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3 years ago

Plzzz help need it right nowww.....

1 answer:

5 0

$2x-5= - 20$

$
2x=-20+5
 
$ $Add \ 5 \ to \ both \ sides
$

$2x=-15

 
$

$x=- \dfrac{15}{2}$ $Divide \ both \ sides \ by \ 2$

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Find the common ratio for the geometric sequence defined by the formula: an=40(2‾√)n−1 a n = 40 ( 2 ) n − 1

WITCHER [35]

The ratio of the geometric sequence 40 $2^{n-1}$ is 2.

Given that geometric sequence is 40* $2^{n-1}$ and we have to find the common ratio of all the terms.

Geometric sequence is a sequence in which all the terms have a common ratio.

Nth termof a GP is a $r^{n-1}$ in which a is first term and r is common ratio.

Geometric sequence=40* $2^{n-1}$

We have to first find the first term, second term and third term of a geometric progression.

First term=40* $2^{1-1}$

=40* $2^{0}$

=40*1

=40

Second term=40* $2^{2-1}$

=40* $2^{1}$

=40*2

=80

Third term=40* $2^{3-1}$

=40* $2^{2}$

=40*4

=160

Ratio of first two terms=80/40=2

Ratio of next two terms=160/80=2

Hence the common ratio of geometric sequence is 2.

Learn more about geometric progression at brainly.com/question/12006112

#SPJ1

4 0

1 year ago

1. Which of the following is the slope-intercept form of 6x + 2y = 28?

Andru [333]

Answer:

2nd option

Step-by-step explanation:

The equation of a line in slope- intercept form is

y = mx + c

Given

6x + 2y = 28 ( subtract 6x from both sides )

2y = - 6x + 28 ( divide the terms by 2 )

y = - 3x + 14 ← in slope- intercept form

6 0

2 years ago

Read 2 more answers

What is the constant term of the quotient when 2x^3-3x^2 + 4x-2 is divided by x^2-x+2?

AlladinOne [14]

Answer:

-1

Step-by-step explanation:

See the attachment for the polynomial long division. The constant in the quotient is -1.

_____

Here, there is a remainder of -x. If there were no remainder the constant in the quotient is the ratio of the constant in the dividend to the constant in the divisor: -2/2 = -1.

That could be a first guess in a "guess and check" solution approach.

<em>Guess</em>: first term of binomial quotient is (2x^3)/x^2 = 2x; last term of binomial quotient is -2/2 = -1. So, the quotient is guessed to be (2x -1).

<em>Check</em>: (2x -1)(x^2 -x +2) = 2x^3 -3x^2 +5x -2

Subtracting this from the actual dividend gives a remainder of -x. This has a lower degree than the divisor, so no further adjustment of the quotient is required.