For what value of a is the equation an identity? a(2x + 3) = 9x + 15 + x

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

3 years ago

(5pg)3=125pg. Is wrong or correct

irakobra [83]

Answer:

125 p^3 g^3

Step-by-step explanation:

(5pg)^3

We know that (ab)^c = a^c * b^c

5^3 p^3 g^3

125 p^3 g^3

4 0

3 years ago

A) 6 haircuts at Chez Diva costs $210<br> B) One haircut at the Hair Factory costs $20

Tpy6a [65]

Answer:36

Step-by-step explanation: 210 divided by 6

7 0

3 years ago

You have purchased a mobile home for $20,000 which depreciates at 10% per year.

iragen [17]

Answer:

C. Three

Step-by-step explanation:

We want to solve the equation for n:

14,580 = 20,000(9/10)^n

14580/20000 = 729/1000 = (9/10)^n

(9/10)^3 = (9/10)^n . . . . . . write the left side as a cube

3 = n . . . . . . equate exponents

After year 3, the value will be $14,580.

_____

You can use logarithms to find n:

log(0.729) = n×log(0.9) . . . . . . taking the log of the 2nd line above

log(0.729)/log(0.9) = n = 3

8 0

3 years ago

The quotient of a number and 14 is 8. Find the number.

mafiozo [28]

Answer:

The number is 112

Step-by-step explanation:

Let the number be 'x'

$\sf \dfrac{x}{14}=8\\\\\text{Multiply both sides by 14}\\\\ \dfrac{x}{14}*14=8*14\\\\$

5 0

3 years ago

Read 2 more answers