Mathematics, 28.02.2020 07:26 julih74

In a pond, the area that is covered by algae doubles each week. When the
algae first spotted the area it covered was about 12.5 square meters.
a. Find the area, in square meters, covered by algae 10 days after it was spotted. Show your
reasoning.

b. Explain why we can find the area covered by algae 1 day after it was spotted by multiplying
12.5 by

(∆ that is part of the question)

.

Answers: 1

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Answers

Answer from: barry14201

Following are the solution to the given points:

$\to A = 12.5 e^{kt}$

Where

t= Number of weeks

$\to 12.5 \ e^k= 2 (12.5)\\\\ \to k=\log 2\\\\$

For point a)

$\to A = 12.5 e^{(\log 2) t}\\\\$

$=12.5 e^{(\log 2) \frac{10}{7} }\ \ sq\cdot m\\\\ = 33.65\ \ square \ meters$

For point b)

when $t = \frac{1}{7},$ then

$\to A = 12.5 e^{\log 2 \cdot \frac{1}{7}}\\\\$

$=12.5(1.104)\\\\$

$\to A=12.5 e^{\frac{1}{7} \log 2}\\\\$

$=12.5 e^{\log 2^{\frac{1}{7}}}\\\\=12.5 \ \sqrt[7]{2} \\\\$

Therefore, the final answer is " $33.65 m^2 \ and \ (12.5) \cdot \sqrt[7]{2}\ \m^2\\\\$ ".

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Answer from: dejanaej6

33.6 square meters

Step-by-step explanation:

a. The area doubles each week, starting with 12.5 square meters.

12.5, 25, 50, 100, 200, ...

We can model this as a geometric sequence, where the first term is 12.5 and the common ratio is 2. The nth term of the sequence is:

aₙ = a₀ (r)^n

aₙ = 12.5 (2)^n

This is the area after n weeks. We can find the daily area by dividing the exponent by 7.

a = 12.5 (2)^(d/7)

where d is the number of days.

So the area after 10 days is:

a = 12.5 (2)^(10/7)

a ≈ 33.6

b. Using exponent properties, we can rewrite the equation as:

a = 12.5 (2)^(d/7)

a = 12.5 (2^(1/7))^d

So after 1 day, the area increases by a factor of 2^(1/7) (seventh root of 2).

Answer from: Quest

the system of inequity is

$y\leq -x+4$

$3y> x$

step-by-step explanation:

from the given graph it is noticed that we have two related lines. one is solid and another is dashed.

if a line passing through two points, then the equation of line is

$y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)$

the solid line passing through the points (0,4) and (4,0).

$y-4=\frac{0-4}{4-0}(x-0)$

$y=-x+4$

the related equation of solid line is $y=-x+4$ .

similarly, find the related equation of dashed line. the dashed liune passing through the points (0,0) and (3,1).

$y-0=\frac{0-1}{3-0}(x-0)$

$3y=x$

the related equation of dashed line is $3y=x$

from the given graph it is noticed that the point (0,1) lies on the shaded region, therefore the point (0,1) must be satisfied by the related equations.

check the related equations by the point (0,1).

$1=-0+4$

$1=4$

this statement is true if the sign of inequality is < instead of equality. since it is an solid line therefore the sign of inequality will be $\leq$ . the required inequality is