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3 years ago

What is the 6th term in the sequence below? 4,-8, 16, -32,

Mathematics

2 answers:

AURORKA [14]3 years ago

4 0

Answer:

−128

Step-by-step explanation:

BartSMP [9]3 years ago

4 0

Assuming the pattern continues...

Answer:

$128$

Step-by-step explanation:

We see that this is a geometric series with common ratio $-2$ .

So, the 6th term is the 4th term times $(-2)^2=2^2=4$ .

So, the answer is $4(-32)=\boxed{128}$ and we're done!

Note:

The closed form for this geometric series is $4*(-2)^{n-1}$ , which means the nth term is always $4*(-2)^{n-1}$ .

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(3x+1)/(3x+2) = (x+2)/(x-5)

jonny [76]

Answer:

x = -9/22 or -0.409 to the nearest thousandth.

Step-by-step explanation:

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2 years ago

What is the equation of the line that is perpendicular to y, =, 1 over 3 x, 4. And contains the point negative 2, negative 5,?.

Anit [1.1K]

The equation of the line is, $\rm y = 3x + 1$ .

Given that,

The equation of the line is,

$\rm y = \dfrac{1}{3} x+4\\$

And contains the point (-2, -5).

We have to determine,

What is the equation of the line that is perpendicular to the given line?

According to the question,

The equation of the line is,

$\rm = \dfrac{1}{3}x + 4\\$

On comparing with the standard equation of the line y = mx +c.

The slope of the line $m_1$ is 1/3.

When two lines are perpendicular the relation between these slopes is,

$\rm m_1\times m_1 = {-1}\\\\\dfrac{-1}{3} \times m_2 = -1\\\\-1 \times m_2 = -1 \times 3\\\\-m_2 = -3\\\\m_2 = 3$

And line contains the point (-2, -5).

Then,

$\rm y = mx +c \\\\-5 = 3 (-2) + c\\\\-5 = -6+c \\\\c = 6-5 \\\\c = 1$

Therefore,

The equation of the line that is perpendicular to the given line and contains the point (-2, -5) is,

$\rm y = mx +c \\\\y = 3x +1$

Hence, The required equation of the line is, $\rm y = 3x + 1$ .

For more details refer to the link given below.

brainly.com/question/14388443

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2 years ago

What is the precent of 2 is 8?

Vlad [161]

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