All categories

3 years ago

Please help no links or random stuff!!

Mathematics

1 answer:

Dmitrij [34]3 years ago

3 0

Answer:

See explanation. I added a photo and tried to explain everything. Please tell me in the comments if I explained it well and if handwriting was ok :)

Step-by-step explanation:

You might be interested in

What is the slope of the line that passes through the points (6, 4) and (3, 8)?

baherus [9]

Slope: (y2-y1)/(x2-x1)
(8-4)/(3-6) = 4/-3
The slope is -4/3

4 0

3 years ago

What's the value of 8 + (-7)

Anastasy [175]

Answer:

Step-by-step explanation:

8 0

3 years ago

Write an equation in slope-intercept form of the line that passes through the given point and is parallel to the graph of the gi

marin [14]

Answer:

The Answer is: y = -3x - 26

Step-by-step explanation:

Given point: (-6, -8)

Given equation:

y = -3x + 3, the slope m = -3

A parallel line has the same slope, m = 3. Use the point slope form and substitute in the point and the slope:

y - y1 = m(x - x1)

y - (-8) = -3(x - (-6))

y + 8 = -3x - 18

y = -3x - 18 - 8

y = -3x - 26

Proof - use the point (-6, -8):

f(x) = -3x -26

f(-6) = -3(-6) - 26

= 18 - 26 = -8, giving (-6, -8)

3 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

Simplify (7a+10b)-(4a-6b).<br> a. 3a2+16b<br> b. 3a+4b<br> c. 3a+16b<br> d. 11a+4b

Daniel [21]

(7a + 10b) - (4a - 6b)
(7a - 4a) + (10b - (-6b))
3a + (10b + 6b)
3a + 16b

4 0

3 years ago