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3 years ago

Math question down below

Mathematics

1 answer:

lozanna [386]3 years ago

3 0

Answer:

315

Step-by-step explanation:

42 is 84 percent of 50. If you apply this to 375, you get 315 because 84 percent of 375 is 315

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Animal populations are not capable of unrestricted growth because of limited habitat and food supplies. Under such conditions th

trasher [3.6K]

Answer:

(a) 100 fishes

(b) t = 10: 483 fishes

t = 20: 999 fishes

t = 30: 1168 fishes

(c)

$P(\infty) = 1200$

Step-by-step explanation:

Given

$P(t) =\frac{d}{1+ke^-{ct}}$

$d = 1200\\k = 11\\c=0.2$

Solving (a): Fishes at t = 0

This gives:

$P(0) =\frac{1200}{1+11*e^-{0.2*0}}$

$P(0) =\frac{1200}{1+11*e^-{0}}$

$P(0) =\frac{1200}{1+11*1}$

$P(0) =\frac{1200}{1+11}$

$P(0) =\frac{1200}{12}$

$P(0) = 100$

Solving (a): Fishes at t = 10, 20, 30

$t = 10$

$P(10) =\frac{1200}{1+11*e^-{0.2*10}} =\frac{1200}{1+11*e^-{2}}\\\\P(10) =\frac{1200}{1+11*0.135}=\frac{1200}{2.485}\\\\P(10) =483$

$t = 20$

$P(20) =\frac{1200}{1+11*e^-{0.2*20}} =\frac{1200}{1+11*e^-{4}}\\\\P(20) =\frac{1200}{1+11*0.0183}=\frac{1200}{1.2013}\\\\P(20) =999$

$t = 30$

$P(30) =\frac{1200}{1+11*e^-{0.2*30}} =\frac{1200}{1+11*e^-{6}}\\\\P(30) =\frac{1200}{1+11*0.00247}=\frac{1200}{1.0273}\\\\P(30) =1168$

Solving (c): $\lim_{t \to \infty} P(t)$

In (b) above.

Notice that as t increases from 10 to 20 to 30, the values of $e^{-ct}$ decreases

This implies that:

${t \to \infty} = {e^{-ct} \to 0}$

So:

The value of P(t) for large values is:

$P(\infty) = \frac{1200}{1 + 11 * 0}$

$P(\infty) = \frac{1200}{1 + 0}$

$P(\infty) = \frac{1200}{1}$

$P(\infty) = 1200$

5 0

3 years ago

Is (2, 2) a solution of y < 4x - 6?<br> Choose 1 answer<br> yes or no

zalisa [80]

No, (2,2) is not solution of y < 4x - 6

Step-by-step explanation:

To solve above linear problem, we use standard method of Back- Substitution.

First of all, A point (2,2) only becomes solution of equation, if it satisfies equation $y < 4x - 6$ .

Here, the $y < 4x - 6$ seems to be like a linear equation.

In this Equation, using Back- Substitution method,

Point (2,2) : first digit 2 corresponds to x- coordinate and second digit 2 corresponds to y-coordinate.

$y < 4x - 6.....(1) \\\ x= 2 .....(2) \\\ y= 2 .....(3)$

By putting value of x and y in equation (1),

Equation becomes,

$2 < 4 \times 2 - 6$

$2 < 8 - 6$

$2 < 2$ .... not true mathematically.

Finally, the result came is not true. Therefore, (2,2) is not solution for equation

$y < 4x - 6$

4 0

3 years ago

√-90=ai√b what is the value of a and b?

OlgaM077 [116]

Answer:

what is this topic in math?

3 0

3 years ago

1. From the statement select the related given statement.

solong [7]

1. Line l; point P not on l.( Take a line I and mark point P outside it or on the line.So from point P there are infinite number of lines out of which only one line is parallel to line I. Suppose you are taking point P on line I, from that point P also infinite number of lines can be drawn but only one line will be coincident or parallel to line I.

2. Plane R is parallel to plane S; Plane T cuts planes R and S.(Imagine you are sitting inside a room ,consider two walls opposite to each other as two planes R and S and floor on which you are sitting as third plane T ,so R and S are parallel and plane T is cutting them so in this case their lines of intersect .But this is not possible in each and every case, suppose R and S planes are parallel to each other and Plane T cuts them like two faces of a building and third plane T is stairs or suppose it is in slanting position i.e not parallel to R and S so in this case also lines of intersection will be parallel.

3. △ABC with midpoints M and N.( As you know if we take a triangle ABC ,the mid points of sides AB and AC being M and N, so the line joining the mid point of two sides of a triangle is parallel to third side and is half of it.

4.Point B is between points A and C.( Take a line segment AC. Mark any point B anywhere on the line segment AC. Three possibilities arises

(i) AB > BC (ii) AB < BC (iii) AB = BC

Since A, B,C are collinear .So in each case $AB + BC =AC$