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3 years ago

Find the slope of the line that contains (3, −6) and (−1, −9).

Mathematics

2 answers:

Zigmanuir [339]3 years ago

8 0

Answer:

Step-by-step explanation:

nikklg [1K]3 years ago

3 0

Answer:

<h2>The answer is option D</h2>

Step-by-step explanation:

The slope of a line given two points can be found by using the formula

$m = \frac{ y_2 - y _ 1}{x_ 2 - x_ 1} \\$

From the question we have

$m = \frac{ - 9 - - 6}{ - 1 - 3} = \frac{ - 9 + 6}{ - 4} = \frac{ - 3}{ - 4} = \frac{3}{4} \\$

We have the final answer as

$\frac{3}{4} \\$

Hope this helps you

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What is the product? 2x+8 x2 + 3x-4 x(x-4)(x-1) 20x+4) 2 (x+4)(x-4) 2x(x-1) 2x(x+4)

guajiro [1.7K]

=−640x10+1280x9+19904x8−40728x7−144488x6+323904x5−162304x4+1024x3+2048x2

step by step

(2x+8x2+3x−4x(x−4)(x−1)(20)x+4)(2)(x+4)(x−4)(2)x(x−1)(2)x(x+4)

=((2x+8x2+3x−4x(x−4)(x−1)(20)x+4)(2)(x+4)(x−4)(2)x(x−1)(2)x)(x+4)

=((2x+8x2+3x−4x(x−4)(x−1)(20)x+4)(2)(x+4)(x−4)(2)x(x−1)(2)x)(x)+((2x+8x2+3x−4x(x−4)(x−1)(20)x+4)(2)(x+4)(x−4)(2)x(x−1)(2)x)(4)

=−640x10+3840x9+4544x8−58904x7+91128x6−40608x5+128x4+512x3−2560x9+15360x8+18176x7−235616x6+364512x5−162432x4+512x3+2048x2

=−640x10+1280x9+19904x8−40728x7−144488x6+323904x5−162304x4+1024x3+2048x2

7 0

3 years ago

Read 2 more answers

Determine whether the lines given in each box are parallel, perpendicular, or neither

andrey2020 [161]

Answer/Step-by-sep explanation:

To determine whether the lines given in each box are parallel, perpendicular, or neither, take the following simple steps:

1. Ensure the equations for both lines being compared are in the slope-intercept form, y = mx + b. Where m is the slope.

2. If both lines have the same slope value, m, then both lines are parallel.

3. If the slope of one line is the negative reciprocal of the other, then both lines are perpendicular. That is, x = -1/x.

4. If the slope of both lines are not the same, nor the negative reciprocal of each other, then they are neither parallel nor perpendicular.

1. y = 3x - 7 and y = 3x + 1.

Both have the same slope value of 3. Therefore, they are parallel.

2. ⬜ $y = -\frac{2}{5}x + 3$ and $y = \frac{2}{5}x + 8$

The slope of both lines are not the same, nor is the slope of one the negative reciprocal of the other. The slope of one is -⅖ and the slope of the other is ⅖. Therefore, they are neither parallel nor perpendicular.⬜

3. $y = -\frac{1}{4}x$ and $y = 4x - 5$

The slope of the first line, ¼, is the negative reciprocal of the slope of the second line, 4.

Therefore, they are perdendicular.

4. 2x + 7y = 28 and 7x - 2y = 4.

Rewrite both equations in the slope-intercept form, y = mx + b.

2x + 7y = 28

7y = -2x + 28

y = -2x/7 + 28/7

y = -²/7 + 4

And

7x - 2y = 4

-2y = -7x + 4

y = -7x/-2 + 4/-2

y = ⁷/2x - 2

The slope of the first line, -²/7, is the negative reciprocal of the slope of the second line, ⁷/2.

Therefore, they are perdendicular.

5.⬜ y = -5x + 1 and x - 5y = 30.

Rewrite the second line equation in the slope-intercept form.

x - 5y = 30

-5y = -x + 30

y = -2x/-5 + 30/-5

y = ⅖x - 6

The slope of both lines are not the same, nor is the slope of one the negative reciprocal of the other. The slope of one is -5 and the slope of the other is ⅖. therefore, they are neither parallel nor perpendicular.⬜

6.⬜ 3x + 2y = 8 and 2x + 3y = -12.

Rewrite both line equations in the slope-intercept form.

3x + 2y = 8

2y = -3x + 8

y = -3x/2 + 8/2

y = -³/2x + 4

And

2x + 3y = -12

3y = -2x -12

y = -2x/3 - 12/3

y = -⅔x - 4

The slope of both lines are not the same, nor is the slope of one the negative reciprocal of the other. The slope of one is -³/2 and the slope of the other is -⅔ therefore, they are neither parallel nor perpendicular.⬜

7. y = -4x - 1 and 8x + 2y = 14.

Rewrite the equation of the second line in the slope-intercept form.

8x + 2y = 14

2y = -8x + 14

y = -8x/2 + 14/2

y = -4x + 7

Both have the same slope value of -4. Therefore, they are parallel.

8.⬜ x + y = 7 and x - y = 9.

Rewrite the equation of both lines in the slope-intercept form.

x + y = 7

y = -x + 7

And

x - y = 9

-y = -x + 9

y = -x/-1 + 9/-1

y = x - 9

The slope of both lines are not the same, nor is the slope of one the negative reciprocal of the other. The slope of one is -1, and the slope of the other is 1, therefore, they are neither parallel nor perpendicular.⬜

9. y = ⅓x + 9 And x - 3y = 3

Rewrite the equation of the second line.

x - 3y = 3

-3y = -x + 3

y = -x/-3 + 3/-3

y = ⅓x - 1

Both have the same slope value of ⅓. Therefore, they are parallel.

10.⬜ 4x + 9y = 18 and y = 4x + 9

Rewrite the equation of the first line.

4x + 9y = 18

9y = -4x + 18

y = -4x/9 + 18/9

y = -⁴/9x + 2

The slope of both lines are not the same, nor is the slope of one the negative reciprocal of the other. The slope of one is -⁴/9, and the slope of the other is 4, therefore, they are neither parallel nor perpendicular.⬜

11.⬜ 5x - 10y = 20 and y = -2x + 6

Rewrite the equation of the first line.

5x - 10y = 20

-10y = -5x + 20

y = -5x/-10 + 20/-10

y = ²/5x - 2

The slope of both lines are not the same, nor is the slope of one the negative reciprocal of the other. The slope of one is ⅖, and the slope of the other is -2, therefore, they are neither parallel nor perpendicular.⬜

12. -9x + 12y = 24 and y = ¾x - 5

Rewrite the equation of the first line.

-9x + 12y = 24

12y = 9x + 24

y = 9x/12 + 24/12

y = ¾x + 2

Both have the same slope value of ¾. Therefore, they are parallel.

5 0

3 years ago

Can anyone help me integrate :

worty [1.4K]

Rewrite the second factor in the numerator as

$2x^2+6x+1=2(x+2)^2-2(x+2)-3$

Then in the entire integrand, set $x+2=\sqrt3\sec t$ , so that $\mathrm dx=\sqrt3\sec t\tan t\,\mathrm dt$ . The integral is then equivalent to

$\displaystyle\int\frac{(\sqrt3\sec t-2)(6\sec^2t-2\sqrt3\sec t-3)}{\sqrt{(\sqrt3\sec t)^2-3}}(\sqrt3\sec t)\,\mathrm dt$
$=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{\sqrt{\sec^2t-1}}\,\mathrm dt$
$=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{\sqrt{\tan^2t}}\,\mathrm dt$
$=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{|\tan t|}\,\mathrm dt$

Note that by letting $x+2=\sqrt3\sec t$ , we are enforcing an invertible substitution which would make it so that $t=\mathrm{arcsec}\dfrac{x+2}{\sqrt3}$ requires $0\le t$ or $\dfrac\pi2$ . However, $\tan t$ is positive over this first interval and negative over the second, so we can't ignore the absolute value.

So let's just assume the integral is being taken over a domain on which $\tan t>0$ so that $|\tan t|=\tan t$ . This allows us to write

$=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{\tan t}\,\mathrm dt$
$=\displaystyle\int(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\csc t\,\mathrm dt$

We can show pretty easily that

$\displaystyle\int\csc t\,\mathrm dt=-\ln|\csc t+\cot t|+C$
$\displaystyle\int\sec t\csc t\,\mathrm dt=-\ln|\csc2t+\cot2t|+C$
$\displaystyle\int\sec^2t\csc t\,\mathrm dt=\sec t-\ln|\csc t+\cot t|+C$
$\displaystyle\int\sec^3t\csc t\,\mathrm dt=\frac12\sec^2t+\ln|\tan t|+C$

which means the integral above becomes

$=3\sqrt3\sec^2t+6\sqrt3\ln|\tan t|-18\sec t+18\ln|\csc t+\cot t|-\sqrt3\ln|\csc2t+\cot2t|-6\ln|\csc t+\cot t|+C$
$=3\sqrt3\sec^2t-18\sec t+6\sqrt3\ln|\tan t|+12\ln|\csc t+\cot t|-\sqrt3\ln|\csc2t+\cot2t|+C$

Back-substituting to get this in terms of $x$ is a bit of a nightmare, but you'll find that, since $t=\mathrm{arcsec}\dfrac{x+2}{\sqrt3}$ , we get

$\sec t=\dfrac{x+2}{\sqrt3}$
$\sec^2t=\dfrac{(x+2)^2}3$
$\tan t=\sqrt{\dfrac{x^2+4x+1}3}$
$\cot t=\sqrt{\dfrac3{x^2+4x+1}}$
$\csc t=\dfrac{x+2}{\sqrt{x^2+4x+1}}$
$\csc2t=\dfrac{(x+2)^2}{2\sqrt3\sqrt{x^2+4x+1}}$

etc.

3 0

3 years ago

You begin the month with $5,550 of inventory. You purchase $13,000 of stock during the month. You end the month with $2,333 of i

Zanzabum

Answer:

The correct answer is $16217

Step-by-step explanation:

You do 5550 + 13000 and get that answer and subtract 2333

3 0

3 years ago

Rhombus EFGH is shown. What is the measure of ∠HGJ?

MaRussiya [10]

Answer:

∠HGJ = 35°

Step-by-step explanation:

Although the question is somewhat incomplete, I have tried to reconstruct the question as much as possible to aid understanding (attached is the pictorial representation of the rhombus).

Firstly, it is important to note that rhombuses are parallelograms which have all sides equal and whose diagonals are perpendicular to each other. In addition to this, their diagonals bisect one other.

Mathematically, |EF| = |FG| = |GH| = |HE|

Another property which rhombuses have is that their opposite angles are equal.

Mathematically, ∠E = ∠G, ∠H = ∠F

The adjacent angles in a rhombus are supplementary. This means that the sum of angles closest to each other is 180°.

Mathematically, ∠E + ∠F = 180°, ∠F + ∠G = 180°, ∠G + ∠H = 180°,

∠H + ∠E = 180°

The question gave us ∠GHE and asked us to find ∠HGJ (denoted as θ in the picture attached). Note that ∠GHE means ∠H. That is, ∠GHE ⇒ ∠H

∠GHE or ∠H = 110°

From the properties of rhombuses (earlier stated), we know that adjacent angles in a rhombus are supplementary

Mathematically,

∠GHE + ∠FGH ⇒ 110° + ∠FGH = 180° ⇒ ∠FGH = 180° - 110°

∠FGH or ∠G = 70°

To calculate for ∠HGJ, we divide ∠FGH by 2, we have:

∠HGJ = ∠FGH ÷ 2 = 70° ÷ 2

∠HGJ = 35°

3 0

3 years ago