Answer:
67
Step-by-step explanation:
Since the diagonals in a rectangle are congruent, AC=BD and
7x+18=10x-3
We need to separate x -- to do that, we can first subtract 7x from both sides, resulting in
18=3x-3
Next, add 3 to both sides to get
21=3x
Divide both sides by 3 to get
x=7
Then, we just plug x into 10x-3 to get BD = 7*10-3 = 67
![\bf log_{{ a}}(xy)\implies log_{{ a}}(x)+log_{{ a}}(y) \\ \quad \\ % Logarithm of rationals \\ \quad \\ % Logarithm of exponentials log_{{ a}}\left( x^{{ b}} \right)\implies {{ b}}\cdot log_{{ a}}(x)\\\\ -----------------------------\\\\ log_b(xy^2z^{-6}\implies log_b(x)+log_b(y^2)+log_b(z^{-6}) \\\\\\log_b(x)+2log_b(y)-6log_b(z)](https://tex.z-dn.net/?f=%5Cbf%20log_%7B%7B%20%20a%7D%7D%28xy%29%5Cimplies%20log_%7B%7B%20%20a%7D%7D%28x%29%2Blog_%7B%7B%20%20a%7D%7D%28y%29%0A%5C%5C%20%5Cquad%20%5C%5C%0A%25%20Logarithm%20of%20rationals%0A%5C%5C%20%5Cquad%20%5C%5C%0A%25%20Logarithm%20of%20exponentials%0Alog_%7B%7B%20%20a%7D%7D%5Cleft%28%20x%5E%7B%7B%20%20b%7D%7D%20%5Cright%29%5Cimplies%20%7B%7B%20%20b%7D%7D%5Ccdot%20%20log_%7B%7B%20%20a%7D%7D%28x%29%5C%5C%5C%5C%0A-----------------------------%5C%5C%5C%5C%0Alog_b%28xy%5E2z%5E%7B-6%7D%5Cimplies%20log_b%28x%29%2Blog_b%28y%5E2%29%2Blog_b%28z%5E%7B-6%7D%29%0A%5C%5C%5C%5C%5C%5Clog_b%28x%29%2B2log_b%28y%29-6log_b%28z%29)
now, the one below that, which is equivalent to that? well, just look above it
I hope this helps you
radius=diameter/2=22/2=11
Area =pi.r^2
Area =3,14.11^2
Area =379.94
Answer:
% Remaining![= [1-(1/2)^{\frac{t}{2.6}}]x100](https://tex.z-dn.net/?f=%20%3D%20%5B1-%281%2F2%29%5E%7B%5Cfrac%7Bt%7D%7B2.6%7D%7D%5Dx100%20)
And replacing the value t =5.5 hours we got:
% Remaining![= [1-(1/2)^{\frac{5.5}{2.6}}]x100 =76.922\%](https://tex.z-dn.net/?f=%20%3D%20%5B1-%281%2F2%29%5E%7B%5Cfrac%7B5.5%7D%7B2.6%7D%7D%5Dx100%20%3D76.922%5C%25)
Step-by-step explanation:
Previous concepts
The half-life is defined "as the amount of time it takes a given quantity to decrease to half of its initial value. The term is most commonly used in relation to atoms undergoing radioactive decay, but can be used to describe other types of decay, whether exponential or not".
Solution to the problem
The half life model is given by the following expression:
![A(t) = A_o (1/2)^{\frac{t}{h}}](https://tex.z-dn.net/?f=A%28t%29%20%3D%20A_o%20%281%2F2%29%5E%7B%5Cfrac%7Bt%7D%7Bh%7D%7D)
Where A(t) represent the amount after t hours.
represent the initial amount
t the number of hours
h=2.6 hours the half life
And we want to estimate the % after 5.5 hours. On this case we can begin finding the amount after 5.5 hours like this:
![A(5.5) = A_o (1/2)^{\frac{5.5}{2.6}}](https://tex.z-dn.net/?f=A%285.5%29%20%3D%20A_o%20%281%2F2%29%5E%7B%5Cfrac%7B5.5%7D%7B2.6%7D%7D)
Now in order to find the percentage relative to the initial amount w can use the definition of relative change like this:
% Remaining = ![\frac{|A_o - A_o(1/2)^{\frac{5.5}{2.6}}|}{A_o} x100](https://tex.z-dn.net/?f=%5Cfrac%7B%7CA_o%20-%20A_o%281%2F2%29%5E%7B%5Cfrac%7B5.5%7D%7B2.6%7D%7D%7C%7D%7BA_o%7D%20x100)
We can take common factor
and we got:
% Remaining![= [1-(1/2)^{\frac{t}{2.6}}]x100](https://tex.z-dn.net/?f=%20%3D%20%5B1-%281%2F2%29%5E%7B%5Cfrac%7Bt%7D%7B2.6%7D%7D%5Dx100%20)
And replacing the value t =5.5 hours we got:
% Remaining ![= [1-(1/2)^{\frac{5.5}{2.6}}]x100 =76.922\%](https://tex.z-dn.net/?f=%3D%20%5B1-%281%2F2%29%5E%7B%5Cfrac%7B5.5%7D%7B2.6%7D%7D%5Dx100%20%3D76.922%5C%25)