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34kurt
2 years ago
12

A The length of a rectangle is 4 m more

Mathematics
1 answer:
Lady bird [3.3K]2 years ago
8 0

Given :

  • The length of a rectangle is 4m more than the width.
  • The area of the rectangle is 45m²

⠀

To Find :

  • The length and width of the rectangle.

⠀

Solution :

We know that,

\qquad { \pmb{ \bf{Length \times Width = Area_{(rectangle)}}}}\:

So,

Let's assume the length of the rectangle as x and the width will be (x – 4).

⠀

Now, Substituting the given values in the formula :

\qquad \sf \: { \dashrightarrow x  \times  (x - 4) = 45 }

\qquad \sf \: { \dashrightarrow {x}^{2}  - 4x = 45 }

\qquad \sf \: { \dashrightarrow {x}^{2}  - 4x - 45 = 0 }

\qquad \sf \: { \dashrightarrow {x}^{2}    - 9x+ 5 x - 45 = 0 }

\qquad \sf \: { \dashrightarrow x(x - 9) + 5(x - 9) = 0 }

\qquad \sf \: { \dashrightarrow (x  - 9) (x  + 5) = 0 }

\qquad \sf \: { \dashrightarrow x = 9, \: \: x =  - 5}

⠀

Since, The length can't be negative, so the length will be 9 which is positive.

⠀

\qquad { \pmb{ \bf{ Length _{(rectangle)} = 9\:m}}}\:

\qquad { \pmb{ \bf{ Width _{(rectangle)} = 9 - 4=5\: m}}}\:

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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

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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.

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2 years ago
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Answer:

Sandra need to score at least <u>56%</u> in her fifth test so that her average is 80%.

Step-by-step explanation:

Given:

First 4 test scores = 87%, 92%, 76%,89%

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Let the minimum score she needs to make in fifth test be 'x'.

Total number of test = 5

Now we know that;

Average is equal to sum of all the scores in the test divided by number of test.

framing in equation form we get;

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Subtracting both side by 344 we get;

344+x-344=400-344\\\\x=56\%

Hence Sandra need to score at least <u>56%</u> in her fifth test so that her average is 80%.

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