All categories

2 years ago

15* Que Equations of lins - Part

Mathematics

1 answer:

Phantasy [73]2 years ago

4 0

There is not a given equation nor ordered pair, so it is impossible to answer this question. I apologise.

You might be interested in

(Y-1/4)-(y-4/5)=(y+4/10)-1

Arte-miy333 [17]

Answer:

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

What is the length of BC¯¯¯¯¯ ?

svlad2 [7]

Hello!

The length of side BC can be found by using the Law of Sines. This law states,

a / sin A = b / sin B = c / sin C.

Given:

Angle A = 51.2 degrees

Angle B = 60.3 degrees

Side AC (or B) = 21 cm

Side BC (or A) = ? cm

Let's plug in these values into the formula to find the missing side length.

a / sin (51.2) = 21 / sin (60.3)

a / sin (51.2) = 24.1759... (multiply both sides by sin(51.2)

a = 24.1759 · sin (51.2)

a = 18.841...

18.841 can be rounded to 19.

Therefore, the length of side BC is equal to 19 centimeters.

5 0

2 years ago

Rewrite the following integral in spherical coordinates.

lora16 [44]

In cylindrical coordinates, we have $r^2=x^2+y^2$ , so that

$z = \pm \sqrt{2-r^2} = \pm \sqrt{2-x^2-y^2}$

correspond to the upper and lower halves of a sphere with radius $\sqrt2$ . In spherical coordinates, this sphere is $\rho=\sqrt2$ .

$1 \le r \le \sqrt2$ means our region is between two cylinders with radius 1 and $\sqrt2$ . In spherical coordinates, the inner cylinder has equation

$x^2+y^2 = 1 \implies \rho^2\cos^2(\theta) \sin^2(\phi) + \rho^2\sin^2(\theta) \sin^2(\phi) = \rho^2 \sin^2(\phi) = 1 \\\\ \implies \rho^2 = \csc^2(\phi) \\\\ \implies \rho = \csc(\phi)$

This cylinder meets the sphere when

$x^2 + y^2 + z^2 = 1 + z^2 = 2 \implies z^2 = 1 \\\\ \implies \rho^2 \cos^2(\phi) = 1 \\\\ \implies \rho^2 = \sec^2(\phi) \\\\ \implies \rho = \sec(\phi)$

which occurs at

$\csc(\phi) = \sec(\phi) \implies \tan(\phi) = 1 \implies \phi = \dfrac\pi4+n\pi$

where $n\in\Bbb Z$ . Then $\frac\pi4\le\phi\le\frac{3\pi}4$ .

The volume element transforms to

$dx\,dy\,dz = r\,dr\,d\theta\,dz = \rho^2 \sin(\phi) \, d\rho \, d\theta \, d\phi$

Putting everything together, we have

$\displaystyle \int_0^{2\pi} \int_1^{\sqrt2} \int_{-\sqrt{2-r^2}}^{\sqrt{2-r^2}} r \, dz \, dr \, d\theta = \boxed{\int_0^{2\pi} \int_{\pi/4}^{3\pi/4} \int_{\csc(\phi)}^{\sqrt2} \rho^2 \sin(\phi) \, d\rho \, d\phi \, d\theta} = \frac{4\pi}3$

4 0

1 year ago

Solve the inequality (x + 4)(x - 1) < 0.

Rzqust [24]

Answer: −4<x<1

Step-by-step explanation: plz mark brainliest

8 0

3 years ago

Samantha is solving a

Licemer1 [7]

the correct answer for this problem: x= -1/3y + 2

Let’s solve for x
3x + y= 6

Add -y to both sides
3x + y + -y = 6+ -y
3x = -y + 6

Divide both sides by 3
3x/3 = -y + 6/ 3

And so your answer is at the top hope this helps.

5 0

2 years ago