The quotient of a number is at most -6. Write an inequality to represent this sentence
hmm...are you sure there isn't anything missing in this question?
oh my bad
you could at least have a statement as let's say the quotient of a number and 6.. if this is all the problem say then if x is the quotient, \[x \le -6\]
The quotient of a number and 8 is at most -6. Write an inequality to represent this sentence
oh okay. So what is a quotient? @Schoolninja7
dividing
right. So if the number. "n", is divided by 8, we would have this: \(\Large\frac{x}{8}\) Make sense so far?
i think its x/8 <=-6
you mean \(\Large\frac{x}{8}\) \(\le\)-6 ???
thank you
am i right
correct!!! \(\color{green}{\checkmark}\)
nice job c:
how do i upvote you guys thanks
just click best response for the person who you think helped most you can also fan people, and they can help you in the future
can u help me with another problem
\[3(x+1)+2<11\]
yeah sure. do you want to solve for x?
solve the inequality then i graph the solution on a number line i kno the number line is going to be an open circle going towards the left
wait dont solve that im sorry i switched up the problem i need helpwith i already solved this one and came u with 2
just solve like a normal equation: 3(x+1)+2\(\gt\)11 3x+3+2\(\gt\)11 3x+5\(\gt\)11 -5 -5 ----------- 3x\(\gt\)6 /3 /3 x\(\gt\)2
oh whoops lol alright
\[5t-2(t+2)\ge8\] is the roblem same rules
try to solve it based on how I solved the last problem. Remember, forget about the inequality; treat it like a normal equation and solve for t.
@Schoolninja7 hello?
t=0.57 but it feels wrong
hmmm...yeah that's not correct. Here's the process.
5t-2(t+2)\(\ge\)8 1.) first, distribute! 5t-2t-4\(\ge\)8 2.) then, add like terms 3t-4\(\ge\)8 3.) then, add 4 to both sides 3t-4\(\ge\)8 +4 +4 ------- 3t\(\ge\)12 4.) Lastly, divide 3 on both sides 3t\(\ge12\) /3 /3 ----- x\(\ge\)4
Does that make sense?
@Schoolninja7
yea tryna figure out how you got to it one min im stuck on the first distrubting part
oh i see what i did rong i distrubuted wrong thanks
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lol thanks
no problem
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