If f(x) = 5x 40, what is f(x) when x = –5?

Answer:

Likely to get it.

Step-by-step explanation

Do you mean 1% or 1? If 1 then it is likely to pick that.

If it is 1% then it's the opposite (unlikely) to pick that.

Hopes this makes sense!

Answer:

The approximate time that it will take Mary to save R40000 is 29 years

Step-by-step explanation:

A= p(1+r/n)^nt

t = log(A/P) / n[log(1 + r/n)]

A = R40000

n = 4

P = R400

r = 16%

t = log(A/P) / n[log(1 + r/n)]

= log(40,000 / 400) / 4{log(1+0.16/4)}

= log(100) / 4{log(1+0.04)}

= 2 / 4{log(1.04)}

= 2 / 4(0.0170)

= 2 / 0.068

= 29.41

t = 29.41 years

Approximately 29 years

The approximate time that it will take Mary to save R40000 is 29 years

Answer:

60 miles can leatherback sea turtle travel in 10 hours .

Step-by-step explanation:

As given

A leatherback sea turtle can travel at a speed of 48 miles in 8 hours.

i.e

48 miles = 8 hours

A leatherback sea turtle can travel in one hour.

1hour = 6 miles

Now calculate for the 10 hours .

A leatherback sea turtle  travel distance in 10 hours = 6 × 10

                                                                                       = 60 miles

Therefore 60 miles can leatherback sea turtle travel in 10 hours .

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! : )

Also, you can copy and paste into a browser that understands LaTeX for a better view.