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olga2289 [7]
2 years ago
6

The number p and 2/3 are additive inverses.

Mathematics
1 answer:
vlada-n [284]2 years ago
6 0

By knowing that p and 2/3 are additive inverses, we will see that p = -2/3

So two numbers A and B are additive inverses if:

A + B = 0

Here the two numbers are p and 2/3, so we have:

2/3 + p = 0

Solving this for p gives:

p = -2/3

Then the value of p is just -2/3.

Now, I can't "drag and drop" the labels in the given image, so I will post a little graph of where each point should be:

If you want to learn more about additive inverses, you can read:

brainly.com/question/16681279

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In a G.P the difference between the 1st and 5th term is 150, and the difference between the
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Answer:

Either \displaystyle \frac{-1522}{\sqrt{41}} (approximately -238) or \displaystyle \frac{1522}{\sqrt{41}} (approximately 238.)

Step-by-step explanation:

Let a denote the first term of this geometric series, and let r denote the common ratio of this geometric series.

The first five terms of this series would be:

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First equation:

a\, r^4 - a = 150.

Second equation:

a\, r^3 - a\, r = 48.

Rewrite and simplify the first equation.

\begin{aligned}& a\, r^4 - a \\ &= a\, \left(r^4 - 1\right)\\ &= a\, \left(r^2 - 1\right) \, \left(r^2 + 1\right) \end{aligned}.

Therefore, the first equation becomes:

a\, \left(r^2 - 1\right) \, \left(r^2 + 1\right) = 150..

Similarly, rewrite and simplify the second equation:

\begin{aligned}&a\, r^3 - a\, r\\ &= a\, \left( r^3 - r\right) \\ &= a\, r\, \left(r^2 - 1\right) \end{aligned}.

Therefore, the second equation becomes:

a\, r\, \left(r^2 - 1\right) = 48.

Take the quotient between these two equations:

\begin{aligned}\frac{a\, \left(r^2 - 1\right) \, \left(r^2 + 1\right)}{a\cdot r\, \left(r^2 - 1\right)} = \frac{150}{48}\end{aligned}.

Simplify and solve for r:

\displaystyle \frac{r^2+ 1}{r} = \frac{25}{8}.

8\, r^2 - 25\, r + 8 = 0.

Either \displaystyle r = \frac{25 - 3\, \sqrt{41}}{16} or \displaystyle r = \frac{25 + 3\, \sqrt{41}}{16}.

Assume that \displaystyle r = \frac{25 - 3\, \sqrt{41}}{16}. Substitute back to either of the two original equations to show that \displaystyle a = -\frac{497\, \sqrt{41}}{41} - 75.

Calculate the sum of the first five terms:

\begin{aligned} &a + a\cdot r + a\cdot r^2 + a\cdot r^3 + a \cdot r^4\\ &= -\frac{1522\sqrt{41}}{41} \approx -238\end{aligned}.

Similarly, assume that \displaystyle r = \frac{25 + 3\, \sqrt{41}}{16}. Substitute back to either of the two original equations to show that \displaystyle a = \frac{497\, \sqrt{41}}{41} - 75.

Calculate the sum of the first five terms:

\begin{aligned} &a + a\cdot r + a\cdot r^2 + a\cdot r^3 + a \cdot r^4\\ &= \frac{1522\sqrt{41}}{41} \approx 238\end{aligned}.

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Answer:

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If you win, you get 1 + 6 = 7 dollar, and if you lose, you lose 1 dollar.

Therefore the expected amount out of this game is:

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