Ask question

Ask question

All categories

3 years ago

12

What is the standard form of the equation of the line through (6, –3) with a slope of 2/3? a. 2x – 3y = 21 b. 3x – 2y = –21 c. –

2x + 3y = 24 d. 3x – 2y = 24

1 answer:

vekshin13 years ago

3 0

Answer:

i think its b

Step-by-step explanation:

im so sorry if im wrong

You might be interested in

Please help will mark brainliest (random answers will be reported)

maxonik [38]

I believe its B
i hope this helps :)

3 0

3 years ago

Read 2 more answers

Two hundred people responded to a survey. Of those who had green eyes, 7 had blonde hair, 9 had brown hair, and 2 had red hair.

cestrela7 [59]

Answer:

18

Step-by-step explanation:

5 0

3 years ago

when simplifying the expression y=(2x(x-3)(x-3))/(x-1)(x-3) do all of the x-3 s get cancelled or just one in the numerator and o

Ksju [112]

Answer:

Just one of the ones in the numerator and the one in the denominator

Step-by-step explanation:

7 0

3 years ago

Find the components of the vertical force Bold Upper FFequals=left angle 0 comma negative 10 right angle0,−10 in the directions

quester [9]

Solution :

Let $v_0$ be the unit vector in the direction parallel to the plane and let $F_1$ be the component of F in the direction of $v_0$ and $F_2$ be the component normal to $v_0$ .

Since, $|v_0| = 1,$

$(v_0)_x=\cos 60^\circ= \frac{1}{2}$

$(v_0)_y=\sin 60^\circ= \frac{\sqrt 3}{2}$

Therefore, $v_0 = \left$

From figure,

$|F_1|= |F| \cos 30^\circ = 10 \times \frac{\sqrt 3}{2} = 5 \sqrt3$

We know that the direction of $F_1$ is opposite of the direction of $v_0$ , so we have

$F_1 = -5\sqrt3 v_0$

$=-5\sqrt3 \left$

$= \left$

The unit vector in the direction normal to the plane, $v_1$ has components :

$(v_1)_x= \cos 30^\circ = \frac{\sqrt3}{2}$

$(v_1)_y= -\sin 30^\circ =- \frac{1}{2}$

Therefore, $v_1=\left< \frac{\sqrt3}{2}, -\frac{1}{2} \right>$

From figure,

$|F_2 | = |F| \sin 30^\circ = 10 \times \frac{1}{2} = 5$

∴ $F_2 = 5v_1 = 5 \left< \frac{\sqrt3}{2}, - \frac{1}{2} \right>$

$=\left$

Therefore,

$F_1+F_2 = \left< -\frac{5\sqrt3}{2}, -\frac{15}{2} \right> + \left< \frac{5 \sqrt3}{2}, -\frac{5}{2} \right>$

$= = F$

3 0

2 years ago

A rectangular room is 6.8 meters longer than it is wide, and perimeter is 29.6 meters. Find the dimensions of the room.

GuDViN [60]

Answer:

dimensions = 10.8 X 4

Step-by-step explanation:

Breadth/ width of room = x

Length of room = 6.8 + x

perimeter = 2 ( L + B )

29.6 = 2 ( 6.8 + x + x )

29.6/2 = 6.8 + 2x

14.8 - 6.8 = 2x

8 = 2x

8/2=x

4=x

Length of room = 6.8 + x = 6.8 + 4 = 10.8 m

Breadth/ width of room = x = 4m

8 0

3 years ago