I found this!!!!
The scientist can use these two measurements to calculate the distance between the Sun and the shooting star by applying one of the trigonometric functions: Cosine of an angle.
- The scientist can substitute these measurements into cos\alpha=\frac{adjacent}{hypotenuse}cosα=
hypotenuse
adjacent
and solve for the distance between the Sun and the shooting star (which would be the hypotenuse of the righ triangle).
Step-by-step explanation:
You can observe in the figure attached that "AC" is the distance between the Sun and the shooting star.
Knowing the distance between the Earth and the Sun "y" and the angle x°, the scientist can use only these two measurements to calculate the distance between the Sun and the shooting star by applying one of the trigonometric functions: Cosine of an angle.
This is:
cos\alpha=\frac{adjacent}{hypotenuse}cosα=
hypotenuse
adjacent
In this case:
\begin{gathered}\alpha=x\°\\\\adjacent=BC=y\\\\hypotenuse=AC\end{gathered}
α=x\°
adjacent=BC=y
hypotenuse=AC
Therefore, the scientist can substitute these measurements into cos\alpha=\frac{adjacent}{hypotenuse}cosα=
hypotenuse
adjacent
, and solve for the distance between the Sun and the shooting star "AC":
cos(x\°)=\frac{y}{AC}cos(x\°)=
AC
y
AC=\frac{y}{cos(x\°)}AC=
cos(x\°)
y
(4,7)(0,7)
notice how ur points have the same y values....this means this is a horizontal line with a slope of 0.
equation is : y = 7...or y = 0x + 7...but we need it in standard form...
0x + y = 7 <== standard form
Answer:
45% of students got hamburgers :)
Step-by-step explanation:
Answer:
the maximum concentration of the antibiotic during the first 12 hours is 1.185
at t= 2 hours.
Step-by-step explanation:
We are given the following information:
After an antibiotic tablet is taken, the concentration of the antibiotic in the bloodstream is modeled by the function where the time t is measured in hours and C is measured in ![\mu g/mL](https://tex.z-dn.net/?f=%5Cmu%20g%2FmL)
![C(t) = 8(e^{(-0.4t)}-e^{(-0.6t)})](https://tex.z-dn.net/?f=C%28t%29%20%3D%208%28e%5E%7B%28-0.4t%29%7D-e%5E%7B%28-0.6t%29%7D%29)
Thus, we are given the time interval [0,12] for t.
- We can apply the first derivative test, to know the absolute maximum value because we have a closed interval for t.
- The first derivative test focusing on a particular point. If the function switches or changes from increasing to decreasing at the point, then the function will achieve a highest value at that point.
First, we differentiate C(t) with respect to t, to get,
![\frac{d(C(t))}{dt} = 8(-0.4e^{(-0.4t)}+ 0.6e^{(-0.6t)})](https://tex.z-dn.net/?f=%5Cfrac%7Bd%28C%28t%29%29%7D%7Bdt%7D%20%3D%208%28-0.4e%5E%7B%28-0.4t%29%7D%2B%200.6e%5E%7B%28-0.6t%29%7D%29)
Equating the first derivative to zero, we get,
![\frac{d(C(t))}{dt} = 0\\\\8(-0.4e^{(-0.4t)}+ 0.6e^{(-0.6t)}) = 0](https://tex.z-dn.net/?f=%5Cfrac%7Bd%28C%28t%29%29%7D%7Bdt%7D%20%3D%200%5C%5C%5C%5C8%28-0.4e%5E%7B%28-0.4t%29%7D%2B%200.6e%5E%7B%28-0.6t%29%7D%29%20%3D%200)
Solving, we get,
![8(-0.4e^{(-0.4t)}+ 0.6e^{(-0.6t)}) = 0\\\displaystyle\frac{e^{-0.4}}{e^{-0.6}} = \frac{0.6}{0.4}\\\\e^{0.2t} = 1.5\\\\t = \frac{ln(1.5)}{0.2}\\\\t \approx 2](https://tex.z-dn.net/?f=8%28-0.4e%5E%7B%28-0.4t%29%7D%2B%200.6e%5E%7B%28-0.6t%29%7D%29%20%3D%200%5C%5C%5Cdisplaystyle%5Cfrac%7Be%5E%7B-0.4%7D%7D%7Be%5E%7B-0.6%7D%7D%20%3D%20%5Cfrac%7B0.6%7D%7B0.4%7D%5C%5C%5C%5Ce%5E%7B0.2t%7D%20%3D%201.5%5C%5C%5C%5Ct%20%3D%20%5Cfrac%7Bln%281.5%29%7D%7B0.2%7D%5C%5C%5C%5Ct%20%5Capprox%202)
At t = 0
![C(0) = 8(e^{(0)}-e^{(0)}) = 0](https://tex.z-dn.net/?f=C%280%29%20%3D%208%28e%5E%7B%280%29%7D-e%5E%7B%280%29%7D%29%20%3D%200)
At t = 2
![C(2) = 8(e^{(-0.8)}-e^{(-1.2)}) = 1.185](https://tex.z-dn.net/?f=C%282%29%20%3D%208%28e%5E%7B%28-0.8%29%7D-e%5E%7B%28-1.2%29%7D%29%20%3D%201.185)
At t = 12
![C(12) = 8(e^{(-4.8)}-e^{(-7.2)}) = 0.059](https://tex.z-dn.net/?f=C%2812%29%20%3D%208%28e%5E%7B%28-4.8%29%7D-e%5E%7B%28-7.2%29%7D%29%20%3D%200.059)
Thus, the maximum concentration of the antibiotic during the first 12 hours is 1.185
at t= 2 hours.