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1 year ago

When one organism gains a benefit at the expense of a second organism and causes them harm is called:

Mathematics

2 answers:

Eddi Din [679]1 year ago

8 0

Answer:

Step-by-step explanation:

lyudmila [28]1 year ago

5 0

When one organism gains a benefit at the expense of a second organism and causes them harm is called:

A. Parasitism

B. Mutualism

C. Commensalism

D. Predation

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Define negative integer. Give an example of a negative integer and then give its opposite

sineoko [7]

An integer is a number and it's opposite.

A negative integer goes to the left of zero on the number line.

An example of a negative integer and it's opposite is:
negative integer: -10
opposite of the negative integer (the positive integer or additive inverse): 10

3 0

3 years ago

If A and B are independent events with P(A) = 0.5 and P(B) = 0.5, then P(A ∩ B) a. is 1.00. b. is 0.5. c. is 0.00. d. None of th

erastova [34]

Answer:

The answer is (d) "None of the these alternatives are correct"

Step-by-step explanation:

If two events A and B are independent, the probability of the intersection $P(A\cap B)$ is defined as:

$P(A\cap B)=P(A)\cdot P(B)$

Therefore, in the exercise:

$P(A\cap B)=P(A)\cdot P(B)=0.5\cdot 0.5\\P(A\cap B)=0.25$

which give (d) "None of the these alternatives are correct"

3 0

3 years ago

The equation of a line is y = -4x + 10<br> What is the y-intercept of the line

Vikentia [17]

Answer:

Step-by-step explanation:

the formula y = mx +c

the y is obviously, y.

m, also means the gradient has a value of -4

c is the y-intercept, so the value of c is 10.

4 0

2 years ago

Write the slope-intercept form of the equation of the line that is perpendicular to AB and passes through Point X. Show all work

aleksandr82 [10.1K]

Answer:

Equation of line is y=(12/5)x+2

Step-by-step explanation:

The slope of line AB is -5/12. The line passing X is perpendicular to line AB and hence have a slope of 12/5. The slope intercept form is given by y=mx+c.

Now, point X satisfies the equation. Plugging in the slope of the line we end up with

y=(12/5)*x+c, now to find c

-10=(12/5)*(-5)+c, c=2

Equation of line is y=(12/5)x+2

7 0

3 years ago

Solve these recurrence relations together with the initial conditions given. a) an= an−1+6an−2 for n ≥ 2, a0= 3, a1= 6 b) an= 7a

8_murik_8 [283]

Answer:

a) 3/5·((-2)^n + 4·3^n)
b) 3·2^n - 5^n
c) 3·2^n + 4^n
d) 4 - 3 n
e) 2 + 3·(-1)^n
f) (-3)^n·(3 - 2n)
g) ((-2 - √19)^n·(-6 + √19) + (-2 + √19)^n·(6 + √19))/√19

Step-by-step explanation:

These homogeneous recurrence relations of degree 2 have one of two solutions. Problems a, b, c, e, g have one solution; problems d and f have a slightly different solution. The solution method is similar, up to a point.

If there is a solution of the form $a[n]=r^n$ , then it will satisfy ...

$r^n=c_1\cdot r^{n-1}+c_2\cdot r^{n-2}$

Rearranging and dividing by $r^{n-2}$ , we get the quadratic ...

$r^2-c_1r-c_2=0$

The quadratic formula tells us values of r that satisfy this are ...

$r=\dfrac{c_1\pm\sqrt{c_1^2+4c_2}}{2}$

We can call these values of r by the names r₁ and r₂.

Then, for some coefficients p and q, the solution to the recurrence relation is ...

$a[n]=pr_1^n+qr_2^n$

We can find p and q by solving the initial condition equations:

$\left[\begin{array}{cc}1&1\\r_1&r_2\end{array}\right] \left[\begin{array}{c}p\\q\end{array}\right] =\left[\begin{array}{c}a[0]\\a[1]\end{array}\right]$

These have the solution ...

$p=\dfrac{a[0]r_2-a[1]}{r_2-r_1}\\\\q=\dfrac{a[1]-a[0]r_1}{r_2-r_1}$

_____

Using these formulas on the first recurrence relation, we get ...

$c_1=1,\ c_2=6,\ a[0]=3,\ a[1]=6\\\\r_1=\dfrac{1+\sqrt{1^2+4\cdot 6}}{2}=3,\ r_2=\dfrac{1-\sqrt{1^2+4\cdot 6}}{2}=-2\\\\p=\dfrac{3(-2)-6}{-5}=\dfrac{12}{5},\ q=\dfrac{6-3(3)}{-5}=\dfrac{3}{5}\\\\a[n]=\dfrac{3}{5}(-2)^n+\dfrac{12}{5}3^n$

The rest of (b), (c), (e), (g) are solved in exactly the same way. A spreadsheet or graphing calculator can ease the process of finding the roots and coefficients for the given recurrence constants. (It's a matter of plugging in the numbers and doing the arithmetic.)

_____

For problems (d) and (f), the quadratic has one root with multiplicity 2. So, the formulas for p and q don't work and we must do something different. The generic solution in this case is ...

$a[n]=(p+qn)r^n$

The initial condition equations are now ...

$\left[\begin{array}{cc}1&0\\r&r\end{array}\right] \left[\begin{array}{c}p\\q\end{array}\right] =\left[\begin{array}{c}a[0]\\a[1]\end{array}\right]$

and the solutions for p and q are ...

$p=a[0]\\\\q=\dfrac{a[1]-a[0]r}{r}$

Using these formulas on problem (d), we get ...

$c_1=2,\ c_2=-1,\ a[0]=4,\ a[1]=1\\\\r=\dfrac{2+\sqrt{2^2+4(-1)}}{2}=1\\\\p=4,\ q=\dfrac{1-4(1)}{1}=-3\\\\a[n]=4-3n$

And for problem (f), we get ...

$c_1=-6,\ c_2=-9,\ a[0]=3,\ a[1]=-3\\\\r=\dfrac{-6+\sqrt{6^2+4(-9)}}{2}=-3\\\\p=3,\ q=\dfrac{-3-3(-3)}{-3}=-2\\\\a[n]=(3-2n)(-3)^n$

_____

<em>Comment on problem g</em>

Yes, the bases of the exponential terms are conjugate irrational numbers. When the terms are evaluated, they do resolve to rational numbers.

6 0

3 years ago