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3 years ago

What is the next term in this geometric sequence?

Mathematics

1 answer:

olga_2 [115]3 years ago

8 0

Your sequence is positive negative positive negative. The next term is o

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Find the roots of the function f(x) = 3x4 - 48.

Andrew [12]

Answer:

$x=\pm 2\text{ or } x = \pm 2i$

Step-by-step explanation:

We are given:

$0=3x^4-48$

Factor:

$0=3(x^4-16)$

Difference of Two Squares:

$0=3(x^2-4)(x^2+4)$

Zero Product Property:

$(x^2-4)=0\text{ or } x^2+4=0$

Solve:

$x^2=4\text{ or } x^2=-4$

Take the square root. Since we are taking an even-root, it requires plus/minus:

$x=\pm 2\text{ or } x = \pm 2i$

6 0

3 years ago

Marty is making curtains. He buys 6 yards of fabric that costs p dollars per yard and 8 yards of fabric that costs q dollars per

irina [24]

8q and 6p because terms are the sets in between + - / and *

7 0

3 years ago

Is 1/2gallon less than 2quarts or equal to 2quartS or more than 2 quarts

Andrew [12]

Answer: equal to

Step-by-step explanation:

There are 4 quarts in a gallon

6 0

3 years ago

Read 2 more answers

Find the partial fraction decomposition of the rational expression with prime quadratic factors in the denominator

SpyIntel [72]

$\dfrac{5x^4-7x^3-12x^2+6x+21}{(x-3)(x^2-2)^2}=\dfrac{a_1}{x-3}+\dfrac{a_2x+a_3}{x^2-2}+\dfrac{a_4x+a_5}{(x^2-2)^2}$
$\implies 5x^4-7x^3-12x^2+6x+21=a_1(x^2-2)^2+(a_2x+a_3)(x-3)(x^2-2)+(a_4x+a_5)(x-3)$

When $x=3$ , you're left with

$147=49a_1\implies a_1=\dfrac{147}{49}=3$

When $x=\sqrt2$ or $x=-\sqrt2$ , you're left with

$\begin{cases}17-8\sqrt2=(\sqrt2a_4+a_5)(\sqrt2-3)&\text{for }x=\sqrt2\\17+8\sqrt2=(-\sqrt2a_4+a_5)(-\sqrt2-3)\end{cases}\implies\begin{cases}-5+\sqrt2=\sqrt2a_4+a_5\\-5-\sqrt2=-\sqrt2a_4+a_5\end{cases}$

Adding the two equations together gives $-10=2a_5$ , or $a_5=-5$ . Subtracting them gives $2\sqrt2=2\sqrt2a_4$ , $a_4=1$ .

Now, you have

$5x^4-7x^3-12x^2+6x+21=3(x^2-2)^2+(a_2x+a_3)(x-3)(x^2-2)+(x-5)(x-3)$
$5x^4-7x^3-12x^2+6x+21=3x^4-11x^2-8x+27+(a_2x+a_3)(x-3)(x^2-2)$
$2x^4-7x^3-x^2+14x-6=(a_2x+a_3)(x-3)(x^2-2)$

By just examining the leading and lagging (first and last) terms that would be obtained by expanding the right side, and matching these with the terms on the left side, you would see that $a_2x^4=2x^4$ and $a_3(-3)(-2)=6a_3=-6$ . These alone tell you that you must have $a_2=2$ and $a_3=-1$ .

So the partial fraction decomposition is

$\dfrac3{x-3}+\dfrac{2x-1}{x^2-2}+\dfrac{x-5}{(x^2-2)^2}$

7 0

4 years ago

What is the midpoint of (3,8) and (6,4)

Andrews [41]

(4.5,6) that is what i got.

4 0

3 years ago