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2 years ago

The expression below describes the number of organisms in a population of bacteria

Mathematics

1 answer:

In-s [12.5K]2 years ago

3 0

Answer:

The answer is the length of time in one time interval

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What does the 2 represent in 2/4

scZoUnD [109]

It represents 1/2, 50% or half of the equation.
it’s also the numerator, tells how many pieces you have from the whole.

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1 year ago

How do you write the algebraic expression: x=-2.5 like i mean the answer

Ganezh [65]

Answer:

2.5 × t = 2.5 t

Step-by-step explanation:

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2 years ago

What is the apparent solution to the system of equations?

Alexus [3.1K]

X = 3
y = -3

You will see this using any graphing tool at your disposal.

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3 years ago

Find the x-intercepts of the quadratic funct<br> f(x) = x2 – 5x – 14

Trava [24]

Answer:

x = - 2, x = 7

Step-by-step explanation:

To find the x- intercepts let f(x) = 0, that is

x² - 5x - 14 = 0 ← in standard form

(x - 7)(x + 2) = 0 ← in factored form

Equate each factor to zero and solve for x

x - 7 = 0 ⇒ x = 7

x + 2 = 0 ⇒ x = - 2

7 0

3 years ago

Find the volume of the solid.

dmitriy555 [2]

In Cartesian coordinates, the region (call it $R$ ) is the set

$R = \left\{(x,y,z) ~:~ x\ge0 \text{ and } y\ge0 \text{ and } 2 \le z \le 4-x^2-y^2\right\}$

In the plane $z=2$ , we have

$2 = 4 - x^2 - y^2 \implies x^2 + y^2 = 2 = \left(\sqrt2\right)^2$

which is a circle with radius $\sqrt2$ . Then we can better describe the solid by

$R = \left\{(x,y,z) ~:~ 0 \le x \le \sqrt2 \text{ and } 0 \le y \le \sqrt{2 - x^2} \text{ and } 2 \le z \le 4 - x^2 - y^2 \right\}$

so that the volume is

$\displaystyle \iiint_R dV = \int_0^{\sqrt2} \int_0^{\sqrt{2-x^2}} \int_2^{4-x^2-y^2} dz \, dy \, dx$

While doable, it's easier to compute the volume in cylindrical coordinates.

$\begin{cases} x = r \cos(\theta) \\ y = r\sin(\theta) \\ z = \zeta \end{cases} \implies \begin{cases}x^2 + y^2 = r^2 \\ dV = r\,dr\,d\theta\,d\zeta\end{cases}$

Then we can describe $R$ in cylindrical coordinates by

$R = \left\{(r,\theta,\zeta) ~:~ 0 \le r \le \sqrt2 \text{ and } 0 \le \theta \le\dfrac\pi2 \text{ and } 2 \le \zeta \le 4 - r^2\right\}$

so that the volume is

$\displaystyle \iiint_R dV = \int_0^{\pi/2} \int_0^{\sqrt2} \int_2^{4-r^2} r \, d\zeta \, dr \, d\theta \\\\ ~~~~~~~~ = \frac\pi2 \int_0^{\sqrt2} \int_2^{4-r^2} r \, d\zeta\,dr \\\\ ~~~~~~~~ = \frac\pi2 \int_0^{\sqrt2} r((4 - r^2) - 2) \, dr \\\\ ~~~~~~~~ = \frac\pi2 \int_0^{\sqrt2} (2r-r^3) \, dr \\\\ ~~~~~~~~ = \frac\pi2 \left(\left(\sqrt2\right)^2 - \frac{\left(\sqrt2\right)^4}4\right) = \boxed{\frac\pi2}$

3 0

1 year ago