Solve for x. -ax + 2b > 8 x < the quantity negative 2 times b plus 8 all over negative a x > the quantity negative 2 times b plus 8 all over negative a x < the quantity 2 times b minus 8 all over negative a x > the quantity negative 2 times b minus 8 all over a
those are the choices
one minute
what you do is try to get x by itself \[-ax+2b>8\] \[-2b -ax -2b > 8 - 2b\] subtract 2b from both sides \[\frac{ -ax }{ -a}< \frac{ -2b+8 }{ -a }\] divide both sides by -a, if you divide by a negative number, you switch the sign from > to < which results in \[x<\frac{ -2b+8 }{ -a}\] being your answer
tell me if you misunderstand or dont see how i did something
yep nsellers23 is correct
thats what I got too
Not totally understand somewhat thanks
well i can explain it better if you tell me which part confused you
Can u help me with one more well actually 2
sure lets see it
ill try and explain it in more depth if i can
can u help with this
well someone please correct me if i am wrong but i believe all you have to do is get the r by itself by doing: start: \[C=2\pi r\] step 1: divide by sides by \[2\pi\] to get r by itself. result: \[r=\frac{ C }{ 2\pi }\] If i'm wrong someone, please correct me but to my knowledge is this how its done.
yes that is the correct thing to do. its been confirmed.
Last ? ok
sure as many questions as you need
so im guessing your not really interested in learning this material? what you really want is just the answers?
noooo
i'll tell u what i think it is and you can check it
sure lets try that for this one
I think it is A or D
I leaning towards A though
@nsellers23
im actually getting b because you start with \[\frac{ 9 }{ 2 }(35+b)=A\] and you multiply both sides by 2 from the \[\frac{ 9 }{ 2 }\] then you end up with \[9(35+b)=2A\] now divide by sides by 9 getting you \[35+b=\frac{ 2A }{ 9 }\] now subtract 35 from both sides because its positive, making this final answer \[b=\frac{ 2A }{ 9 }-35\]
make sense?
Ok now i see how u do it thanksssss :) i'm going to make another ? tag u so i can give u another medal is that alright
sure if you need anymore help just tag me again
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