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3 years ago

I would like to know this answer because this is trickey

Mathematics

1 answer:

Mandarinka [93]3 years ago

3 0

Answer:

A: yes

B: 40

C: 1x=40

I'm not really sure about C btw.

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Is y=9x+4 proportional or non proportional

Sauron [17]

the answer is it is proportional

5 0

3 years ago

Find the volume of the region between the planes x plus y plus 2 z equals 2 and 4 x plus 4 y plus z equals 8 in the first octant

Alex787 [66]

Find the intercepts for both planes.

Plane 1, x + y + 2z = 2:

$y=z=0\implies x=2\implies (2,0,0)$

$x=z=0\implies y=2\implies(0,2,0)$

$x=y=0\implies 2z=2\implies z=1\implies(0,0,1)$

Plane 2, 4x + 4y + z = 8:

$y=z=0\implies4x=8\implies x=2\implies(2,0,0)$

$x=z=0\implies4y=8\impliesy=2\implies(0,2,0)$

$x=y=0\implies z=8\implies(0,0,8)$

Both planes share the same x- and y-intercepts, but the second plane's z-intercept is higher, so Plane 2 acts as the roof of the bounded region.

Meanwhile, in the (x, y)-plane where z = 0, we see the bounded region projects down to the triangle in the first quadrant with legs x = 0, y = 0, and x + y = 2, or y = 2 - x.

So the volume of the region is

$\displaystyle\int_0^2\int_0^{2-x}\int_{\frac{2-x-y}2}^{8-4x-4y}\mathrm dz\,\mathrm dy\,\mathrm dx=\displaystyle\int_0^2\int_0^{2-x}\left(8-4x-4y-\frac{2-x-y}2\right)\,\mathrm dy\,\mathrm dx$

$=\displaystyle\int_0^2\int_0^{2-x}\left(7-\frac72(x+y)\right)\,\mathrm dy\,\mathrm dx=\int_0^2\left(7(2-x)-\frac72x(2-x)-\frac74(2-x)^2\right)\,\mathrm dx$

$=\displaystyle\int_0^2\left(7-7x+\frac74 x^2\right)\,\mathrm dx=\boxed{\frac{14}3}$

3 0

2 years ago

A parking lot charges $3 to park a car for the first hour and $2 per hour after that. If you use more than one parking space, th

Vikki [24]

Given, a parking lot charges $3 for first hour and $2 per hour after that.

So for t hours, the parking lot charges $3 for the first hour and after first hour there is $(t-1)$ hours left.

So for $(t-1)$ hours it will charge $2 per hour.

The charges for $(t-1)$ hours = $ $2(t-1)$ .

Total charges for t hours for one car = $ $(3+2(t-1))$

Now for the second car, it will charge 75% of the first car.

So the charges for second car

=$[ $(3+2(t-1))(75/100)$ ]

=$ $0.75(3+2(t-1))$

There are 3 cars. That parking charges for the third car is also 75% of the first car.

So for third car the parking charges are same as for the second car.

Total parking charges for 3 cars

= $ $(3+2(t-1))+(0.75(3+2(t-1))+(0.75(3+2(t-1))$

= $ $(3+2(t-1))+(0.75)(2(3+2(t-1))$

We have got the required answer here.

The correct option is option C.

8 0

3 years ago

Solve the following initial-value problem, showing all work, including a clear general solution as well as the particular soluti

Vikki [24]

Answer:

General Solution is $y=x^{3}+cx^{2}$ and the particular solution is $y=x^{3}-\frac{1}{2}x^{2}$

Step-by-step explanation:

$x\frac{\mathrm{dy} }{\mathrm{d} x}=x^{3}+3y\\\\Rearranging \\\\x\frac{\mathrm{dy} }{\mathrm{d} x}-3y=x^{3}\\\\\frac{\mathrm{d} y}{\mathrm{d} x}-\frac{3y}{x}=x^{2}$

This is a linear diffrential equation of type

$\frac{\mathrm{d} y}{\mathrm{d} x}+p(x)y=q(x)$ ..................(i)

here $p(x)=\frac{-2}{x}$

$q(x)=x^{2}$

The solution of equation i is given by

$y\times e^{\int p(x)dx}=\int e^{\int p(x)dx}\times q(x)dx$

we have $e^{\int p(x)dx}=e^{\int \frac{-2}{x}dx}\\\\e^{\int \frac{-2}{x}dx}=e^{-2ln(x)}\\\\=e^{ln(x^{-2})}\\\\=\frac{1}{x^{2} } \\\\\because e^{ln(f(x))}=f(x)]\\\\Thus\\\\e^{\int p(x)dx}=\frac{1}{x^{2}}$