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lbvjy [14]
3 years ago
5

Answer correctly for brainliest +10 points!

Mathematics
1 answer:
Natalka [10]3 years ago
7 0
The top right corner is the table that represents a linear function because when x goes up by 1 y continuously goes down by 2 and stays consistent
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31+3
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Midpoint is the center point on the line segment <br> True or false?
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3 years ago
In a G.P the difference between the 1st and 5th term is 150, and the difference between the
liubo4ka [24]

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:

  • a,
  • a\cdot r,
  • a \cdot r^2,
  • a \cdot r^3,
  • a \cdot r^4.

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}.

4 0
2 years ago
PLEASE PLEASE HELP I WILL MARK BRAINLIEST!!!! HELPPP
Mars2501 [29]
I will go with B.

How the data spread out is measured by standard deviation. IQR is <span>measure of variability.</span>
7 0
3 years ago
intrun depozit sunt 240 t de cereale in prima zi se scoate a cincea parte din cantitateiar a doua zi se scoate o sesime din rest
Alex73 [517]
Cat is out on the first day: 
<span>240: 5 = 48 t </span>

<span>What remains after you take it out on day 48 t </span>

<span>240-48 = 192 t </span>

<span>Cat is out in the next </span>
<span>192 6 = 32t </span>

<span>What remains after you take it out 32 t of the day? </span>
<span>192-32 = </span>160t<span>
</span>
7 0
3 years ago
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