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3 years ago

Pleease help me with this math work

Mathematics

1 answer:

OleMash [197]3 years ago

6 0

The first one because the left bound is included [ and the right one is not )

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Please help me with this urgenttlyyyyyy

Kazeer [188]

Answer: Just drag each box over to the other box

Step-by-step explanation:

just how i said

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3 years ago

4 cats each ate 1/4 cup of canned food and 1/4 cup of dry food. how much food did they eat altogether

VashaNatasha [74]

Answer: 1 cup canned and 1 cup dry, 2cups put together

Step-by-step explanation:

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3 years ago

How do i solve the question in the image provided ?

Mars2501 [29]

Answer:

0.19+0.46i and 0.19-0.46i

Step-by-step explanation:

$\neq (-b+\sqrt{b^2-4ac} )/2a\\and\\(-b-\sqrt{b^2-4ac} )/2a\\\\a=8\\b=3\\c=2\\(-3+\sqrt{3^2-4*8*2} )/(2*8)\\= - 0.1875+0.463512i\\(-3-\sqrt{3^2-4*8*2} )/(2*8) = -0.1875-0.463512i$

hope its right

6 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

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

7 0

3 years ago

Write the missing decimal so the each pair adds to 1

oee [108]

Answer:

top one is 0.6 bottom one is 0.1

7 0

3 years ago