Please help, it’s urgent

Mathematics

1 answer:

WINSTONCH [101]2 years ago

4 0

Answer:

Hey buddy, here is your answer. Hope it helps you.

Step-by-step explanation:

As angle CAB is a linear pair and is 109 degrees. We need to simply subtract it by 180. So 180-109=71. So angle CAE is 71 degrees.

You might be interested in

What is the approximate distance from Denver to Chicago? Use a proportional relationship to solve the problem. Show your work.

Juli2301 [7.4K]

Answer:

$d = 1002.6\,mi$

Step-by-step explanation:

Approximately 5 units are equal to 557 miles and the scaled distance between Denver to Chicago is 9 units. The real distance can be estimated by simple rule of three:

$d = \frac{9}{5}\times (557\,mi)$

$d = 1002.6\,mi$

7 0

3 years ago

Plz help this question out

Elden [556K]

Answer:

bad in this..............

3 0

2 years ago

After an antibiotic tablet is taken, the concentration of the antibiotic in the bloodstream is modelled by the function C(t)=8(e

Alexxx [7]

Answer:

the maximum concentration of the antibiotic during the first 12 hours is 1.185 $\mu g/mL$ 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 $\mu g/mL$

$C(t) = 8(e^{(-0.4t)}-e^{(-0.6t)})$

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,

$\frac{d(C(t))}{dt} = 8(-0.4e^{(-0.4t)}+ 0.6e^{(-0.6t)})$

Equating the first derivative to zero, we get,

$\frac{d(C(t))}{dt} = 0\\\\8(-0.4e^{(-0.4t)}+ 0.6e^{(-0.6t)}) = 0$

Solving, we get,

$8(-0.4e^{(-0.4t)}+ 0.6e^{(-0.6t)}) = 0\\\displaystyle\frac{e^{-0.4}}{e^{-0.6}} = \frac{0.6}{0.4}\\\\e^{0.2t} = 1.5\\\\t = \frac{ln(1.5)}{0.2}\\\\t \approx 2$