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3 years ago

6x + 11 = 21 Help I’m on a quiz and this question is so confusing

Mathematics

1 answer:

Westkost [7]3 years ago

3 0

Answer:

Step by Step Solution

Rearrange:

Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation :

6*x+11-(21)=0

Step by step solution :

STEP

Pulling out like terms

1.1 Pull out like factors :

6x - 10 = 2 • (3x - 5)

Equation at the end of step

STEP

Equations which are never true:

2.1 Solve : 2 = 0

This equation has no solution.

A a non-zero constant never equals zero.

Solving a Single Variable Equation:

2.2 Solve : 3x-5 = 0

Add 5 to both sides of the equation :

3x = 5

Divide both sides of the equation by 3:

x = 5/3 = 1.667

One solution was found :

x = 5/3 = 1.667

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3 years ago

What is the explicit rule for this geometric sequence? a1=-15;an=1/5•an-1

Ad libitum [116K]

Answer:

Option A is correct.

$a_n =-15 (\frac{1}{5})^{n-1}$

Step-by-step explanation:

Given: $a_1 = -15$ ; $a_n = \frac{1}{5} \cdot a_{n-1}$

for n = 2

$a_2 = \frac{1}{5} \cdot a_{2-1}$ = $\frac{1}{5}\cdot a_1 = \frac{1}{5} \cdot (-15)$

for n =3

$a_3 = \frac{1}{5} \cdot a_{3-1}$ = $\frac{1}{5} \cdota_2= (\frac{1}{5})^2\cdot (-15)$

Simlarly , for n =4

$a_3 = \frac{1}{5} \cdot a_{4-1}$ = $\frac{1}{5}\cdot a_3= (\frac{1}{5})^3\cdot (-15)$

and so on...

Common ratio(r) states that for a geometric sequence or geometric series, the common ratio is the ratio of a term to the previous term.

we have; $r = \frac{a_2}{a_1} = \frac{1}{5}$

Now; by recursive formula for geometric series:

$a_n = a_1 r^{n-1}$

where $a_1$ is the first term and r is the common ratio:

Substitute the value of $a_1 = -15$ and $r = \frac{1}{5}$

we have;

$a_n =-15 (\frac{1}{5})^{n-1}$

therefore, the explicit rule for the geometric sequence is; $a_n =-15 (\frac{1}{5})^{n-1}$

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3 years ago