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2 years ago

Scientists released 10 birds into a new habitat in year 0. Each year, there were

Mathematics

2 answers:

Natali [406]2 years ago

8 0

Answer:

f(x) = $3^{x}$ + 10

Step-by-step explanation:

Year one would be 10 + 3

Year two would be (3*3)+10

year three would be (3*3*3)+10

Which easily simplifies into $3^{x}$ + 10.

IgorLugansk [536]2 years ago

8 0

Answer:

The function can be written in Matlab program to compute the current number of birds in the habitat after x number or years.

f(x)=3ˣ*10

Step-by-step explanation:

From the question, in year zero we have 10 birds. This number increased to 30 on the first year.

0 Year 3⁰*10 = 10 Birds

1st Year 3¹*10 = 30 Birds

2nd Year 3²*10 = 90 Birds

3rd Year 3³*10 = 270 Birds

4th Year 3⁴*10 = 810 Birds

xth Year 3ˣ*10 = Current number of Birds

f(x) = 3ˣ*10

See below the MatLab function to solve the problem

function NumOfBirds=currentbirds(x);%function to compute current number of birds in x number of years

prompt='enter number of years in digits: ';% This prompt you to input the number of years

x=input(prompt);

NNOB=3^(x)*10;% This compute the new number of birds in the habitat

b=['The current number of birds is: ',num2str(NNOB)];

disp(b)% this display the result of the computation

SEE BELOW RESULT OF THE FUNCTION WHEN TESTED

>> currentbirds

enter number of years in digits: 3

The current number of birds is: 270

>> currentbirds

enter number of years in digits: 5

The current number of birds is: 2430

>> currentbirds

enter number of years in digits: 4

The current number of birds is: 810

>> currentbirds

enter number of years in digits: 10

The current number of birds is: 590490

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Jenna mixed two shades of purple paint. For Purple People Eater, she used 3 pints of red paint for every 5 pints of blue paint.

emmasim [6.3K]

Answer:

$\frac{12}{35}$

Step-by-step explanation:

Given :

→ For Purple People Eater, she used 3 pints of red paint for every 5 pints of blue paint.

i.e, The ratio of red paint to blue paint is 3:5

→ For Purple Majesty, she used a blue to red paint ratio of 7 to 4.

i.e, The ratio of red paint to blue paint is 4:7

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$=\frac{7}{4}$

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Multiply numerator and denominator by 4 in $\frac{3}{5}$

$\Rightarrow\frac{3\times 4}{5\times4}=\frac{12}{20}$

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The graphs of the quadratic functions f(x) = 6 – 10x2 and g(x) = 8 – (x – 2)2 are provided below. Observe there are TWO lines si

natta225 [31]

Answer:

a) y = 7.74*x + 7.5

b) y = 1.148*x + 6.036

Step-by-step explanation:

Given:

f(x) = 6 - 10*x^2

g(x) = 8 - (x-2)^2

Find:

(a) The line simultaneously tangent to both graphs having the LARGEST slope has equation

(b) The other line simultaneously tangent to both graphs has equation,

Solution:

- Find the derivatives of the two functions given:

f'(x) = -20*x

g'(x) = -2*(x-2)

- Since, the derivative of both function depends on the x coordinate. We will choose a point x_o which is common for both the functions f(x) and g(x). Point: ( x_o , g(x_o)) Hence,

g'(x_o) = -2*(x_o -2)

- Now compute the gradient of a line tangent to both graphs at point (x_o , g(x_o) ) on g(x) graph and point ( x , f(x) ) on function f(x):

m = (g(x_o) - f(x)) / (x_o - x)

m = (8 - (x_o-2)^2 - 6 + 10*x^2) / (x_o - x)

m = (8 - (x_o^2 - 4*x_o + 4) - 6 + 10*x^2)/(x_o - x)

m = ( 8 - x_o^2 + 4*x_o -4 -6 +10*x^2) /(x_o - x)

m = ( -2 - x_o^2 + 4*x_o + 10*x^2) /(x_o - x)

- Now the gradient of the line computed from a point on each graph m must be equal to the derivatives computed earlier for each function:

m = f'(x) = g'(x_o)

- We will develop the first expression:

m = f'(x)

( -2 - x_o^2 + 4*x_o + 10*x^2) /(x_o - x) = -20*x

Eq 1. (-2 - x_o^2 + 4*x_o + 10*x^2) = -20*x*x_o + 20*x^2

And,

m = g'(x_o)

( -2 - x_o^2 + 4*x_o + 10*x^2) /(x_o - x) = -20*x

-2 - x_o^2 + 4*x_o + 10*x^2 = -2(x_o - 2)(x_o - x)

Eq 2 -2 - x_o^2 + 4*x_o+ 10*x^2 = -2(x_o^2 - x_o*(x + 2) + 2*x)

- Now subtract the two equations (Eq 1 - Eq 2):

-20*x*x_o + 20*x^2 + 2*x_o^2 - 2*x_o*(x + 2) + 4*x = 0

-22*x*x_o + 20*x^2 + 2*x_o^2 - 4*x_o + 4*x = 0

- Form factors: 20*x^2 - 20*x*x_o - 2*x*x_o + 2*x_o^2 - 4*x_o + 4*x = 0

20*x*(x - x_o) - 2*x_o*(x - x_o) + 4*(x - x_o) = 0

(x - x_o)(20*x - 2*x_o + 4) = 0

x = x_o , x_o = 10x + 2

- For x_o = 10x + 2 ,

(g(10*x + 2) - f(x))/(10*x + 2 - x) = -20*x

(8 - 100*x^2 - 6 + 10*x^2)/(9*x + 2) = -20*x

(-90*x^2 + 2) = -180*x^2 - 40*x

90*x^2 + 40*x + 2 = 0

- Solve the quadratic equation above:

x = -0.0574, -0.387

- Largest slope is at x = -0.387 where equation of line is:

y - 4.502 = -20*(-0.387)*(x + 0.387)

y = 7.74*x + 7.5

- Other tangent line:

y - 5.97 = 1.148*(x + 0.0574)

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