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

- anonymous

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

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- schrodinger

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- R

Where is the problem?

- anonymous

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?

- anonymous

you know pre-cal?

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## More answers

- R

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

- anonymous

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

- anonymous

no is 0.05

- anonymous

lfcmaniac,, what goes after your last step?

- anonymous

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

- anonymous

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 ?

- anonymous

oh yeah

- anonymous

why is the x negative?
is the adsolute value

- anonymous

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

- anonymous

what does the solution interval represents?

- R

|x-1.14|<0.05
-0.05

- anonymous

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

- anonymous

lfc you a teacher?

- anonymous

R

- anonymous

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

- anonymous

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

- anonymous

the <_ is nothing but \[\le\] isn't it ?
and no i aint a teacher, i am just a 12th grade student :P

- anonymous

yeah

- R

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

- anonymous

nothing chnge much really,

- anonymous

i just need to change the c, to <-

- anonymous

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

- anonymous

may come handy in the future?

- anonymous

use the Equation button underneath the typing space my friend

- anonymous

equation button ?

- anonymous

never used it before

- anonymous

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

- anonymous

where is the button?

- anonymous

?

- anonymous

you know R?

- R

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

- anonymous

thx

- anonymous

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

- anonymous

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

- R

Review this problem very carefully as I think there may be an error. I come up with the answer: 1.09

- R

Yes, if I can.

- anonymous

yeah, i got the exact same thing

- anonymous

should i write it to
1.09<_x
1.19>_x

- anonymous

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.

- anonymous

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

- R

O.K let me study it a bit

- anonymous

ok

- R

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?

- anonymous

no

- R

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

- anonymous

is ok.

- anonymous

you try

- R

Good luck with your studies and this "test"

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