Can someone help me with this please, I would really appreciate it, thank you!
(Geometry)

Answer:

squareroot 18, 9, 4pi, 14

b - squareroot of 4 is 2and the squareroot of 9 is 3. 5 is between 4 and 9, meaning it must be between 2 and 3. Also a negative so on the left of the 0.

c - sqrt of 1 is 1 and sqrt of 4 is 2.  between 1 and 2. closer to 2 (because 3 is closer to 4) than to 1.

j=irrational cant be in fraction form

k=rational is in raction form

l= rational it equals 4 which is an integer

m=irrational because it has pi in it

for the graph...

j=3.46

k=1.88

l=4

m=4.71

Step-by-step explanation:

Use a calculator or think about perfect squares to determine answers.

rational can be a fraction or integer, irrational has pi or a decimal.

Answer:

3+(-4)

Step-by-step explanation:

i got it right

Answer:

In conclusion, the only possible outcome is $61^\circ,$ $61^\circ,$ $61^\circ,$ $178^\circ,$ and $179^\circ$.

Step-by-step explanation:

Okay, so let's just dive in head on. Since we know that all the angles in a pentagon must add up to $540^{\circ}$ and that there are $5$ angles in a pentagon, we know that $61^\circ$ is the third angle,  $c$, of the pentagon. We also know that $a^\circ,$ $b^\circ,$ $c^\circ,$ $d^\circ,$ and $e^\circ,$ are all less than $180$. We know that in a regular pentagon all angles are $108^\circ$, however, the median angle is $61^\circ$ so we know that this is not a regular pentagon.

Now, since the median of our pentagon is $61^\circ$, the other numbers would center around $61$. With this information, we can figure out many solutions. However, there is one very important piece of information we almost forgot- the mode! What this means is, you cannot have an answer like $60^\circ,$ $61^\circ,$ $61^\circ,$ $179^\circ,$ and $179^\circ$ since there is only one mode.

Now let's figure out what the mode is. Is it $61$, or is it another number? Let's explore the possibilities of the mode being $61.$ If the mode is $61,$ it could either be $b$ or $d$. Let's first think about it being $b$. This would mean that the data set is $a^\circ,$ $61^\circ,$ $61^\circ,$ $d^\circ,$ and $e^\circ.$ The numbers would still need to add up to $540,$ so let's subtract $122$ (the two $61$'s) from $540$ to see how many more degrees we still need. We would get $418$. This means that $a,$ $d,$ and $e$ added together is $418$. If it is true that $b$ is $61,$ this would mean that $a, \leq61, 61, d, \leq e.$ If this is true, there could only be one possibility. This would be $61^\circ,$ $61^\circ,$ $61^\circ,$ $178^\circ,$ and $179^\circ$. If we changed $a$ to $60$, then there would be two modes. $a$ can't be $59$ since then $e$ would be $180$. $a$ also can't be any higher than $61$ since then it would not be $a$ at all. So basically, if $b$ were $61$, then the data set could only be $61^\circ,$ $61^\circ,$ $61^\circ,$ $178^\circ,$ and $179^\circ$.

But what if $d$ were $61?$ Then the data set would be $a, \leq b, 61, 61, \leq e.$ It would not be possible. This is because the highest number $e$ can be is $179.$. If this is, then we still have $239^\circ$ left to go. $a$ and $b$ would have to be greater than $61$, and this would not be possible because then it would not be $a$ and $b$ at all.  

Okay, we're almost done. What if the mode isn't $61$ at all, but a whole different number? This would either mean that $a=b$ or that $d=e$. If $d=e$ and $d=179,$ this means that $a$ and $b$ would have to both be $60.5$. We can't have two modes, and $b$ could not be $61$ because we can't have two modes. If $d$ were smaller, like $178$, then $a+b$ would need to be $123$ and this is not possible since that would be over the median of $61$. $d$ cannot be larger since that would go over the max of $179$.  

If $a=b$, let's think about if $a$ were $60$. $d+e$ would need to equal 359, and once again we can't have two modes, and $d$ could not be $179$ because $e$ cannot be $180$. If $a$ were smaller, like $59$, then $d+e$ would need to be $361$ and this is not possible since that would be over the max of $179$. $a$ cannot be larger since that would exceed the median of $61$.  

In conclusion, the only possible outcome is $61^\circ,$ $61^\circ,$ $61^\circ,$ $178^\circ,$ and $179^\circ$.

Make sure you understand! : )

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