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Why is 5x4 greater than 5 x 1/4 Explain.<br> I Need This Answer by today

bearhunter [10]

Because 5x4 is 20 but 5x1/4 is 1,25.

5 x 1 = 5... / 4 = 1,25
Or ... -> 1/4 = 0,25 * 5 = 1,25

;)

8 0

2 years ago

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Find the next two term in this sequence

MariettaO [177]

Answer: 0.5 and 0.25

Step-by-step explanation: You have to divide the number by 2 to get to the next term.

7 0

3 years ago

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Solve the following application problem. Be sure to show your system of equations and the steps to solve the problem.

ololo11 [35]

Answer:

boat: 6 mph
current: 2 mph

Step-by-step explanation:

The relationship between time, speed, and distance is ...

speed = distance/time

For boat speed b and current speed c, the speed downstream is ...

b +c = (16 mi)/(2 h) = 8 mi/h

The speed upstream is ...

b -c = (16 mi)/(4 h) = 4 mi/h

Adding the two equations eliminates the c term:

2b = 12 mi/h

b = 6 mi/h . . . . . divide by 2

Solving the second equation for c, we get ...

c = b -4 mi/h = 6 mi/h -4 mi/h = 2 mi/h

The speed of the boat in still water is 6 mi/h; the current is 2 mi/h.

6 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,

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