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shtirl [24]
2 years ago
6

A colony contains 1500 bacteria. The population increases at a rate of 115% each hour. If x represents the number of

Mathematics
2 answers:
miss Akunina [59]2 years ago
5 0

Answer:

f(x) = 1500(2.15)^x

Step-by-step explanation:

Let the function that represents the population of bacteria after x hours is,

f(x)=ab^x

For x = 0, f(x) = 1500,

1500=a(1+r)^0

1500=a

Now, the population increases at a rate of 115% each hour,

So, the population after 1 hour = (100+115)% of 1500 = 215% of 1500 = 3225,

That is, for x = 1, f(x) = 3225,

3225 =ab

3225=1500(b)

\implies b =2.15

Hence, the function that represents the given scenario is,

f(x)=1500(2.15)^x

kari74 [83]2 years ago
3 0

Answer:

Step-by-step explanation:

A colony contains 1500 bacteria. The population increases at a rate of 115% each hour. If x represents the number of hours elapsed, which function represents the scenario?

f(x) = 1500(1.15)x  

f(x) = 1500(115)x

f(x) = 1500(2.15)x

f(x) = 1500(215)x

The answer to this problem is a f(x) = 1500(2.15)x

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The volume of the region R bounded by the x-axis is: \mathbf{\iint_R(x^2+y^2)dA = \int ^{tan^{-1}(4)}_{0} \int^{\frac{2}{cos \theta}}_{0} \ r^3 dr d\theta}

<h3>What is the volume of the solid (R) on the X-axis?</h3>

If the axis of revolution is the boundary of the plane region and the cross-sections are parallel to the line of revolution, we may use the polar coordinate approach to calculate the volume of the solid.

From the given graph:

The given straight line passes through two points (0,0) and (2,8). Thus, the equation of the straight line becomes:

\mathbf{y-y_1 = \dfrac{y_2-y_1}{x_2-x_1}(x-x_1)}

here:

  • (x₁, y₁) and (x₂, y₂) are two points on the straight line

Suppose we assign (x₁, y₁) = (0, 0) and (x₂, y₂) = (2, 8)  from the graph, we have:

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y = 4x

Now, our region bounded by the three lines are:

  • y = 0
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Similarly, the change in polar coordinates is:

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  • y = rsinθ

where;

  • x² + y² = r²  and dA = rdrdθ

Therefore;

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  • ⇒ r = 0   to   r = 2/cosθ
  •    θ = 0  to    θ = tan⁻¹ (4)

Then:

\mathbf{\iint_R(x^2+y^2)dA = \int ^{tan^{-1}(4)}_{0} \int^{\frac{2}{cos \theta}}_{0} \ r^2 (rdr d\theta )}

\mathbf{\iint_R(x^2+y^2)dA = \int ^{tan^{-1}(4)}_{0} \int^{\frac{2}{cos \theta}}_{0} \ r^3 dr d\theta}

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As a student, you are able to earn extra money by assisting your neighbors with odd jobs. If you charged $10.25 an hour for your
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Answer:

821.95

Step-by-step explanation:

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A theater group made appearances in two cities. The hotel charge before tax in the second city was $1000 lower than in the first
Artyom0805 [142]

Answer:

Let x = the charge in 1st city before taxes

Let y = the charge in 2nd city before taxes

 

 

Set up equation before taxes.

 

y = x - 1500        eq1

 

 

Set up equation for total tax paid.

 

0.065x + 0.06y = 378.75         eq2

 

 

Substitute eq1 into eq2.

 

0.065x + 0.06(x - 1500) = 378.75

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Substitute this value of x into eq1.

 

y = 3750 - 1500

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The hotel charge in city one is $3750 and the hotel charge in city two is $2250

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