Question

can anybody help me with this problem?
is pre-cacul

Answer 1

Where is the problem?

Answer 2

i have a question,
here it goes,
A butcher's scale is accurate to within 0.05 pound. A sirloin steak weights 1.14 pounds on this scale. Solve the inequality: lx-1.14l<_0.05. What does the solution interval represent?

Answer 3

you know pre-cal?

Answer 4

Some, thinking. the absolute value of the quantity x-1.14 must be less than ??? is that a -0.05????

Answer 5

|x-1.14| <_ 0.05
=> -(x-1.14) <_ 0.05
or (x-1.14) <_ 0.05
solve it up!

Answer 6

no is 0.05

Answer 7

lfcmaniac,, what goes after your last step?

Answer 8

R, the inequality: is alraedy their, is given. lx-1.14l<_0.05

Answer 9

see what you now have is
-x + 1.14 <_ 0.05
=> 1.14 - 0.05 <_ x
=> 1.09 <_ x
or
x<_ 0.05-1.14
=> x<_ -1.09
because of the negative sign the inequality gets reversed,
=> x _> 1.09

get that ?

Answer 10

oh yeah

Answer 11

why is the x negative?
is the adsolute value

Answer 12

Worldboy, \[|x| => \pm x\] ?

Answer 13

what does the solution interval represents?

Answer 14

|x-1.14|<0.05
-0.05<x-1.14<0.05
-0.05+1.14<x-1.14+1.14<0.05+1.14
1.09<x<1.19

Answer 15

the solution interval represents that x is any number greater or equal to 1.09

Answer 16

lfc you a teacher?

Answer 17

R

Answer 18

ther is a mistake, becuse is suppost to be <_, equal sign

Answer 19

is it going to be the same result, wih with the <_ sign?

Answer 20

the <_ is nothing but \[\le\] isn't it ?

and no i aint a teacher, i am just a 12th grade student :P

Answer 21

yeah

Answer 22

Is the problem \[|x-1.14|\le0.05\]

Answer 23

nothing chnge much really,

Answer 24

i just need to change the c, to <-

Answer 25

maniac, how you did the sign, equal or greater?

Answer 26

may come handy in the future?

Answer 27

use the Equation button underneath the typing space my friend

Answer 28

equation button ?

Answer 29

never used it before

Answer 30

give it a try and you'll get the knack of it soon

Answer 31

where is the button?

Answer 32

?

Answer 33

you know R?

Answer 34

Click on the Equation and then you will get a separate box to enter equations and a menu of mathematical symbols, and operators.

Answer 35

thx

Answer 36

omg, i going save this site as my favorite, i am going help the people who i can help.

Answer 37

R, you help me with another problem, i am having problem?

Answer 38

Review this problem very carefully as I think there may be an error. I come up with the answer: 1.09<x<1.19\[1.09\le x \le 1.19\] Rework it and see what you get.

Answer 39

Yes, if I can.

Answer 40

yeah, i got the exact same thing

Answer 41

should i write it to
1.09<_x
1.19>_x

Answer 42

look rthe other problem i need help in, it goes like this
i have a question i don't quite understand, i am preparing for a exam.

A 100-gallon mixture of citrus extract and water is 3% citrus extract, and it needs to be diluted. The function f(x)=3/100+x represents the percentrage (in decimal form) of the miture that is citrus extract when x gallons of water are added.
Find the domain of f(x). Then, find the percentage of citrus extract in the mixture after 50 gallons of water are added to it.

Answer 43

i think the prof gave me a problem silimiar to this, i going check my notes

Answer 44

O.K let me study it a bit

Answer 45

ok

Answer 46

I don't quite understand the function, but there is 100 gallons of mixture of which is 3% citrus or 3 gallons of citrus 97 gallons of water. It looks like they add 50 gallons of water (no more citrus) giving 150 gallolns of mixture of which only 3 gallons is citrus (since no more was added) So now the percerntage is 3/150 or only 2% citrus. Do you have the answer?

Answer 47

no

Answer 48

It appears the domain of x could be the set of real numbers. What bothers me is the function seems to have a constant of .03 rather than attaining a 3 % mixture. It has been over 30 years since I had Calculus and I need to brush up on it. I am afraid I am not too much help to you at this time WorldBoy

Answer 49

is ok.

Answer 50

you try

Answer 51

Good luck with your studies and this "test"