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3 years ago

What is the answer to 2b + 8- 5b + 3 = -13 + 8b - 5

2 answers:

8 0

B= 2 7/11

EXPLANATION : 2b-5b = -3b & 8+3 =11 , that’s leaves you with 3b+11=-13+8b-5 . do it to the other side . -13-5=-18 . leaves you with the equation -3b+11=-18+8b !!
move the variable to the left side & the constant to the right . which leads you to -3b-8b=-18-11 !! combine like terms & that takes you to -11b=-29 !! Divide both sides by -11 which makes B = 29/11 which is equivalent to 2 7/11. or 2.63

Alborosie3 years ago

7 0

Answer:

2 8/11 = b

Step-by-step explanation:

2b + 8 - 5b + 3 = -13 + 8b - 5

combine like terms

11 - 3b = -18 + 8b

add 3b

11 = -18 + 11b

add 18

29 = 11b

divide by 11

2 8/11 = b

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See explanation

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$f'(x)=2x$

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$b=0\\ \\a=2$

Therefore,

$SA=2\pi \int\limits^2_0 x^2\sqrt{1+(2x)^2} \, dx=2\pi \int\limits^2_0 x^2\sqrt{1+4x^2} \, dx$

Apply substitution

$x=\dfrac{1}{2}\tan u\\ \\dx=\dfrac{1}{2}\cdot \dfrac{1}{\cos ^2 u}du$

Then

$SA=2\pi \int\limits^2_0 x^2\sqrt{1+4x^2} \, dx=2\pi \int\limits^{\arctan(4)}_0 \dfrac{1}{4}\tan^2u\sqrt{1+\tan^2u} \, \dfrac{1}{2}\dfrac{1}{\cos^2u}du=\\ \\=\dfrac{\pi}{4}\int\limits^{\arctan(4)}_0 \tan^2u\sec^3udu=\dfrac{\pi}{4}\int\limits^{\arctan(4)}_0(\sec^3u+\sec^5u)du$

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$\int\limits^{\arctan(4)}_0 \sec^3udu=2\sqrt{17}+\dfrac{1}{2}\ln (4+\sqrt{17})\\ \\ \int\limits^{\arctan(4)}_0 \sec^5udu=\dfrac{1}{8}(-(2\sqrt{17}+\dfrac{1}{2}\ln(4+\sqrt{17})))+17\sqrt{17}+\dfrac{3}{4}(2\sqrt{17}+\dfrac{1}{2}\ln (4+\sqrt{17}))$

Hence,

$SA=\pi \dfrac{-\ln(4+\sqrt{17})+132\sqrt{17}}{32}$

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