Answer:
A) -2
Step-by-step explanation:
The form is indeterminate at x=0, so L'Hopital's rule applies. The resulting form is also indeterminate at x=0, so a second application is required.
Let f(x) = x·sin(x); g(x) = cos(x) -1
Then f'(x) = sin(x) +x·cos(x), and g'(x) = -sin(x).
We still have f'(0)/g'(0) = 0/0 . . . . . indeterminate.
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Differentiating numerator and denominator a second time gives ...
f''(x) = 2cos(x) -sin(x)
g''(x) = -cos(x)
Then f''(0)/g''(0) = 2/-1 = -2
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I like to start by graphing the expression to see if that is informative as to what the limit should be. The graph suggests the limit is -2, as we found.
Answer:
3. 16
Explanation:
Factor the numerator and denominator and cancel the common factors.
Answer:
Your answer is absolutely correct
Step-by-step explanation:
The work would be as follows:
![\int _0^{\sqrt{\pi }}4x^3\cos \left(x^2\right)dx,\\\\\mathrm{Take\:the\:constant\:out}:\quad \int a\cdot f\left(x\right)dx=a\cdot \int f\left(x\right)dx\\=> 4\cdot \int _0^{\sqrt{\pi }}x^3\cos \left(x^2\right)dx\\\\\mathrm{Apply\:u-substitution:}\:u=x^2\\=> 4\cdot \int _0^{\pi }\frac{u\cos \left(u\right)}{2}du\\\\\mathrm{Apply\:Integration\:By\:Parts:}\:u=u,\:v'=\cos \left(u\right)\\=> 4\cdot \frac{1}{2}\left[u\sin \left(u\right)-\int \sin \left(u\right)du\right]^{\pi }_0\\\\](https://tex.z-dn.net/?f=%5Cint%20_0%5E%7B%5Csqrt%7B%5Cpi%20%7D%7D4x%5E3%5Ccos%20%5Cleft%28x%5E2%5Cright%29dx%2C%5C%5C%5C%5C%5Cmathrm%7BTake%5C%3Athe%5C%3Aconstant%5C%3Aout%7D%3A%5Cquad%20%5Cint%20a%5Ccdot%20f%5Cleft%28x%5Cright%29dx%3Da%5Ccdot%20%5Cint%20f%5Cleft%28x%5Cright%29dx%5C%5C%3D%3E%204%5Ccdot%20%5Cint%20_0%5E%7B%5Csqrt%7B%5Cpi%20%7D%7Dx%5E3%5Ccos%20%5Cleft%28x%5E2%5Cright%29dx%5C%5C%5C%5C%5Cmathrm%7BApply%5C%3Au-substitution%3A%7D%5C%3Au%3Dx%5E2%5C%5C%3D%3E%204%5Ccdot%20%5Cint%20_0%5E%7B%5Cpi%20%7D%5Cfrac%7Bu%5Ccos%20%5Cleft%28u%5Cright%29%7D%7B2%7Ddu%5C%5C%5C%5C%5Cmathrm%7BApply%5C%3AIntegration%5C%3ABy%5C%3AParts%3A%7D%5C%3Au%3Du%2C%5C%3Av%27%3D%5Ccos%20%5Cleft%28u%5Cright%29%5C%5C%3D%3E%204%5Ccdot%20%5Cfrac%7B1%7D%7B2%7D%5Cleft%5Bu%5Csin%20%5Cleft%28u%5Cright%29-%5Cint%20%5Csin%20%5Cleft%28u%5Cright%29du%5Cright%5D%5E%7B%5Cpi%20%7D_0%5C%5C%5C%5C)
![\int \sin \left(u\right)du=-\cos \left(u\right)\\=> 4\cdot \frac{1}{2}\left[u\sin \left(u\right)-\left(-\cos \left(u\right)\right)\right]^{\pi }_0\\\\\mathrm{Simplify\:}4\cdot \frac{1}{2}\left[u\sin \left(u\right)-\left(-\cos \left(u\right)\right)\right]^{\pi }_0:\quad 2\left[u\sin \left(u\right)+\cos \left(u\right)\right]^{\pi }_0\\\\\mathrm{Compute\:the\:boundaries}:\quad \left[u\sin \left(u\right)+\cos \left(u\right)\right]^{\pi }_0=-2\\=> 2(-2) = - 4](https://tex.z-dn.net/?f=%5Cint%20%5Csin%20%5Cleft%28u%5Cright%29du%3D-%5Ccos%20%5Cleft%28u%5Cright%29%5C%5C%3D%3E%204%5Ccdot%20%5Cfrac%7B1%7D%7B2%7D%5Cleft%5Bu%5Csin%20%5Cleft%28u%5Cright%29-%5Cleft%28-%5Ccos%20%5Cleft%28u%5Cright%29%5Cright%29%5Cright%5D%5E%7B%5Cpi%20%7D_0%5C%5C%5C%5C%5Cmathrm%7BSimplify%5C%3A%7D4%5Ccdot%20%5Cfrac%7B1%7D%7B2%7D%5Cleft%5Bu%5Csin%20%5Cleft%28u%5Cright%29-%5Cleft%28-%5Ccos%20%5Cleft%28u%5Cright%29%5Cright%29%5Cright%5D%5E%7B%5Cpi%20%7D_0%3A%5Cquad%202%5Cleft%5Bu%5Csin%20%5Cleft%28u%5Cright%29%2B%5Ccos%20%5Cleft%28u%5Cright%29%5Cright%5D%5E%7B%5Cpi%20%7D_0%5C%5C%5C%5C%5Cmathrm%7BCompute%5C%3Athe%5C%3Aboundaries%7D%3A%5Cquad%20%5Cleft%5Bu%5Csin%20%5Cleft%28u%5Cright%29%2B%5Ccos%20%5Cleft%28u%5Cright%29%5Cright%5D%5E%7B%5Cpi%20%7D_0%3D-2%5C%5C%3D%3E%202%28-2%29%20%3D%20-%204)
Hence proved that your solution is accurate.
Answer:
≈419.4 cm.
Step-by-step explanation:
To solve this problem, you will need to set up an exponential graph as shown:
y= 40 ·
The 40 represents the initial height of the tree, while (1.6) represents the annual growth rate. 'x' represents the amount of years elapsed.
We can simply solve this by substituting 5 for x:
y= 40· 
y ≈ 419.4 cm.
Answer:
17) #1 = 5
#2 = 49
18) Median= 50.5
Range = 88
Step-by-step explanation:
17) #1 - There are 5 3’s on the leaf side.
#2 - The number that is shown the most times is 49
18) There are 30 numbers in the stem and leaf plot, so the median falls between numbers 15 and 16. In this plot, those numbers are 50 and 51. So the median is 50.5.
The data ranges from 10 to 98. 98-10=88
