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larisa [96]
3 years ago
9

Samantha invested $880 in an account paying an interest rate of 3.1% compounded

Mathematics
1 answer:
irakobra [83]3 years ago
8 0

Answer: 11.8

Step-by-step explanation:

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If there are 2 blue, 5 yellow, and 3
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2/10 because you add 2 + 5 + 3 and put it over how many blues you have.
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Please give me the answer to the problem.
Mandarinka [93]
She made 3 sandwiches because 5-2 3/4 = 2 1/4.
2 1/4 as an improper fraction=(4*2)+1=9/4
9/4 / 3/4=3.
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Which equation represents the scenario?
BlackZzzverrR [31]

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<em>30(30 - x) = 540 </em>

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A = l w

<u><em>30(30 - x) = 540</em></u>

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3 years ago
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Please help, will give brainliest!
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Step-by-step explanation:

In parallelogram, diagonals bisect each other

DP = IP

7x - 8 = 3x

7x       = 3x + 8

7x - 3x = 8

      4x = 8

       x = 8/4

     x = 2

NP = YP

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6 0
3 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

At t = 0

C(0) = 8(e^{(0)}-e^{(0)}) = 0

At t = 2

C(2) = 8(e^{(-0.8)}-e^{(-1.2)}) = 1.185

At t = 12

C(12) = 8(e^{(-4.8)}-e^{(-7.2)}) = 0.059

Thus, the maximum concentration of the antibiotic during the first 12 hours is 1.185 \mu g/mL at t= 2 hours.

4 0
3 years ago
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