy (anonymous):

so my answer is h = -7

OpenStudy (anonymous):

so for example \(h(3)=4,f(5)=2,h(7)=0\) etc

OpenStudy (anonymous):

because I divided the (-1/8) OHH i seee

OpenStudy (anonymous):

no you had it right the first time

OpenStudy (anonymous):

i think i have confused you you had it right at the beginning

OpenStudy (anonymous):

oh where u said h(-1/8) = 7(1/8) final answer

OpenStudy (anonymous):

\[h\left(-\frac{1}{8}\right)=7-\left(-\frac{1}{8}\right)=7+\frac{1}{8}=\frac{57}{8}\]

OpenStudy (anonymous):

i meant the mixed number \(7\tfrac{1}{8}\) as in seven and one eighth sorry to confuse you

OpenStudy (anonymous):

oh gotcha so it should look like this h(-1/8) = 57/8 and now do I divide the (-1/8) to the 57/8 ?

OpenStudy (anonymous):

no now you are done

OpenStudy (anonymous):

just leave the answer like that no simplifying

OpenStudy (anonymous):

omg i see what u did nvm I got it my final answer is 57/8 right ? no h = 57/8

OpenStudy (anonymous):

is this my final answer?

OpenStudy (anonymous):

h = 57/8

OpenStudy (anonymous):

?

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