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2 years ago

why is the number of producers usually much greater than the number of other organisms in an ecosystem?

Mathematics

1 answer:

Anon25 [30]2 years ago

5 0

Answer:

The producers are simple organisms. The producers are consumed by the consumers in a food chain which may be hetrotrophs like herbivores as well as omnivores. The density or population of the producers is expected to be more so as to feed the large population of the organisms at the higher trophic levels.

Step-by-step explanation:

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Write and solve an equation to find the value of x and a missing angle in the triangle below.

Arturiano [62]

Answer:

x = 27°, ∠JLK = 55°

Step-by-step explanation:

From the diagram in the question above,

The exterior angle of a triangle is equal to the sum of the two opposite side

3x+13 = 39+(2x+1)

3x+13 = 40+2x

collect like terms and solve for x

3x-2x = 40-13

x = 27°.

∠JKL = 2x+1

Substitute the value of x

∠JKL = 2(27)+1

∠JKL = 54+1

∠JKL = 55°

5 0

2 years ago

C=2(3.14)r; c=14(3.14) solve for r

tekilochka [14]

Answer:

c=2(3.14)r

given c=14(3.14), then r

make r the subject formula and substitute given c=14(3.14)

r=c/2(3.14)

r=14(3.14)/2(3.14)

r=7

Step-by-step explanation:

7 0

3 years ago

Solve the simultaneous equations y=3x+2 and y=x+6

kogti [31]

Answer:

x=2

Step-by-step explanation:

y=3x+2

y=x+6

Since you're stating that, they both equal because equation is equated to equation 2 because of the Y.

Therefore, substitute into both equations as shown:

3x+2=x+6

3x-x=6-2 (bring 2 to the other side by subtracting 2 as well as x)

2x=4

2x/2=4/2 (dividing 2 on both sides to get rid off 2 in 2x)

x=2.

5 0

2 years ago

Find the missing angle.

Alina [70]

Answer:

sorry bro just wanted the points but thanks have a great day

Step-by-step explanation:

3 0

3 years ago

Surface integrals using an explicit description. Evaluate the surface integral \iint_{S}^{}f(x,y,z)dS using an explicit represen

Jobisdone [24]

Parameterize $S$ by the vector function

$\vec r(x,y)=x\,\vec\imath+y\,\vec\jmath+f(x,y)\,\vec k$

so that the normal vector to $S$ is given by

$\dfrac{\partial\vec r}{\partial x}\times\dfrac{\partial\vec r}{\partial y}=\left(\vec\imath+\dfrac{\partial f}{\partial x}\,\vec k\right)\times\left(\vec\jmath+\dfrac{\partial f}{\partial y}\,\vec k\right)=-\dfrac{\partial f}{\partial x}\vec\imath-\dfrac{\partial f}{\partial y}\vec\jmath+\vec k$

with magnitude

$\left\|\dfrac{\partial\vec r}{\partial x}\times\dfrac{\partial\vec r}{\partial y}\right\|=\sqrt{\left(\dfrac{\partial f}{\partial x}\right)^2+\left(\dfrac{\partial f}{\partial y}\right)^2+1}$

In this case, the normal vector is

$\dfrac{\partial\vec r}{\partial x}\times\dfrac{\partial\vec r}{\partial y}=-\dfrac{\partial(8-x-2y)}{\partial x}\,\vec\imath-\dfrac{\partial(8-x-2y)}{\partial y}\,\vec\jmath+\vec k=\vec\imath+2\,\vec\jmath+\vec k$

with magnitude $\sqrt{1^2+2^2+1^2}=\sqrt6$ . The integral of $f(x,y,z)=e^z$ over $S$ is then

$\displaystyle\iint_Se^z\,\mathrm d\Sigma=\sqrt6\iint_Te^{8-x-2y}\,\mathrm dy\,\mathrm dx$

where $T$ is the region in the $x,y$ plane over which $S$ is defined. In this case, it's the triangle in the plane $z=0$ which we can capture with $0\le x\le8$ and $0\le y\le\frac{8-x}2$ , so that we have

$\displaystyle\sqrt6\iint_Te^{8-x-2y}\,\mathrm dx\,\mathrm dy=\sqrt6\int_0^8\int_0^{(8-x)/2}e^{8-x-2y}\,\mathrm dy\,\mathrm dx=\boxed{\sqrt{\frac32}(e^8-9)}$

5 0

3 years ago