Hope this attached file will help u.
To know if Diagonals are perpendicular we need to know two line's slope
If two lines are pependicular their slope m1×m2=-1
For if they are making rhombus two opposite equal sides so u need to know the distancebetween the points.
Answer:
It diverges.
Step-by-step explanation:
We are given the inetegral: ![\int\limits^{\infty}_2 \frac{1}{x} (\ln x)^2 dx](https://tex.z-dn.net/?f=%5Cint%5Climits%5E%7B%5Cinfty%7D_2%20%5Cfrac%7B1%7D%7Bx%7D%20%28%5Cln%20x%29%5E2%20dx)
![\int\limits^{\infty}_2 \frac{1}{x} (\ln x)^2 dx=\int\limits^{\infty}_2 (\ln x)^2 d(\ln x)=\\\\=\lim_{t \to \infty} \int\limits^t_2 (\ln x)^2d(\ln x)=\lim_{t \to \infty} \frac{(\ln t)^3}{3} |^t_2=\infty-\frac{(\ln 2)^3}{3} =\infty](https://tex.z-dn.net/?f=%5Cint%5Climits%5E%7B%5Cinfty%7D_2%20%5Cfrac%7B1%7D%7Bx%7D%20%28%5Cln%20x%29%5E2%20dx%3D%5Cint%5Climits%5E%7B%5Cinfty%7D_2%20%28%5Cln%20x%29%5E2%20d%28%5Cln%20x%29%3D%5C%5C%5C%5C%3D%5Clim_%7Bt%20%5Cto%20%5Cinfty%7D%20%5Cint%5Climits%5Et_2%20%28%5Cln%20x%29%5E2d%28%5Cln%20x%29%3D%5Clim_%7Bt%20%5Cto%20%5Cinfty%7D%20%5Cfrac%7B%28%5Cln%20t%29%5E3%7D%7B3%7D%20%7C%5Et_2%3D%5Cinfty-%5Cfrac%7B%28%5Cln%202%29%5E3%7D%7B3%7D%20%3D%5Cinfty)
So it is divergent.
Answer:
![(a)-\dfrac{11\sqrt{5}}{25} \\(b) -\dfrac{2\sqrt{5}}{25} \\(c)\dfrac{11}{2}](https://tex.z-dn.net/?f=%28a%29-%5Cdfrac%7B11%5Csqrt%7B5%7D%7D%7B25%7D%20%5C%5C%28b%29%20-%5Cdfrac%7B2%5Csqrt%7B5%7D%7D%7B25%7D%20%5C%5C%28c%29%5Cdfrac%7B11%7D%7B2%7D)
Step-by-step explanation:
![\tan \alpha =\dfrac12, \pi < \alpha< \dfrac{3 \pi}{2}](https://tex.z-dn.net/?f=%5Ctan%20%5Calpha%20%3D%5Cdfrac12%2C%20%5Cpi%20%3C%20%5Calpha%3C%20%5Cdfrac%7B3%20%5Cpi%7D%7B2%7D)
Therefore:
![\alpha$ is in Quadrant III](https://tex.z-dn.net/?f=%5Calpha%24%20is%20in%20Quadrant%20III)
Opposite = -1
Adjacent =-2
Using Pythagoras Theorem
![Hypotenuse^2=Opposite^2+Adjacent^2\\=(-1)^2+(-2)^2=5\\Hypotenuse=\sqrt{5}](https://tex.z-dn.net/?f=Hypotenuse%5E2%3DOpposite%5E2%2BAdjacent%5E2%5C%5C%3D%28-1%29%5E2%2B%28-2%29%5E2%3D5%5C%5CHypotenuse%3D%5Csqrt%7B5%7D)
Therefore:
![\sin \alpha =-\dfrac{1}{\sqrt{5}}\\\cos \alpha =-\dfrac{2}{\sqrt{5}}](https://tex.z-dn.net/?f=%5Csin%20%5Calpha%20%3D-%5Cdfrac%7B1%7D%7B%5Csqrt%7B5%7D%7D%5C%5C%5Ccos%20%5Calpha%20%3D-%5Cdfrac%7B2%7D%7B%5Csqrt%7B5%7D%7D)
Similarly
![\cos \beta =\dfrac35, \dfrac{3 \pi}{2}](https://tex.z-dn.net/?f=%5Ccos%20%5Cbeta%20%3D%5Cdfrac35%2C%20%5Cdfrac%7B3%20%5Cpi%7D%7B2%7D%3C%5Cbeta%3C2%5Cpi%5C%5C%5Cbeta%20%24%20is%20in%20Quadrant%20IV%20%28x%20is%20negative%2C%20y%20is%20positive%29%2C%20therefore%3A%5C%5CAdjacent%3D%24-3%5C%5C%24Hypotenuse%3D5%5C%5COpposite%3D4%20%20%28Using%20Pythagoras%20Theorem%29)
![\sin \beta =\dfrac{4}{5}\\\tan \beta =-\dfrac{4}{3}](https://tex.z-dn.net/?f=%5Csin%20%5Cbeta%20%3D%5Cdfrac%7B4%7D%7B5%7D%5C%5C%5Ctan%20%5Cbeta%20%3D-%5Cdfrac%7B4%7D%7B3%7D)
(a)
![\cos(\alpha + \beta)=\cos\alpha\cos\beta-\sin \alpha\sin \beta\\](https://tex.z-dn.net/?f=%5Ccos%28%5Calpha%20%2B%20%5Cbeta%29%3D%5Ccos%5Calpha%5Ccos%5Cbeta-%5Csin%20%5Calpha%5Csin%20%5Cbeta%5C%5C)
![=-\dfrac{2}{\sqrt{5}}\cdot \dfrac{3}{5}-(-\dfrac{1}{\sqrt{5}})(\dfrac{4}{5})\\=-\dfrac{2\sqrt{5}}{5}\cdot \dfrac{3}{5}+\dfrac{\sqrt{5}}{5}\cdot\dfrac{4}{5}\\=-\dfrac{2\sqrt{5}}{25}](https://tex.z-dn.net/?f=%3D-%5Cdfrac%7B2%7D%7B%5Csqrt%7B5%7D%7D%5Ccdot%20%5Cdfrac%7B3%7D%7B5%7D-%28-%5Cdfrac%7B1%7D%7B%5Csqrt%7B5%7D%7D%29%28%5Cdfrac%7B4%7D%7B5%7D%29%5C%5C%3D-%5Cdfrac%7B2%5Csqrt%7B5%7D%7D%7B5%7D%5Ccdot%20%5Cdfrac%7B3%7D%7B5%7D%2B%5Cdfrac%7B%5Csqrt%7B5%7D%7D%7B5%7D%5Ccdot%5Cdfrac%7B4%7D%7B5%7D%5C%5C%3D-%5Cdfrac%7B2%5Csqrt%7B5%7D%7D%7B25%7D)
(b)
![\sin(\alpha + \beta)=\sin\alpha\cos\beta+\cos \alpha\sin \beta](https://tex.z-dn.net/?f=%5Csin%28%5Calpha%20%2B%20%5Cbeta%29%3D%5Csin%5Calpha%5Ccos%5Cbeta%2B%5Ccos%20%5Calpha%5Csin%20%5Cbeta)
![\sin(\alpha + \beta)=\sin\alpha\cos\beta+\cos \alpha\sin \beta\\=-\dfrac{1}{\sqrt{5}}\cdot\dfrac35+(-\dfrac{2}{\sqrt{5}})(\dfrac{4}{5})\\=-\dfrac{\sqrt{5}}{5}\cdot\dfrac35-\dfrac{2\sqrt{5}}{5}\cdot\dfrac{4}{5}\\=-\dfrac{11\sqrt{5}}{25}](https://tex.z-dn.net/?f=%5Csin%28%5Calpha%20%2B%20%5Cbeta%29%3D%5Csin%5Calpha%5Ccos%5Cbeta%2B%5Ccos%20%5Calpha%5Csin%20%5Cbeta%5C%5C%3D-%5Cdfrac%7B1%7D%7B%5Csqrt%7B5%7D%7D%5Ccdot%5Cdfrac35%2B%28-%5Cdfrac%7B2%7D%7B%5Csqrt%7B5%7D%7D%29%28%5Cdfrac%7B4%7D%7B5%7D%29%5C%5C%3D-%5Cdfrac%7B%5Csqrt%7B5%7D%7D%7B5%7D%5Ccdot%5Cdfrac35-%5Cdfrac%7B2%5Csqrt%7B5%7D%7D%7B5%7D%5Ccdot%5Cdfrac%7B4%7D%7B5%7D%5C%5C%3D-%5Cdfrac%7B11%5Csqrt%7B5%7D%7D%7B25%7D)
(c)
![\tan(\alpha + \beta)=\dfrac{\sin(\alpha + \beta)}{\sin(\alpha + \beta)}=-\dfrac{11\sqrt{5}}{25} \div -\dfrac{2\sqrt{5}}{25} =\dfrac{11}{2}](https://tex.z-dn.net/?f=%5Ctan%28%5Calpha%20%2B%20%5Cbeta%29%3D%5Cdfrac%7B%5Csin%28%5Calpha%20%2B%20%5Cbeta%29%7D%7B%5Csin%28%5Calpha%20%2B%20%5Cbeta%29%7D%3D-%5Cdfrac%7B11%5Csqrt%7B5%7D%7D%7B25%7D%20%5Cdiv%20-%5Cdfrac%7B2%5Csqrt%7B5%7D%7D%7B25%7D%20%3D%5Cdfrac%7B11%7D%7B2%7D)
The thousandth place is 4, which is less than 5 ⇒ Discard
339.4749 ≈ 339.47
Answer: 339.47
Answer:
4.9
Step-by-step explanation:
Let us first find the frequency from the cumulative frequency.
Starting number is same.
To find the next number subtract the previous number from that number.
i.e first number = 0.15
Second number = 0.45 – 0.15 = 0.3
Third number = 0.60 – 0.45 = 0.15
Fourth number = 0.85 – 0.60 = 0.25
Fifth number = 1.00 – 0.85 = 0.15
Value Frequency Value × Frequency
1 0.15 0.15
3 0.3 0.9
5 0.15 0.75
7 0.25 1.75
9 0.15 1.35
Number of frequency = 1
Total value = 0.15 + 0.9 + 0.75 + 1.75 + 1.35 = 4.9
Mean =
= 4.9