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3 years ago

Factor completely by grouping: 6x² + 2x + 9x +3 is

Mathematics

2 answers:

masha68 [24]3 years ago

8 0

Answer:

(2x+3)(x+1)

Step-by-step explanation:

6x² + 2x + 9x +3

(6x² + 2x) + (9x +3)

2x (x+1) + 3 (x+1)

(2x+3)(x+1)

Pavlova-9 [17]3 years ago

3 0

Answer:

(3x+1) ( 2x+3)

Step-by-step explanation:

6x² + 2x + 9x +3

Make the groups

6x² + 2x + 9x +3

Factor 2x out of the first group and 3 out of the second group

2x( 3x+1) + 3(3 x+1)

Now factor out the common term inside the parentheses

(3x+1) ( 2x+3)

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arsen [322]

we are given

$R(t)=cos(8t)i+sin(8t)j+8ln(cos(t))k$

now, we can find x , y and z components

$x=cos(8t),y=sin(8t),z=8ln(cos(t))$

Arc length calculation:

we can use formula

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now, we can plug these values

$L=\int _0^{\frac{\pi }{4}}\sqrt{(-8sin(8t))^2+(8cos(8t))^2+(-8tan(t))^2} dt$

now, we can simplify it

$L=\int _0^{\frac{\pi }{4}}\sqrt{64+64tan^2(t)} dt$

$L=\int _0^{\frac{\pi }{4}}8\sqrt{1+tan^2(t)} dt$

$L=\int _0^{\frac{\pi }{4}}8\sqrt{sec^2(t)} dt$

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now, we can solve integral

$\int \:8\sec \left(t\right)dt$

$=8\ln \left|\tan \left(t\right)+\sec \left(t\right)\right|$

now, we can plug bounds

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$=8\ln \left(\sqrt{2}+1\right)-0$

so,

$L=8\ln \left(1+\sqrt{2}\right)$ ..............Answer

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