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2 years ago

Help plsss

Mathematics

1 answer:

andrew11 [14]2 years ago

8 0

(b) is the answer.

Step-by-step explanation:

By the Pythagorean Theorem,

A² + B² = C²

Where:

A = Length of side 1

B = Length of side 2

C = Hypotenuse

This rule applies to all right-angled triangles.

The length of the hypotenuse of a right-angled triangle is always the largest value.

Therefore, we can test the answers with the equation above.

(a)

8² + 18² = 20²

64 + 324 = 400

388 ≠ 400

The rule of Pythagorean theorem doesn't work on a, so (a) is not a right-angled triangle.

(b)

12² + 35² = 37²

144 + 1225 = 1369

1369 = 1369

The rule of Pythagorean theorem works here, so (b) is a right-angled triangle.

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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,

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

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

2 years ago

How do you solve: 2(b - 5) + 7b-4

Alinara [238K]

2(b-5)+7b-4
Solve 2(b-5) to get 2b-10
Combine your like terms 2b-10+7b-4
Solve 2b+7b to get 9b
Solve -10+(-4) to get -14
So your answer is
9b-14

4 0

2 years ago

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Situation:A researcher in North America discoversa fossile that contains 65% of its originalamount of C-14..-ktN=NoeNo inital am

zheka24 [161]

SOLUTION

We have been given the equation of the decay as

$\begin{gathered} N=N_0e^{-kt} \\ where\text{ } \\ N_0=initial\text{ amount of C-14 at time t} \\ N=amount\text{ of C-14 at time t = 65\% of N}_0=0.65N_0 \\ k=0.0001 \\ t=time\text{ in years = ?} \end{gathered}$

So we are looking for the time

Plugging the values into the equation, we have

$\begin{gathered} N=N_0e^{-kt} \\ 0.65N_0=N_0e^{-0.0001t} \\ e^{-0.0001t}=\frac{0.65N_0}{N_0} \\ e^{-0.0001t}=0.65 \end{gathered}$

Taking Ln of both sides, we have

$\begin{gathered} ln(e^{-0.000t})=ln(0.65) \\ -0.0001t=ln(0.65) \\ t=\frac{ln(0.65)}{-0.0001} \\ t=4307.82916 \end{gathered}$

Hence the answer is 4308 to the nearest year

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5 months ago

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Nastasia [14]

Answer:45 * t = 2.5 * (1-t)...the equation will have one solution.

Step-by-step explanation:

For this case, the first thing you should know is:

d: v * t

Where,

d: distance

v: speed

t: time

To go to school by bus we have:

d = 45 * t

To return from school we have:

d = 2.5 * (1-t)

how the distance is the same:45 * t = 2.5 * (1-t)

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2 years ago

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The circle graph shows how a family budgets its annual income. If the total annual income is $140,000, what amount is budgeted f

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Answer:

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Step-by-step explanation:

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3 years ago

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