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3 years ago

Tough one!

1 answer:

7 0

Imx->0 (asin2x + b log(cosx))/x4 = 1/2 [0/0 form] ,applying L'Hospital rule ,we get

= > limx->0 (2a*sinx*cosx - (b /cosx)*sinx)/ 4x3 = 1/2 => limx->0 (a*sin2x - b*tanx)/ 4x3 = 1/2 [0/0 form],

applying L'Hospital rule again ,we get,

= > limx->0 (2a*cos2x - b*sec2x) / 12x2 = 1/2

For above limit to exist,Numerator must be zero so that we get [0/0 form] & we can further proceed.

Hence 2a - b =0 => 2a = b ------(A)

limx->0 (b*cos2x - b*sec2x) / 12x2 = 1/2 [0/0 form], applying L'Hospital rule again ,we get,

= > limx->0 b*(-2sin2x - 2secx*secx.tanx) / 24x = 1/2 => limx->0 2b*[-sin2x - (1+tan2x)tanx] / 24x = 1/2

[0/0 form], applying L'Hospital rule again ,we get,

limx->0 2b*[-2cos2x - (sec2x+3tan2x*sec2x)] / 24 = 1/2 = > 2b[-2 -1] / 24 = 1/2 => -6b/24 = 1/2 => b = -2

from (A), we have , 2a = b => 2a = -2 => a = -1

Hence a =-1 & b = -2

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2 years ago

Youre purchasing a new television with a 32" screen. the 32" measurement represents the diagonal measurement of the screen. You

Nonamiya [84]

Answer:

B. 28 inches

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We can use the Pythagorean theorem to solve this problem.

a^2 +b^2 =c^2 where a and b are the legs and c is the hypotenuse (diagonal)

16^2 + b^2 = 32^2

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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

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3 years ago

Round 6.09933418366 to 4 decimal places

Sergio039 [100]

Answer:

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4 years ago

Please someone answer this please

gladu [14]

Option C:

$\frac{3 a^{2} b^{11}}{2 }$ is equivalent to the given expression.

Solution:

Given expression:

$$\frac{-18 a^{-2} b^{5}}{-12 a^{-4} b^{-6}}$

To find which expression is equivalent to the given expression.

$$\frac{-18 a^{-2} b^{5}}{-12 a^{-4} b^{-6}}$

Using exponent rule: $\frac{1}{a^m}=a^{-m}, \ \ \frac{1}{a^{-m}}=a^{m}$

$$=\frac{-18 a^{-2} b^{5}a^{4} b^{6}}{-12 }$

$$=\frac{-18 a^{-2} a^{4} b^{5} b^{6}}{-12 }$

Using exponent rule: ${a^m}\cdot{a^n}=a^{m+n}$

$$=\frac{-18 a^{(-2+4)} b^{(5+6)}}{-12 }$

$$=\frac{-18 a^{2} b^{11}}{-12 }$

Divide both numerator and denominator by the common factor –6.

$$=\frac{3 a^{2} b^{11}}{2 }$

$$\frac{-18 a^{-2} b^{5}}{-12 a^{-4} b^{-6}}=\frac{3 a^{2} b^{11}}{2 }$

Therefore, $\frac{3 a^{2} b^{11}}{2 }$ is equivalent to the given expression.

Hence Option C is the correct answer.

8 0

4 years ago