All categories

3 years ago

(3x – 70) (3y+40) 120 x

Mathematics

2 answers:

enyata [817]3 years ago

4 0

Answer:

3x-70

Step-by-step explanation:

(3y+40 ) 120x is the answer of that question answer

Lelu [443]3 years ago

4 0

120x(3x-70)(3y+40)=
360x 2nd power -8400x)(3y+40)=
1080x 2nd power y+14400x 2nd power-25200xy-336000x

You might be interested in

A table of values of a linear function is shown below. Find the output when the input is n.

Vlad [161]

Answer:

Step-by-step explanation:

output = y

input = x

y=-2x + 3

if you follow the pattern, the next input is 5 which will output -7

3 0

2 years ago

Use the graph to determine the function's domain and range. Explain your answer.

mixas84 [53]

The domain is defined as all the possible x values. The graph extends to the left and to the right without bounds so the domain is All Real Values of x.
It can also be written as (-∞, ∞) This is called interval notation.

Note that the minimum value of f(x) is -4 so the range is [-4, ∞). (All real values of y equal to or greater than -4)

3 0

3 years ago

The mathematics department at a community college collected data for the number of students enrolled in 40 math classes over the

Colt1911 [192]

Answer:

Step-by-step explanation:

6 0

3 years ago

Whats the solution

Sergio039 [100]

Answer:

Step-by-step explanation:

To find the center and the radius, we need to put the equation in the (x-a)^2+(y-b)^2=r^2 [a is the x-coordinate, b is the y-coordinate, and r is the radius].

(x+6)^2+(y+7)^2 - 36 - 49 + 69 = 0

(x+6)^2+(y+7)^2 - 16 = 0

(x+6)^2+(y+7)^2 = 16

(x+6)^2+(y+7)^2 = 4^2

The center is (-6,-7) and the radius is 4.

Hope this helps!

5 0

2 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

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

8 0

1 year ago