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3 years ago

Solve for x in the literal equation x/8-g= a

Mathematics

1 answer:

sweet-ann [11.9K]3 years ago

5 0

Answer:

x= 8a+ 8g

Step-by-step explanation:

x/8 -g = a

x/8 = a+g

× 8

x= 8a+ 8g

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2 years ago

Which of the following statements describes one part of completing the square for x 2 + 4x = 32?

Westkost [7]

SOS:

Answer:

Take the square root of 36 and subtract 2.

Step-by-step explanation:

$x^2+4x=32\\(\frac{1}{2}*4=2; 2^2=4) : x^2+4x+4=36\\\sqrt{(x+2)^2}=\sqrt{36}\\ x+2=±6: -2+6=4, -2-6=-8$

<em>Hope this helps!!</em>

7 0

3 years ago

Find the length of the curve. R(t) = cos(8t) i + sin(8t) j + 8 ln cos t k, 0 ≤ t ≤ π/4

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

$L=\int\limits^a_b {\sqrt{(x')^2+(y')^2+(z')^2} } \, dt$

$x'=-8sin(8t),y=8cos(8t),z=-8tan(t)$

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

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

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

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

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

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

and we get

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

so,

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

5 0

3 years ago