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Roman55 [17]
3 years ago
13

Find the sum of 5m+3n+p,-5+3n

Mathematics
1 answer:
Free_Kalibri [48]3 years ago
6 0
You add the letters with numbers in front for example :
6n +5m +p -5


(The postive and negative 5 dont cancel becayse there isnt a m on the negative)
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It says that there was an error with my answer, but I don't know what, so I screenshotted what I typed and attached them as 4 images below:

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3 years ago
If x- 2 is a factor of x^2 -bx+ b, where b is a constant, what is the value of b
KATRIN_1 [288]
\bf x^2-\underline{b} x+b\implies \stackrel{given}{(x-2)}~(x-y)\impliedby \textit{now let's do some FOIL}\\\\
-------------------------------\\\\
\begin{cases}
-yx-2x=-\underline{b}x\implies y+2=\underline{b}\\
(-2)(-y)=b\implies 2y=b
\end{cases}
\\\\\\
\underline{b}=b\implies y+2=2y\implies 2=y

what is b?   from the first equation in the system, y + 2 = b.
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3 years ago
COMPLETE THE EXERCISE BELOW TO DECOMPOSE 930
PolarNik [594]
I saw the complete problem:

complete the exercise below to decompose 930

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5 0
3 years ago
A detective has interviewed four witnesses to a crime. From the stories of the witnesses the detective has concluded
SVETLANKA909090 [29]

Answer:

Step-by-step explanation:

Step 1:we should turn the “facts” into logical expressions.

Step 2:Assign variables

Our variables are: b (for butler), c (for cook), g (for gardener), h (for handyman)

Step 3 . Let the value of true indicate that the person is telling the truth. For example, b = T means the butler is telling the truth. Then we can translate the statements as follows:

Check attachment

Step 4

We know all of these must hold. So we really want to know when (b→c)∧¬(c∧g)∧¬(¬g∧¬h)∧(h → ¬c) is true.

Step 5

Behold, another truth table task! If you work out the truth table, you see that this statement is only true when:

Check attachment

Step 6

Therefore we can say the butler and cook are definitely lying, but we can not determine if the gardener or handyman are lying. However, there are four variables in this problem. Instead, we can do this with some straight reasoning. Notice that:

If the butler is telling the truth, then by (a) the cook is telling the truth.

If the cook is telling the truth, then by (b) the gardener is lying.

If the gardener is lying, then by (c) the handyman is telling the truth.

If the handyman is telling the truth, they by (d) the cook is lying.

This leads to a contradiction (the cook can not be both telling the truth and lying)! Therefore the butler and cook must be lying. What about the gardener and the handyman? We don’t have enough information to figure out if they are lying. Since the cook is lying, we can’t use (b) to come to any conclusions about the gardener. Even if we are able to conclude that the gardener is telling the truth, we can’t use (c) to come to any conclusions about the handyman.

4 0
3 years ago
The savings account offering which of these APRs and compounding periods offers the best APY?
zzz [600]
\bf \qquad  \qquad  \textit{Annual Yield Formula}
\\\\
~~~~~~~~~~~~\textit{4.0784\% compounded monthly}\\\\
~~~~~~~~~~~~\left(1+\frac{r}{n}\right)^{n}-1
\\\\
\begin{cases}
r=rate\to 4.0784\%\to \frac{4.0784}{100}\to &0.040784\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{monthly, thus twelve}
\end{array}\to &12
\end{cases}
\\\\\\
\left(1+\frac{0.040784}{12}\right)^{12}-1\\\\
-------------------------------\\\\
~~~~~~~~~~~~\textit{4.0798\% compounded semiannually}\\\\


\bf ~~~~~~~~~~~~\left(1+\frac{r}{n}\right)^{n}-1
\\\\
\begin{cases}
r=rate\to 4.0798\%\to \frac{4.0798}{100}\to &0.040798\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{semi-annually, thus twice}
\end{array}\to &2
\end{cases}
\\\\\\
\left(1+\frac{0.040798}{2}\right)^{2}-1\\\\
-------------------------------\\\\


\bf ~~~~~~~~~~~~\textit{4.0730\% compounded daily}\\\\
~~~~~~~~~~~~\left(1+\frac{r}{n}\right)^{n}-1
\\\\
\begin{cases}
r=rate\to 4.0730\%\to \frac{4.0730}{100}\to &0.040730\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{daily, thus 365}
\end{array}\to &365
\end{cases}
\\\\\\
\left(1+\frac{0.040730}{365}\right)^{365}-1
7 0
3 years ago
Read 2 more answers
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