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3 years ago

Lines c and d are perpendicular and if the slope of line c is -4 what is the slope of line d ?

2 answers:

8 0

Perpendicular lines, slope is opposite and reciprocal. so if the slope of line c is -4 then the slope of line d will be 1/4

Hope it helps

siniylev [52]3 years ago

4 0

The answer to your question is 1/4

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Explain how you would graph 3x+9y= -18 without converting to slope-intercept form

Zolol [24]

Answer:

you would solve using x and y intercepts.

Step-by-step explanation:

for the X intercept, Y=0

3x=-18. you divide by three on both sides

x=-6. (-6,0)

the for the y intercept, X=0

+9y=-18. you divide 9 on both sides.

y=-2. (0,-2)

then you graph both and connect the points on the x and y axis.

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3 years ago

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Genrish500 [490]

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3 years ago

Determine how many solutions if correct bainliest

PolarNik [594]

Answer:

No solutions

Step-by-step explanation:

6x+2y=20

-3x-y=-4 (by rewriting)

6/-3 = 2/-1≠20/-4

-2=-2≠-5

So it has no solutions

5 0

3 years ago

123456789101112131415161718

lesya692 [45]

Answer:

the third one is the correct answer

Step-by-step explanation:

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7 0

3 years ago

Find the components of the vertical force Bold Upper Fequalsleft angle 0 comma negative 8 right anglein the directions parallel

nydimaria [60]

Answer with Step-by-step explanation:

We are given that

F=<0,-8>=0i-8j=-8j

$\theta=\frac{\pi}{3}$

The component of force is divided into two direction

1.Along the plane

2.Perpendicular to the plane

1.The vector parallel to the plane will be= $r=cos\frac{\pi}{3}i-sin\frac{\pi}{3}j=\frac{1}{2}i-\frac{\sqrt 3}{2}j$

By using $cos\frac{\pi}{3}=\frac{1}{2},sin\frac{\pi}{3}=\frac{\sqrt 3}{2}$

Force along the plane will be= $\mid F_x\mid=F\cdot r$

Force along the plane will be = $\mid F_x\mid=F\cdot (\frac{1}{2}i-\frac{\sqrt 3}{2}j)=-8j\cdot(\frac{1}{2}i-\frac{\sqrt 3}{2}j)=8\times \frac{\sqrt 3}{2}=4\sqrt 3$ N

By using $i\cdot i=j\cdoty j=k\cdot k=1,i\cdot j=j\cdot k=k\cdot i=j\cdot i=k\cdot j=i\cdot k=0$

Therefore, force along the plane= $\mid F_x\mid(\frac{1}{2}i-\frac{\sqrt 3}{2}j)=4\sqrt 3(\frac{1}{2}i-\frac{\sqrt 3}{2}j)$

2.The vector perpendicular to the plane= $r=-sin\frac{\pi}{3}-cos\frac{\pi}{3}=-\frac{\sqrt 3}{2}i-\frac{1}{2}j$

The force perpendicular to the plane= $\mid F_y\mid=F\cdot r=-8j(-\frac{\sqrt 3}{2}i-\frac{1}{2}j)$

The force perpendicular to the plane= $4$ N

Therefore, $F_y=4(-\frac{\sqrt 3}{2}i-\frac{1}{2}j)$

Sum of two component of force= $F_x+F_y=4\sqrt 3(\frac{1}{2}i-\frac{\sqrt 3}{2}j)+4(-\frac{\sqrt 3}{2}i-\frac{1}{2}j)$

Sum of two component of force= $2\sqrt 3i-6j-2\sqrt3 i-2j=-8j$

Hence,sum of two component of forces=Total force.

6 0

3 years ago