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3 years ago

Please answer. And explain. 20pts

Mathematics

2 answers:

Anon25 [30]3 years ago

8 0

The answer is D

Because 140 divided by 3.50 is 40. Therefore they would have to sell at least 40. (Or greater than/equal too 40)
Please mark as Brainliest

RSB [31]3 years ago

6 0

I believe it's A. Because your going from 140 to 3.50 and divide that to get the last answer

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Which equation represents the vertical asymptote of the graph?

Annette [7]

The curve is y equals 0 from negative x to negative y near x equals negative 8.

If a curve has Vertical Asymptote i.e the line x=p,it is never touched by the given curve.The curve remains almost parallel to the line x=p, till the end.The two i.e a line and curve will never meet each other.

→ x is almost equal to p but not p.

so in the denominator , it is x=-8,

Vertical Asymptote occurs when we put , denominator of curve=0.

so vertical asymptote of curve is : x= -8

8 0

3 years ago

MULTIPLE CHOICE HELP ASAP! Brainliest for correct answer.

N76 [4]

Answer:

An equation of the circle with centre (-2,1) and radius 3 is $\mathbf{(x+2)^2+(x-1)^2=9}$

Option D is correct.

Step-by-step explanation:

Looking at the figure we get centre of circle C (-2,1) and radius of circle r = 3

The equation of circle is of form: $(x-h)^2+(x-k)^2=r^2$ where (h,k) is centre and r is radius.

We have centre C (-2,1) so, we have h = -2 and k = 1

We have radius = 3 so, r = 3

Putting values in the equation and finding the required equation:

$(x-h)^2+(x-k)^2=r^2\\(x-(-2))^2+(x-1)^2=(3)^2\\(x+2)^2+(x-1)^2=9$

So, an equation of the circle with centre (-2,1) and radius 3 is $\mathbf{(x+2)^2+(x-1)^2=9}$

Option D is correct.

7 0

3 years ago

Calculus 3 help please.

Reptile [31]

I assume each path $C$ is oriented positively/counterclockwise.

(a) Parameterize $C$ by

$\begin{cases} x(t) = 4\cos(t) \\ y(t) = 4\sin(t)\end{cases} \implies \begin{cases} x'(t) = -4\sin(t) \\ y'(t) = 4\cos(t) \end{cases}$

with $-\frac\pi2\le t\le\frac\pi2$ . Then the line element is

$ds = \sqrt{x'(t)^2 + y'(t)^2} \, dt = \sqrt{16(\sin^2(t)+\cos^2(t))} \, dt = 4\,dt$

and the integral reduces to

$\displaystyle \int_C xy^4 \, ds = \int_{-\pi/2}^{\pi/2} (4\cos(t)) (4\sin(t))^4 (4\,dt) = 4^6 \int_{-\pi/2}^{\pi/2} \cos(t) \sin^4(t) \, dt$

The integrand is symmetric about $t=0$ , so

$\displaystyle 4^6 \int_{-\pi/2}^{\pi/2} \cos(t) \sin^4(t) \, dt = 2^{13} \int_0^{\pi/2} \cos(t) \sin^4(t) \,dt$

Substitute $u=\sin(t)$ and $du=\cos(t)\,dt$ . Then we get

$\displaystyle 2^{13} \int_0^{\pi/2} \cos(t) \sin^4(t) \, dt = 2^{13} \int_0^1 u^4 \, du = \frac{2^{13}}5 (1^5 - 0^5) = \boxed{\frac{8192}5}$

(b) Parameterize $C$ by

$\begin{cases} x(t) = 2(1-t) + 5t = 3t - 2 \\ y(t) = 0(1-t) + 4t = 4t \end{cases} \implies \begin{cases} x'(t) = 3 \\ y'(t) = 4 \end{cases}$

with $0\le t\le1$ . Then

$ds = \sqrt{3^2+4^2} \, dt = 5\,dt$

and

$\displaystyle \int_C x e^y \, ds = \int_0^1 (3t-2) e^{4t} (5\,dt) = 5 \int_0^1 (3t - 2) e^{4t} \, dt$

Integrate by parts with

$u = 3t-2 \implies du = 3\,dt \\\\ dv = e^{4t} \, dt \implies v = \frac14 e^{4t}$

$\displaystyle \int u\,dv = uv - \int v\,du$

$\implies \displaystyle 5 \int_0^1 (3t-2) e^{4t} \,dt = \frac54 (3t-2) e^{4t} \bigg|_{t=0}^{t=1} - \frac{15}4 \int_0^1 e^{4t} \,dt \\\\ ~~~~~~~~ = \frac54 (e^4 + 2) - \frac{15}{16} e^{4t} \bigg|_{t=0}^{t=1} \\\\ ~~~~~~~~ = \frac54 (e^4 + 2) - \frac{15}{16} (e^4 - 1) = \boxed{\frac{5e^4 + 55}{16}}$

$\begin{cases} x(t) = 3(1-t)+t = -2t+3 \\ y(t) = (1-t)+2t = t+1 \\ z(t) = 2(1-t)+5t = 3t+2 \end{cases} \implies \begin{cases} x'(t) = -2 \\ y'(t) = 1 \\ z'(t) = 3 \end{cases}$

with $0\le t\le1$ . Then

$ds = \sqrt{(-2)^2 + 1^2 + 3^2} \, dt = \sqrt{14} \, dt$

and

$\displaystyle \int_C y^2 z \, ds = \int_0^1 (t+1)^2 (3t+2) \left(\sqrt{14}\,ds\right) \\\\ ~~~~~~~~ = \sqrt{14} \int_0^1 \left(3t^3 + 8t^2 + 7t + 2\right) \, dt \\\\ ~~~~~~~~ = \sqrt{14} \left(\frac34 t^4 + \frac83 t^3 + \frac72 t^2 + 2t\right) \bigg|_{t=0}^{t=1} \\\\ ~~~~~~~~ = \sqrt{14} \left(\frac34 + \frac83 + \frac72 + 2\right) = \boxed{\frac{107\sqrt{14}}{12}}$