PLEASE HELP ASAP!! WILL MARK BAINLIEST

The function f(x) = 20(3)^x represents the growth of a fish population every year in a local lake. Jesse wants to manipulate the

valentina_108 [34]

We need the exponents law

$a^{bx}=(a^b)^x$
So if
a=3
b=1/2

then
$3^{1/2*x}=(3^{1/2})^x$
where x represent every half a year.

3 0

3 years ago

Read 2 more answers

The upper-left coordinates on a rectangle are (-6,0)(−6,0)left parenthesis, minus, 6, comma, 0, right parenthesis, and the upper

nikklg [1K]

Answer:

See attachment for rectangle

Step-by-step explanation:

Given

$A = (-6,0)$

$B = (-4,0)$

$Area = 12$

Required

Draw the rectangle

First, we calculate the distance between A and B using distance formula;

$AB = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}$

So, we have:

$AB = \sqrt{(-6 - -4)^2 + (0- 0)^2}$

$AB = \sqrt{(-2)^2 + (0)^2}$

$AB = \sqrt{4 + 0}$

$AB = \sqrt{4}$

$AB = 2$

The above represents the length of the triangle.

Next, calculate the width using:

$Length * Width = Area$

$2* Width = 12$

Divide both sides by 2

$Width = 6$

This implies that, the width of the rectangle is 6 units.

We have:

$A = (-6,0)$

$B = (-4,0)$

Since A and B are at the upper left and right, then the ther two points are below.

6 units below each of the above point are:

$C = (-6,0-6)\\$ => $C = (-6,-6)$

$D = (-4,0-6)$ => $D = (-4,-6)$

Hence, the points of the rectangle are:

$A = (-6,0)$

$B = (-4,0)$

$C = (-6,-6)$

$D = (-4,-6)$

<em>See attachment for rectangle</em>

6 0

3 years ago

The length of some fish are modeled by a von Bertalanffy growth function. For Pacific halibut, this function has the form L(t) =

Dafna1 [17]

Answer:

a) L'(t) = 34.416*e^(-0.18*t)

b) L'(0) = 34 cm/yr , L'(1) =29 cm/yr , L'(6) =12 cm/yr

c) t = 10 year

Step-by-step explanation:

Given:

- The length of fish grows with time. It is modeled by the relation:

L(t) = 200*(1-0.956*e^(-0.18*t))

Where,

L: Is length in centimeter of a fist

t: Is the age of the fish in years.

Find:

(a) Find the rate of change of the length as a function of time

(b) In this part, give you answer to the nearest unit. At what rate is the fish growing at age: t = 0 , t = 1, t = 6

c) When will the fish be growing at a rate of 6 cm/yr? (nearest unit)

Solution:

- The rate of change of length of a fish as it ages each year can be evaluated by taking a derivative of the Length L(t) function with respect to x. As follows:

dL(t)/dt = d(200*(1-0.956*e^(-0.18*t))) / dt

dL(t)/dt = 34.416*e^(-0.18*t)

- Then use the above relation to compute:

L'(t) = 34.416*e^(-0.18*t)

L'(0) = 34.416*e^(-0.18*0) = 34 cm/yr

L'(1) = 34.416*e^(-0.18*1) = 29 cm/yr

L'(6) = 34.416*e^(-0.18*6) = 12 cm/yr

- Next, again use the derived L'(t) to determine the year when fish is growing at a rate of 6 cm/yr:

6 cm/yr = 34.416*e^(-0.18*t)

e^(0.18*t) = 34.416 / 6

0.18*t = Ln(34.416/6)

t = Ln(34.416/6) / 0.18

t = 10 year

7 0

3 years ago

ANSWER PLS?<br> I DONT GET IT

lisabon 2012 [21]

Answer:

the First one

Step-by-step explanation:

The First represent a straight line that Is a Linear function while the other One represent hyperbole, exponential and parabola that are not Linear function

4 0

1 year ago

Y

Luba_88 [7]

Answer:

a

Step-by-step explanation:

$y\leq\frac{1}{2}x+2 \\\\1\leq \frac{1}{2}(0)+2\\\\1\leq 0+2\\\\1\leq 2$

This statement is true

Hope this helps!

3 0

2 years ago