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galina1969 [7]
3 years ago
5

Write the area as a sum and as a product

Mathematics
1 answer:
galina1969 [7]3 years ago
3 0

Answer:

0+5+20= 25 the answer is 25

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MCR3U1 Culminating 2021.pdf
11111nata11111 [884]

Answer:

(a) y = 350,000 \times (1 + 0.07132)^t

(b) (i) The population after 8 hours is 607,325

(ii) The population after 24 hours is 1,828,643

(c) The rate of increase of the population as a percentage per hour is 7.132%

(d) The doubling time of the population is approximately, 10.06 hours

Step-by-step explanation:

(a) The initial population of the bacteria, y₁ = a = 350,000

The time the colony grows, t = 12 hours

The final population of bacteria in the colony, y₂ = 800,000

The exponential growth model, can be written as follows;

y = a \cdot (1 + r)^t

Plugging in the values, we get;

800,000 = 350,000 \times (1 + r)^{12}

Therefore;

(1 + r)¹² = 800,000/350,000 = 16/7

12·㏑(1 + r) = ㏑(16/7)

㏑(1 + r) = (㏑(16/7))/12

r = e^((㏑(16/7))/12) - 1 ≈ 0.07132

The  model is therefore;

y = 350,000 \times (1 + 0.07132)^t

(b) (i) The population after 8 hours is given as follows;

y = 350,000 × (1 + 0.07132)⁸ ≈ 607,325.82

By rounding down, we have;

The population after 8 hours, y = 607,325

(ii) The population after 24 hours is given as follows;

y = 350,000 × (1 + 0.07132)²⁴ ≈ 1,828,643.92571

By rounding down, we have;

The population after 24 hours, y = 1,828,643

(c) The rate of increase of the population as a percentage per hour =  r × 100

∴   The rate of increase of the population as a percentage = 0.07132 × 100 = 7.132%

(d) The doubling time of the population is the time it takes the population to double, which is given as follows;

Initial population = y

Final population = 2·y

The doubling time of the population is therefore;

2 \cdot y = y \times (1 + 0.07132)^t

Therefore, we have;

2·y/y =2 = (1 + 0.07132)^t

t = ln2/(ln(1 + 0.07132)) ≈ 10.06

The doubling time of the population is approximately, 10.06 hours.

8 0
3 years ago
WILL MARK BRAINLIEST TO THE BEST ANSWER!!
ruslelena [56]

\dfrac{12^3}{12^7} = 12^{3 - 7} = 12^{-4} = \dfrac{1}{12^4}

Answer: B

8 0
3 years ago
Read 2 more answers
NEED HELP ASAP PLEASE WILL GIVE BRAINLIEST:
ivann1987 [24]

Answer:

It can go 250 m deep

Step-by-step explanation:

4 0
2 years ago
Find the standard equation of the parabola that satisfies the given conditions. Also, find the length of the latus rectum of eac
lakkis [162]

Answer:

The standard parabola

                                y² = -18 x +27

Length of Latus rectum = 4 a = 18

                         

Step-by-step explanation:

<u><em>Explanation:-</em></u>

Given focus : (-3 ,0) ,directrix  : x=6

Let P(x₁ , y₁) be the point on parabola

PM perpendicular to the the directrix L

                          SP² = PM²

                (x₁ +3)²+(y₁-0)²  = (\frac{x_{1}-6 }{\sqrt{1} } )^{2}

              x₁²+6 x₁ +9 + y₁² = x₁²-12 x₁ +36

                          y₁² = -18 x₁ +36 -9

                           y₁² = -18 x₁ +27

The standard parabola

                                y² = -18 x +27

    Length of Latus rectum = 4 a = 4 (18/4) = 18

                         

5 0
3 years ago
Find the area of the shaded region. Round the nearest hundredth if necessary. YZ=14.2m
andreyandreev [35.5K]

Complete Question

The complete question is shown on the first uploaded image

Answer:

1

   A_1  =  67.58 \ in^2

2

   A_2 =415.4 \ ft^2

3

   A_3  =  8.48 \ cm^2

4

  A_4 =  480.38 \ m^2

Step-by-step explanation:

Generally the area of a sector is mathematically represented as

         A =  \frac{\theta}{360} * \pi r^2

Now at r_1  = 11 in and  \theta_1 =  64^o

       A_1 =  \frac{64}{360} * 3.142  * 11^2

       A_1  =  67.58 \ in^2

Now at  r_2  = 20 ft in and  \theta_2  =  119 ^o

       A_2 =  \frac{119}{360} * 3.142 *  20^2

       A_2 =415.4 \ ft^2

Now at  r_3  = 6.5 cm   and  \theta_3 =  23 ^o

     A_3  =  \frac{23}{360} * 3.142 *  6.5 ^2

      A_3  =  8.48 \ cm^2

Now at  r_4  = 14.2 m   and  \theta_4 = 360 -87 =  273 ^o

         A_4 =  \frac{273}{360}  * 3.142 * 14.2^2

          A_4 =  480.38 \ m^2

6 0
3 years ago
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