What is the value of the system determinant for the following system of equations? 6x + y = 4 5x + y - 7 = 0

The end behavior of a polynomial function is the behavior of the graph of as approaches positive infinity or negative infinity. The degree and the leading coefficient of a polynomial function determine the end behavior of the graph.

Degree - 3 (odd);

Leading coefficient - 2 (positive).

Then

• when then
• when then

See attached graph of the function for graphical illustration.

Answer:

x=5

y=0

Step-by-step explanation: -7y + 16x - 80 = 0

x=5

-9y + 4x - 20 = 0

y=0

The distance between these points (0, 2) and (5, 2) will be 5 units. Then the correct option is B.

What is the distance between two points?

It is the measure of distance between the two points and is known as length. The length is measured in meters generally.

Let one point be (x, y) and another point be (h, k).

Then the distance between the points will be

D² = (x - h)² + (y - k)²

The two pints are given below.

(0, 2) and (5, 2)

Then the distance between these points (0, 2) and (5, 2) will be given as,

D² = (5 - 0)² + (2 - 2)²

D² = 5²

D = 5 units

The distance between these points (0, 2) and (5, 2) will be 5 units. Then the correct option is B.

Learn more about the distance between two points here:

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Answer:

B) 5 units

Step-by-step explanation:

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 $179^\circ$.

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