Help ;-;
Hey there!
I think we have to plug in the value of k into that ..right?
\(\color{#0cbb34}{\text{Originally Posted by}}\) @Laylalyssa I think we have to plug in the value of k into that ..right? \(\color{#0cbb34}{\text{End of Quote}}\) Yeah I think so
What do u need help with?
\[\frac{3}{2}+\left(-\frac{5}{2}\right)+\left(-2\right)\] this?
Plug in k = -5/2 \(\dfrac{3}{2} + \left(-\dfrac{5}{2}\right) + (-2)\) Do you know how to add/subtract fractions? \(\dfrac{3}{2} + \left(-\dfrac{5}{2}\right) = \dfrac{3}{2} - \dfrac{5}{2} = ?\)
\(\color{#0cbb34}{\text{Originally Posted by}}\) @YECRYHL What do u need help with? \(\color{#0cbb34}{\text{End of Quote}}\) theres a question posted a screenshot
\(\color{#0cbb34}{\text{Originally Posted by}}\) @Laylalyssa \[\frac{3}{2}+\left(-\frac{5}{2}\right)+\left(-2\right)\] this? \(\color{#0cbb34}{\text{End of Quote}}\) Yeah I believe it should be like that.
Oops, I believe you were referring to me and not the OP.
💀no continue💀
The question is all yours since you feel this intensely about it :)
\(\color{#0cbb34}{\text{Originally Posted by}}\) @AZ The question is all yours since you feel this intensely about it :) \(\color{#0cbb34}{\text{End of Quote}}\) ok den
\(\color{#0cbb34}{\text{Originally Posted by}}\) @AZ Plug in k = -5/2 \(\dfrac{3}{2} + \left(-\dfrac{5}{2}\right) + (-2)\) Do you know how to add/subtract fractions? \(\dfrac{3}{2} + \left(-\dfrac{5}{2}\right) = \dfrac{3}{2} - \dfrac{5}{2} = ?\) \(\color{#0cbb34}{\text{End of Quote}}\) Yeah I kinda remember how to add/sub fractions, I haven’t done it in a while but I think I know how to
You know how ..so can u give it a try
\(\color{#0cbb34}{\text{Originally Posted by}}\) @Laylalyssa You know how ..so can u give it a try \(\color{#0cbb34}{\text{End of Quote}}\) I mean I can try but I think I may get it wrong lol, but I’m gonna try >:)
did u try it or no
do you want to show them how to subtract fractions, just to jog their memory :)
\(\color{#0cbb34}{\text{Originally Posted by}}\) @AZ do you want to show them how to subtract fractions, just to jog their memory :) \(\color{#0cbb34}{\text{End of Quote}}\) sure one sec
So for example 6/4 - 10/4 6/4 - 10/4 6/4 + -10/4 6+-10/4 -4/4 = -1 lmk if that made sense 💀
basically, when you add or subtract, you have to make sure the denominator is the same. If it is, you can just add or subtract the numerator together \(\dfrac{a}{c} + \dfrac{b}{c} = \dfrac{a+b}{c}\) \(\dfrac{x}{z} - \dfrac{y}{z} = \dfrac{x-y}{z}\)
when you're multiplying fractions, the denominator (or the bottom part of the fraction) doesn't matter. You just multiply across. Dividing fractions is a little bit more trickier than that so we'll get to if you have a question that involves that :)
take the lcm and solve
so i dont wan to interfere nobody's work but first of all i have a question the given equation wan be \[\frac{ 3 }{ 2 }+(-k)+(-2)\] where \[k = - \frac{ 5 }{ 2 }\]
yes ?
@FriskIsNice
i ve wrote it right , correct this equation ?
yes, that is what it is
ok i m happy that you r here now bc. then you see given this equation with (-k) and the value of k is with minus too so in this way - case what will result ? @AZ please
a + (-b) = a - b
will result - (-k) what is plus
i understand what you say - but i think it in this way @Laylalyssa opinion pls ?
\(\color{#0cbb34}{\text{Originally Posted by}}\) @jhonyy9 so i dont wan to interfere nobody's work but first of all i have a question the given equation wan be \[\frac{ 3 }{ 2 }+(-k)+(-2)\] where \[k = - \frac{ 5 }{ 2 }\] \(\color{#0cbb34}{\text{End of Quote}}\) in place of k if you substitute the value of k so will result +(-(-5/2)) = +5/2
\(\color{#0cbb34}{\text{Originally Posted by}}\) @jhonyy9 bc not minus k = -5/2 just k = -5/2 \(\color{#0cbb34}{\text{End of Quote}}\)
Good catch! I missed that originally :-)
so really i dont wan interfere your calcule - work but - i think this is in this way right
You are absolutely correct, we all missed the negative in front of k because we got focused on the fact that k was a negative value
\(\color{#0cbb34}{\text{Originally Posted by}}\) @AZ You are absolutely correct, we all missed the negative in front of k because we got focused on the fact that k was a negative value \(\color{#0cbb34}{\text{End of Quote}}\) ty
@Laylalyssa pls
do you understand it now ?
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