Answer:
It’s a ratio
Step-by-step explanation:
Is that the whole question?
Check out the attached image below. I added a variable y to help find x
tan(angle) = opposite/adjacent
Focusing on the smaller 30-60-90 triangle, we see that
tan(60) = 91/y
y*tan(60) = 91
y = 91/tan(60)
y = 52.53887
now focus on the largest triangle and we see that
tan(30) = 91/(x+y)
0.57735 = 91/(x+52.53887)
0.57735*(x+52.53887) = 91
0.57735*x + 0.57735*52.53887 = 91
0.57735*x + 30.3333165945 = 91
0.57735*x + 30.3333165945 - 30.3333165945 = 91 - 30.3333165945
0.57735*x = 60.6666834055
x = 60.6666834055/0.57735
x = 105.07782697757
x = 105.1
So the final answer is approximately 105.1
Answer:
72
Step-by-step explanation:
One way to do this is to use the formula
where
and
are the lengths of the two diagonals.
In the figure,
and
(2 + 10 and 6 + 6).
The area is
square units.
Another way that doesn't depend on learning a formula is to remember that the diagonals of a kite are perpendicular, so the 4 small triangles are right triangles. Two of the smaller triangles have base =2 and height = 6, so their areas are 1/2(2)(6) = 6. Double that--there are 2 congruent small triangles--to get an area of 12.
There are two larger right triangles with base =10 and height = 6, so their area are 1/2(10)(6)=30. There are two of those, so their combined area is 60.
Finally, put the small triangle area and the large triangle area together to get a total of 12 + 60 = 72
Answer:
Given that,
The number of grams A of a certain radioactive substance present at time, in years
from the present, t is given by the formula
![A=45e^{-0.0045(t)}](https://tex.z-dn.net/?f=A%3D45e%5E%7B-0.0045%28t%29%7D)
a) To find the initial amount of this substance
At t=0, we get
![A=45e^{-0.0045(0)}](https://tex.z-dn.net/?f=A%3D45e%5E%7B-0.0045%280%29%7D)
![A=45e^0](https://tex.z-dn.net/?f=A%3D45e%5E0)
We know that e^0=1 ( anything to the power zero is 1)
we get,
![A=45](https://tex.z-dn.net/?f=A%3D45)
The initial amount of the substance is 45 grams
b)To find thehalf-life of this substance
To find t when the substance becames half the amount.
A=45/2
Substitute this we get,
![\frac{45}{2}=45e^{-0.0045(t)}](https://tex.z-dn.net/?f=%5Cfrac%7B45%7D%7B2%7D%3D45e%5E%7B-0.0045%28t%29%7D)
![\frac{1}{2}=e^{-0.0045(t)}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%3De%5E%7B-0.0045%28t%29%7D)
Taking natural logarithm on both sides we get,
![\ln (\frac{1}{2})=-0.0045(t)^{}](https://tex.z-dn.net/?f=%5Cln%20%28%5Cfrac%7B1%7D%7B2%7D%29%3D-0.0045%28t%29%5E%7B%7D)
![(-1)\ln (\frac{1}{2})=0.0045(t)](https://tex.z-dn.net/?f=%28-1%29%5Cln%20%28%5Cfrac%7B1%7D%7B2%7D%29%3D0.0045%28t%29)
![\ln (\frac{1}{2})^{-1}=0.0045(t)](https://tex.z-dn.net/?f=%5Cln%20%28%5Cfrac%7B1%7D%7B2%7D%29%5E%7B-1%7D%3D0.0045%28t%29)
![\ln (2)=0.0045(t)](https://tex.z-dn.net/?f=%5Cln%20%282%29%3D0.0045%28t%29)
![0.6931=0.0045(t)](https://tex.z-dn.net/?f=0.6931%3D0.0045%28t%29)
![t=\frac{0.6931}{0.0045}](https://tex.z-dn.net/?f=t%3D%5Cfrac%7B0.6931%7D%7B0.0045%7D)
![t=154.02](https://tex.z-dn.net/?f=t%3D154.02)
Half-life of this substance is 154.02
c) To find the amount of substance will be present around in 2500 years
Put t=2500
we get,
![A=45e^{-0.0045(2500)}](https://tex.z-dn.net/?f=A%3D45e%5E%7B-0.0045%282500%29%7D)
![A=45e^{-11.25}](https://tex.z-dn.net/?f=A%3D45e%5E%7B-11.25%7D)
![A=45\times0.000013=0.000585](https://tex.z-dn.net/?f=A%3D45%5Ctimes0.000013%3D0.000585)
![A=0.000585](https://tex.z-dn.net/?f=A%3D0.000585)
The amount of substance will be present around in 2500 years is 0.000585 grams