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4 years ago

How do you do this?PLZ HELP

Mathematics

1 answer:

Alisiya [41]4 years ago

5 0

3(x^2+10x+5)-5(x-k)=
3x^2+30x+15-5x+5k=
3x^2+25x+15+5k

for this to be divisible by x every term must include x or get eliminated

the problematic terms are 15 and 5k
to eliminate them they must equal 0 when added:
15+5k=0
5k=-15
k=-3

so A) -3 is the solution

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Find the explicit formula for the geometric sequence cn given below. Note that c1=6.

kodGreya [7K]

The explicit formula for the geometric sequence $c_{n}$ is

$c_{n}=6(\frac{-1}{3})^{n-1}$

Step-by-step explanation:

The formula of the nth term of a geometric sequence is

$a_{n}=a(r)^{n-1}$ , where:

a is the first term of the sequence
r is the common ratio between the consecutive terms

The geometric sequence of $c_{n}$ is 6 , -2 , $\frac{2}{3}$ , $\frac{-2}{9}$ , $\frac{2}{27}$

∵ $c_{1}$ = 6

∵ $r=\frac{c_{2}}{c_{1}}$

∵ $c_{2}$ = -2

∴ $r=\frac{-2}{6}$

- Divide both term by 2 to simplify it

∴ $r=\frac{-1}{3}$

∵ $c_{n}=c_{1}(r)^{n-1}$

- Substitute the value of $c_{1}$ and r in the rule above

∴ $c_{n}=6(\frac{-1}{3})^{n-1}$

The explicit formula for the geometric sequence $c_{n}$ is

$c_{n}=6(\frac{-1}{3})^{n-1}$

Learn more:

You can learn more about the geometric sequence in brainly.com/question/1522572

#LearnwithBrainly

6 0

3 years ago

*WILL GIVE BRAINLIEST FOR BEST ANSWER DONT ANSWER IF YOU DONT KNOW AND NO VIRUS LINKS OR I WILL REPORT YOU!!!*

Digiron [165]

56 is the answer :)

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3 years ago

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Solve x2 + 12x + 6 = 0 using the completing-the-square method.

anygoal [31]

E d g e n u i t y Question: Solve x2 + 12x + 6 = 0 using the completing-the-square method.

(A) x= -6+ V30

(B) x= 6 + V30

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Describe the behavior of the function ppp around its vertical asymptote at x=-2x=−2x, equals, minus, 2.

insens350 [35]

Answer:

$x->-2^{-}, p(x)->-\infty$ and as $x->-2^{+}, p(x)->-\infty$

Step-by-step explanation:

Given

$p(x) = \frac{x^2-2x-3}{x+2}$ -- Missing from the question

Required

The behavior of the function around its vertical asymptote at $x = -2$

$p(x) = \frac{x^2-2x-3}{x+2}$

Expand the numerator

$p(x) = \frac{x^2 + x -3x - 3}{x+2}$

Factorize

$p(x) = \frac{x(x + 1) -3(x + 1)}{x+2}$

Factor out x + 1

$p(x) = \frac{(x -3)(x + 1)}{x+2}$

We test the function using values close to -2 (one value will be less than -2 while the other will be greater than -2)

We are only interested in the sign of the result

----------------------------------------------------------------------------------------------------------

As x approaches -2 implies that:

$x -> -2^{-}$ Say x = -3

$p(x) = \frac{(x -3)(x + 1)}{x+2}$

$p(-3) = \frac{(-3-3)(-3+1)}{-3+2} = \frac{-6 * -2}{-1} = \frac{+12}{-1} = -12$

We have a negative value (-12); This will be called negative infinity

This implies that as x approaches -2, p(x) approaches negative infinity

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Take note of the superscript of 2 (this implies that, we approach 2 from a value less than 2)

As x leaves -2 implies that: $x>-2$

Say x = -2.1

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We have a negative value (-56.1); This will be called negative infinity

This implies that as x leaves -2, p(x) approaches negative infinity

$x->-2^{+}, p(x)->-\infty$

So, the behavior is:

$x->-2^{-}, p(x)->-\infty$ and as $x->-2^{+}, p(x)->-\infty$

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saveliy_v [14]

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