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3 years ago

SLOPE AND PARALLELISM

Mathematics

2 answers:

Lostsunrise [7]3 years ago

6 0

The answer is B maybe try it?

NISA [10]3 years ago

6 0

The answer is B
Hope this help!

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Steps to divide 125.00 by 7.75

kenny6666 [7]

_____
7.75 | 125

we want to get rid of that decimal....so we move it 2 spaces to the right....by doing that, we also have to move the decimal under the division sign by 2 spaces to the right....turning ur problem into :

16.1 R 225 <== ur answer
______
775 |12500.0
775
------- subtract
4750
4650
---------subtract
1000
775
-----------subtract
225

5 0

4 years ago

How to square imaginary numbers

IrinaVladis [17]

<span>1.Take the principle square root of a negative number.
2 Write a complex number in standard form.
3.Add and subtract complex numbers.
4.Multiply complex numbers.
5. Divide complex numbers.</span>

8 0

4 years ago

Read 2 more answers

The measure of angle t is 60 degrees. What is the x-coordinate of the point where the terminal side intersects the unit circle?

Anestetic [448]

Angle t would be equal to cos(60 degrees) which has coordinates (1/2,sqrt3/2) so that answer is 1/2

7 0

3 years ago

Let C be the curve of intersection of the parabolic cylinder x^2 = 2y, and the surface 3z = xy. Find the exact length of C from

Maslowich

I've attached a plot of the intersection (highlighted in red) between the parabolic cylinder (orange) and the hyperbolic paraboloid (blue).

The arc length can be computed with a line integral, but first we'll need a parameterization for $C$ . This is easy enough to do. First fix any one variable. For convenience, choose $x$ .

Now, $x^2=2y\implies y=\dfrac{x^2}2$ , and $3z=xy\implies z=\dfrac{x^3}6$ . The intersection is thus parameterized by the vector-valued function

$\mathbf r(x)=\left\langle x,\dfrac{x^2}2,\dfrac{x^3}6\right\rangle$

where $0\le x\le 4$ . The arc length is computed with the integral

$\displaystyle\int_C\mathrm dS=\int_0^4\|\mathbf r'(x)\|\,\mathrm dx=\int_0^4\sqrt{x^2+\dfrac{x^4}4+\dfrac{x^6}{36}}\,\mathrm dx$

Some rewriting:

$\sqrt{x^2+\dfrac{x^4}4+\dfrac{x^6}{36}}=\sqrt{\dfrac{x^2}{36}}\sqrt{x^4+9x^2+36}=\dfrac x6\sqrt{x^4+9x^2+36}$

Complete the square to get

$x^4+9x^2+36=\left(x^2+\dfrac92\right)^2+\dfrac{63}4$

So in the integral, you can substitute $y=x^2+\dfrac92$ to get

$\displaystyle\frac16\int_0^4x\sqrt{\left(x^2+\frac92\right)^2+\frac{63}4}\,\mathrm dx=\frac1{12}\int_{9/2}^{41/2}\sqrt{y^2+\frac{63}4}\,\mathrm dy$

Next substitute $y=\dfrac{\sqrt{63}}2\tan z$ , so that the integral becomes

$\displaystyle\frac1{12}\int_{9/2}^{41/2}\sqrt{y^2+\frac{63}4}\,\mathrm dy=\frac{21}{16}\int_{\arctan(3/\sqrt7)}^{\arctan(41/(3\sqrt7))}\sec^3z\,\mathrm dz$

This is a fairly standard integral (it even has its own Wiki page, if you're not familiar with the derivation):

$\displaystyle\int\sec^3z\,\mathrm dz=\frac12\sec z\tan z+\frac12\ln|\sec x+\tan x|+C$

So the arc length is

$\displaystyle\frac{21}{32}\left(\sec z\tan z+\ln|\sec x+\tan x|\right)\bigg|_{z=\arctan(3/\sqrt7)}^{z=\arctan(41/(3\sqrt7))}=\frac{21}{32}\ln\left(\frac{41+4\sqrt{109}}{21}\right)+\frac{41\sqrt{109}}{24}-\frac98$

4 0

4 years ago

The formula for the surface area of a cube is A equal 6 s to the power of 2 comma where s is the side length. Cube X has a side

cricket20 [7]

Answer:

Surface area of Cube Y is 3 times greater than the surface area of cube X

Step-by-step explanation:

Surface area of a cube,A = 6s²

Where,

s = side length

Cube X has a side length of 1,

Surface area of cube, X = 6s²

= 6 * 1²

= 6 * 1

= 6

cube Y has a side length of 2

Surface area of cube, Y = 6s²

= 6 * 2²

= 6 * 4

= 24

How much greater is the surface area of cube Y than cube X?

Surface area of Cube Y is 3 times greater than the surface area of cube X

3 0

3 years ago