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2 years ago

Factor the expression 98x^3-50xy^298 x 3 − 50 x y 2

Mathematics

1 answer:

pickupchik [31]2 years ago

6 0

Answer:

euejdnznznnznznsnsnsnjsksjs

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Write the equation for the vertical line that contains point E(-7,7)

Levart [38]

Vertical line are represented by x = a number....that number being the x value in ur set of points

so ur equation is : x = -7

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3 years ago

Find the common difference of the arithmetic sequence 5, 14, 23, ...

vodka [1.7K]

It’s plus 9 so the next number should be 32

7 0

3 years ago

Read 2 more answers

The period 7 (in seconds) of a pendulum is given by T = 2pi√( L divided by 32) where L stands for the length (in feet) of the pe

oksano4ka [1.4K]

Answer:

128 feet

Step-by-step explanation:

$t = 2\pi \sqrt{\frac{l}{32} }$

So now we will substitute the values of T and π into the equation.

$12.56 = 2 \times 3.14 \sqrt{ \frac{l}{32} }$

$12.56 = 6.28 \sqrt{ \frac{l}{32} }$

Divide both sides by 6.28

$2 = \sqrt{ \frac{l}{32} }$

Square both sides.

$4 = \frac{l}{32}$

Multiply both sides by 32

$l = 4 \times 32 \\ l = 128 \: feets$

5 0

1 year ago

Which inequality is equivalent to absolute value x-4 >9

hoa [83]

Answer:

x> 13

Step-by-step explanation:

x-4 >9

Add 4 to each side

x-4+4 >9+4

x> 13

7 0

3 years ago

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

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

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3 years ago