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2 years ago

Tn=n²−n+4 whtatagtaatta

Mathematics

1 answer:

almond37 [142]2 years ago

7 0

Answer:

T=n-1+4/n

n=(n^2-n+4)/T

Step-by-step explanation:

T=n-1+4/n

n=(n^2-n+4)/T

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Which numbers belongs in the box?

EleoNora [17]

Answer:

Step-by-step explanation:

$42 + 120 = 2 + 160$

3 0

2 years ago

Read 2 more answers

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

Quinzel earns $1254 each month. His total deductions are 20% of his pay. How much is deducted from his pay each month ?

vova2212 [387]

20% = 0.20

1254 x 0.20 = 250.80

so he has $250.80 deducted every month

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3 years ago

HELPPP PLEASEEE!!!???

muminat

Answer:

Step-by-step explanation:

I got it lol

6 0

2 years ago

Geometry 1.4 - Segment Addition Postulate & Midpoint

Roman55 [17]

Answer:

Step-by-step explanation:

AB + BC = AC

14 + 3x-4 = 4x+4

Combine like terms

10 +3x = 4x+4

Subtract 3x from each side

10+3x-3x = 4x+4-3x

10 = x+4

Subtract 4 from each side

10-4 =x-4+4

6 =x

We want AC

AC = 4x+4 = 4*6+4 = 24+4 = 28

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2 years ago