Answer:
![\cos(\alpha +\beta)=\frac{2}{3}(1-\frac{\sqrt{5}}{5})](https://tex.z-dn.net/?f=%5Ccos%28%5Calpha%20%2B%5Cbeta%29%3D%5Cfrac%7B2%7D%7B3%7D%281-%5Cfrac%7B%5Csqrt%7B5%7D%7D%7B5%7D%29)
Step-by-step explanation:
Let the hypotenuse of the smaller triangle be h units.
Then; from the Pythagoras Theorem.
![h^2=4^2+2^2](https://tex.z-dn.net/?f=h%5E2%3D4%5E2%2B2%5E2)
![h^2=16+4](https://tex.z-dn.net/?f=h%5E2%3D16%2B4)
![h^2=20](https://tex.z-dn.net/?f=h%5E2%3D20)
![h=\sqrt{20}](https://tex.z-dn.net/?f=h%3D%5Csqrt%7B20%7D)
![h=2\sqrt{5}](https://tex.z-dn.net/?f=h%3D2%5Csqrt%7B5%7D)
From the smaller triangle;
and ![\sin(\alpha)=\frac{2}{2\sqrt{5} }=\frac{1}{\sqrt{5} }](https://tex.z-dn.net/?f=%5Csin%28%5Calpha%29%3D%5Cfrac%7B2%7D%7B2%5Csqrt%7B5%7D%20%7D%3D%5Cfrac%7B1%7D%7B%5Csqrt%7B5%7D%20%7D)
From the second triangle, let the other other shorter leg of the second triangle be s units.
Then;
![s^2+4^2=6^2](https://tex.z-dn.net/?f=s%5E2%2B4%5E2%3D6%5E2)
![s^2+16=36](https://tex.z-dn.net/?f=s%5E2%2B16%3D36)
![s^2=36-16](https://tex.z-dn.net/?f=s%5E2%3D36-16)
![s^2=20](https://tex.z-dn.net/?f=s%5E2%3D20)
![s=\sqrt{20}](https://tex.z-dn.net/?f=s%3D%5Csqrt%7B20%7D)
![s=2\sqrt{5}](https://tex.z-dn.net/?f=s%3D2%5Csqrt%7B5%7D)
![\cos(\beta)=\frac{2\sqrt{5} }{6}=\frac{\sqrt{5} }{3}](https://tex.z-dn.net/?f=%5Ccos%28%5Cbeta%29%3D%5Cfrac%7B2%5Csqrt%7B5%7D%20%7D%7B6%7D%3D%5Cfrac%7B%5Csqrt%7B5%7D%20%7D%7B3%7D)
and
![\sin(\beta)=\frac{4}{6}=\frac{2}{3}](https://tex.z-dn.net/?f=%5Csin%28%5Cbeta%29%3D%5Cfrac%7B4%7D%7B6%7D%3D%5Cfrac%7B2%7D%7B3%7D)
We now use the double angle property;
![\cos(\alpha +\beta)=\cos(\alpha)\cos(\beta) -\sin(\alpha)\sin(\beta)](https://tex.z-dn.net/?f=%5Ccos%28%5Calpha%20%2B%5Cbeta%29%3D%5Ccos%28%5Calpha%29%5Ccos%28%5Cbeta%29%20-%5Csin%28%5Calpha%29%5Csin%28%5Cbeta%29)
we plug in the values to obtain;
![\cos(\alpha +\beta)=\frac{2}{\sqrt{5} }\times \frac{\sqrt{5} }{3}-\frac{1}{\sqrt{5} }\times \frac{2}{3}](https://tex.z-dn.net/?f=%5Ccos%28%5Calpha%20%2B%5Cbeta%29%3D%5Cfrac%7B2%7D%7B%5Csqrt%7B5%7D%20%7D%5Ctimes%20%5Cfrac%7B%5Csqrt%7B5%7D%20%7D%7B3%7D-%5Cfrac%7B1%7D%7B%5Csqrt%7B5%7D%20%7D%5Ctimes%20%5Cfrac%7B2%7D%7B3%7D)
![\cos(\alpha +\beta)=\frac{2}{3}(1-\frac{\sqrt{5}}{5})](https://tex.z-dn.net/?f=%5Ccos%28%5Calpha%20%2B%5Cbeta%29%3D%5Cfrac%7B2%7D%7B3%7D%281-%5Cfrac%7B%5Csqrt%7B5%7D%7D%7B5%7D%29)
Answer:
![\Large \boxed{\mathrm{\bold{D.}} \ f(x) = | 2x^2 + x | \ \mathrm{and} \ g(x) = (x + 1)}](https://tex.z-dn.net/?f=%5CLarge%20%5Cboxed%7B%5Cmathrm%7B%5Cbold%7BD.%7D%7D%20%5C%20f%28x%29%20%3D%20%7C%202x%5E2%20%2B%20x%20%7C%20%5C%20%5Cmathrm%7Band%7D%20%5C%20g%28x%29%20%3D%20%28x%20%2B%201%29%7D)
Step-by-step explanation:
![f(g(x)) = | 2(x + 1)^2 + (x + 1) |](https://tex.z-dn.net/?f=f%28g%28x%29%29%20%3D%20%7C%202%28x%20%2B%201%29%5E2%20%2B%20%28x%20%2B%201%29%20%7C)
The first option :
![f(x) = (x + 1)^2 \ \mathrm{and} \ g(x) = | 2x + 1 |](https://tex.z-dn.net/?f=f%28x%29%20%3D%20%28x%20%2B%201%29%5E2%20%5C%20%5Cmathrm%7Band%7D%20%5C%20g%28x%29%20%3D%20%7C%202x%20%2B%201%20%7C)
![f(g(x))=(|2x+1|+1)^2](https://tex.z-dn.net/?f=f%28g%28x%29%29%3D%28%7C2x%2B1%7C%2B1%29%5E2)
The second option :
![f(x) = (x + 1) \ \mathrm{and} \ g(x) = | 2x^2 + x |](https://tex.z-dn.net/?f=f%28x%29%20%3D%20%28x%20%2B%201%29%20%5C%20%5Cmathrm%7Band%7D%20%5C%20g%28x%29%20%3D%20%7C%202x%5E2%20%2B%20x%20%7C)
![f(g(x))=(|2x^2 +x|+1)](https://tex.z-dn.net/?f=f%28g%28x%29%29%3D%28%7C2x%5E2%20%2Bx%7C%2B1%29)
The third option :
![f(x) = | 2x + 1 | \ \mathrm{and} \ g(x) = (x + 1)^2](https://tex.z-dn.net/?f=f%28x%29%20%3D%20%7C%202x%20%2B%201%20%7C%20%5C%20%5Cmathrm%7Band%7D%20%5C%20g%28x%29%20%3D%20%28x%20%2B%201%29%5E2)
![f(g(x))=|2(x+1)^2 +1 |](https://tex.z-dn.net/?f=f%28g%28x%29%29%3D%7C2%28x%2B1%29%5E2%20%2B1%20%7C)
The fourth option :
![f(x) = | 2x^2 + x | \ \mathrm{and} \ g(x) = (x + 1)](https://tex.z-dn.net/?f=f%28x%29%20%3D%20%7C%202x%5E2%20%2B%20x%20%7C%20%5C%20%5Cmathrm%7Band%7D%20%5C%20g%28x%29%20%3D%20%28x%20%2B%201%29)
![f(g(x))= | 2(x + 1)^2 + (x + 1) |](https://tex.z-dn.net/?f=f%28g%28x%29%29%3D%20%7C%202%28x%20%2B%201%29%5E2%20%2B%20%28x%20%2B%201%29%20%7C)
The slope of a line is rise/run. The formula is y=mx+b. But to find it on a graph you have to pin point 2 coordinate points. Doesn’t matter which ones they are. And then you have to put them in a y2-y1 over x2-x1 and that will give you slope.
The correct answer is D.14.
The interquartile range is the difference between the Upper Quartile and the Lower Quartile of a data set.
By arranging the numbers in the stem and leaf plot from lowest to highest value, you get an even set of data (31,33,35,41,43,46,48,49,49,50). By dividing the set into two, you get a lower half of your data (31,33,35,41,43) and an upper half (46,48,49,49,50). The lower quartile is the median or the midpoint of the lower half of your data while upper quartile is for the upper half. The lower quartiles is 35 and upper quartile is 49. The difference between the two is 14.