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4 years ago

In ΔDEF, the measure of ∠F=90°, the measure of ∠E=34°, and FD = 32 feet. Find the length of EF to the nearest tenth of a foot.

Mathematics

2 answers:

NeX [460]4 years ago

6 0

If I am correct, Side EF would have the length 26.5

Lana71 [14]4 years ago

5 0

Answer: EF = 47.4

Step-by-step explanation:

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4x – 3y=8<br> 2. 5x – 2y = -11

AlladinOne [14]

Answer:

$x = -7$

$y = - 12$

Step-by-step explanation:

The given system is:

4x-3y=8

5x-2y=-11

We make x the subject in the top equation to get:

$x = \frac{3}{4}y + 2$

Plug this expression for x in the bottom equation to get

$5( \frac{3}{4}y + 2) - 2y= - 11$

We expand to obtain

$\frac{15}{4} y + 10 - 2y = - 11$

Multiply through by 4

$15y + 40 - 8y = - 44$

Group similar terms

$15y - 8y = - 44 - 40$

$7y = - 84$

$y = - \frac{84}{7} =-12$

This means

$x = -12\times \frac{3}{4} + 2$

$x = - 7$

6 0

4 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,

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