I'm pretty sure they would show B. Death rates. Hope this helps!
Answer : The value of
is
.
Explanation :
As we are given 6 right angled triangle in the given figure.
First we have to calculate the value of
.
Using Pythagoras theorem in triangle 1 :
![(Hypotenuse)^2=(Perpendicular)^2+(Base)^2](https://tex.z-dn.net/?f=%28Hypotenuse%29%5E2%3D%28Perpendicular%29%5E2%2B%28Base%29%5E2)
![(x_1)^2=(1)^2+(1)^2](https://tex.z-dn.net/?f=%28x_1%29%5E2%3D%281%29%5E2%2B%281%29%5E2)
![x_1=\sqrt{(1)^2+(1)^2}](https://tex.z-dn.net/?f=x_1%3D%5Csqrt%7B%281%29%5E2%2B%281%29%5E2%7D)
![x_1=\sqrt{2}](https://tex.z-dn.net/?f=x_1%3D%5Csqrt%7B2%7D)
Now we have to calculate the value of
.
Using Pythagoras theorem in triangle 2 :
![(Hypotenuse)^2=(Perpendicular)^2+(Base)^2](https://tex.z-dn.net/?f=%28Hypotenuse%29%5E2%3D%28Perpendicular%29%5E2%2B%28Base%29%5E2)
![(x_2)^2=(1)^2+(X_1)^2](https://tex.z-dn.net/?f=%28x_2%29%5E2%3D%281%29%5E2%2B%28X_1%29%5E2)
![(x_2)^2=(1)^2+(\sqrt{2})^2](https://tex.z-dn.net/?f=%28x_2%29%5E2%3D%281%29%5E2%2B%28%5Csqrt%7B2%7D%29%5E2)
![x_2=\sqrt{(1)^2+(\sqrt{2})^2}](https://tex.z-dn.net/?f=x_2%3D%5Csqrt%7B%281%29%5E2%2B%28%5Csqrt%7B2%7D%29%5E2%7D)
![x_2=\sqrt{3}](https://tex.z-dn.net/?f=x_2%3D%5Csqrt%7B3%7D)
Now we have to calculate the value of
.
Using Pythagoras theorem in triangle 3 :
![(Hypotenuse)^2=(Perpendicular)^2+(Base)^2](https://tex.z-dn.net/?f=%28Hypotenuse%29%5E2%3D%28Perpendicular%29%5E2%2B%28Base%29%5E2)
![(x_3)^2=(1)^2+(X_2)^2](https://tex.z-dn.net/?f=%28x_3%29%5E2%3D%281%29%5E2%2B%28X_2%29%5E2)
![(x_3)^2=(1)^2+(\sqrt{3})^2](https://tex.z-dn.net/?f=%28x_3%29%5E2%3D%281%29%5E2%2B%28%5Csqrt%7B3%7D%29%5E2)
![x_3=\sqrt{(1)^2+(\sqrt{3})^2}](https://tex.z-dn.net/?f=x_3%3D%5Csqrt%7B%281%29%5E2%2B%28%5Csqrt%7B3%7D%29%5E2%7D)
![x_3=\sqrt{4}](https://tex.z-dn.net/?f=x_3%3D%5Csqrt%7B4%7D)
Now we have to calculate the value of
.
Using Pythagoras theorem in triangle 4 :
![(Hypotenuse)^2=(Perpendicular)^2+(Base)^2](https://tex.z-dn.net/?f=%28Hypotenuse%29%5E2%3D%28Perpendicular%29%5E2%2B%28Base%29%5E2)
![(x_4)^2=(1)^2+(X_3)^2](https://tex.z-dn.net/?f=%28x_4%29%5E2%3D%281%29%5E2%2B%28X_3%29%5E2)
![(x_4)^2=(1)^2+(\sqrt{4})^2](https://tex.z-dn.net/?f=%28x_4%29%5E2%3D%281%29%5E2%2B%28%5Csqrt%7B4%7D%29%5E2)
![x_4=\sqrt{(1)^2+(\sqrt{4})^2}](https://tex.z-dn.net/?f=x_4%3D%5Csqrt%7B%281%29%5E2%2B%28%5Csqrt%7B4%7D%29%5E2%7D)
![x_4=\sqrt{5}](https://tex.z-dn.net/?f=x_4%3D%5Csqrt%7B5%7D)
Now we have to calculate the value of
.
Using Pythagoras theorem in triangle 5 :
![(Hypotenuse)^2=(Perpendicular)^2+(Base)^2](https://tex.z-dn.net/?f=%28Hypotenuse%29%5E2%3D%28Perpendicular%29%5E2%2B%28Base%29%5E2)
![(x_5)^2=(1)^2+(X_4)^2](https://tex.z-dn.net/?f=%28x_5%29%5E2%3D%281%29%5E2%2B%28X_4%29%5E2)
![(x_5)^2=(1)^2+(\sqrt{5})^2](https://tex.z-dn.net/?f=%28x_5%29%5E2%3D%281%29%5E2%2B%28%5Csqrt%7B5%7D%29%5E2)
![x_5=\sqrt{(1)^2+(\sqrt{5})^2}](https://tex.z-dn.net/?f=x_5%3D%5Csqrt%7B%281%29%5E2%2B%28%5Csqrt%7B5%7D%29%5E2%7D)
![x_5=\sqrt{6}](https://tex.z-dn.net/?f=x_5%3D%5Csqrt%7B6%7D)
Now we have to calculate the value of
.
Using Pythagoras theorem in triangle 6 :
![(Hypotenuse)^2=(Perpendicular)^2+(Base)^2](https://tex.z-dn.net/?f=%28Hypotenuse%29%5E2%3D%28Perpendicular%29%5E2%2B%28Base%29%5E2)
![(x_6)^2=(1)^2+(X_5)^2](https://tex.z-dn.net/?f=%28x_6%29%5E2%3D%281%29%5E2%2B%28X_5%29%5E2)
![(x_6)^2=(1)^2+(\sqrt{6})^2](https://tex.z-dn.net/?f=%28x_6%29%5E2%3D%281%29%5E2%2B%28%5Csqrt%7B6%7D%29%5E2)
![x_6=\sqrt{(1)^2+(\sqrt{6})^2}](https://tex.z-dn.net/?f=x_6%3D%5Csqrt%7B%281%29%5E2%2B%28%5Csqrt%7B6%7D%29%5E2%7D)
![x_6=\sqrt{7}](https://tex.z-dn.net/?f=x_6%3D%5Csqrt%7B7%7D)
Therefore, the value of
is
.
the answer will be d because space is always cold but that doesn't mean that the precipitation is the same temperature
Explanation:
Erosion erodes rocks and such to form sand. Deposition moves the sand to another place ro form the dunes.
Answer:
Option B is correct.
Explanation:
Infiltration depends greatly on the features of the soil. If the soil has excessive clay or mud, water will not find a way into the groundwater system and will remain as surface water.
An alluvial fan is a perfect location for this since there will be little or no resistance to the flowing water. The streams are already a connection with the groundwater system, and falling water will flow directly to it.