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hodyreva [135]
3 years ago
13

Put this equation into slope-intercept form.5x – 4y = 12​

Mathematics
2 answers:
Klio2033 [76]3 years ago
8 0

Answer:

y = 5/4x - 3

Step-by-step explanation:

Slope-intercept form is y = mx + b, where m is the slope and b is the y-intercept.

First, isolate variable y on the left side of the equation.

- 4y = - 5x + 12

Now, divide both sides by - 4 to solve for y.

y = 5/4x - 3

larisa [96]3 years ago
7 0

Step-by-step explanation: In slope-intercept or <em>y = mx + b</em> form, the <em>y</em> is by itself on the left side of the equation.

So our first task in this problem is to get

<em>y</em> by itself on the left side of the equation.

First subtract 5x from both sides to get -4y = -5x + 12.

I put the x term first because that's how it is in y = mx + b form.

Now divide both sides by -4 to get <em>y = 5/4x - 3</em>.

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Step-by-step explanation:

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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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