All categories

3 years ago

How do you solve this? |6x|+3=21

Mathematics

2 answers:

Sliva [168]3 years ago

6 0

Answer:

Step-by-step explanation:

1. 21= 6x+3

2. Subtract the 3 from one side which will cancel it out. Then go to the other side of the equation and subtract 3 from 21. The equation will now look like this- 19= 6x

3. Now divide by 6 on one side to cancel it out. Then go to the other side and divide 19 by 6.

Answer: X = 3.16666666667

<u>Two things to remember when doing these equations</u>

- Think of it as trying to get x by itself

- If you add, subtract, divide, or multiply on one side of the equation always remember to do the same thing to the other side.

Lady_Fox [76]3 years ago

4 0

6x plus 3 =21
21-3=18 divided by 6 equals 3
X=3

You might be interested in

Solve the equation for x.

horrorfan [7]

Answer:

X= -1

Step-by-step explanation:

$(\frac{1}{3} )(6x)+(\frac{1}{3} )(-15)=(\frac{1}{2} )(10x)+(\frac{1}{2} )(-4)$

(Distribute)

2x+−5=5x+−2

2x−5=5x−2

Step 2: Subtract 5x from both sides.

2x−5−5x=5x−2−5x

−3x−5=−2

Step 3: Add 5 to both sides.

−3x−5+5=−2+5

−3x=3

Step 4: Divide both sides by -3.

Which = -1

Hope this helps :)

6 0

3 years ago

What is the absolute value for: <br> |40|

evablogger [386]

The absolute value is 40.

8 0

3 years ago

Read 2 more answers

Let a=(1,2,3,4), b=(4,3,2,1) and c=(1,1,1,1) be vectors in R4. Part (a) [4 points]: Find (a⋅2c)b+||−3c||a. Part (b) [6 points]:

love history [14]

Solution :

Given :

a = (1, 2, 3, 4) , b = ( 4, 3, 2, 1), c = (1, 1, 1, 1) ∈ $R^4$

a). (a.2c)b + ||-3c||a

Now,

(a.2c) = (1, 2, 3, 4). 2 (1, 1, 1, 1)

= (2 + 4 + 6 + 6)

= 20

-3c = -3 (1, 1, 1, 1)

= (-3, -3, -3, -3)

||-3c|| = $\sqrt{(-3)^2 + (-3)^2 + (-3)^2 + (-3)^2 }$

$=\sqrt{9+9+9+9}$

$=\sqrt{36}$

= 6

Therefore,

(a.2c)b + ||-3c||a = (20)(4, 3, 2, 1) + 6(1, 2, 3, 4)

= (80, 60, 40, 20) + (6, 12, 18, 24)

= (86, 72, 58, 44)

b). two vectors $\vec A$ and $\vec B$ are parallel to each other if they are scalar multiple of each other.

i.e., $\vec A=r \vec B$ for the same scalar r.

Given $\vec p$ is parallel to $\vec a$ , for the same scalar r, we have

$\vec p = r (1,2,3,4)$

$\vec p = (r,2r,3r,4r)$ ......(1)

Let $\vec q = (q_1,q_2,q_3,q_4)$ ......(2)

Now given $\vec p$ and $\vec q$ are perpendicular vectors, that is dot product of $\vec p$ and $\vec q$ is zero.

$q_1r + 2q_2r + 3q_3r + 4q_4r = 0$

$q_1 + 2q_2 + 3q_3 + 4q_4 = 0$ .......(3)

Also given the sum of $\vec p$ and $\vec q$ is equal to $\vec b$ . So

$\vec p + \vec q = \vec b$

$(r,2r,3r,4r) + (q_1+q_2+q_3+q_4)=(4, 3,2,1)$

∴ $q_1 = 4-r , \ q_2=3-2r, \ q_3 = 2-3r, \ q_4=1-4r$ ....(4)

Putting the values of $q_1,q_2,q_3,q_4$ in (3),we get

$r=\frac{2}{3}$

So putting this value of r in (4), we get

$\vec p =\left( \frac{2}{3}, \frac{4}{3}, 2, \frac{8}{3} \right)$

$\vec q =\left( \frac{10}{3}, \frac{5}{3}, 0, \frac{-5}{3} \right)$

These two vectors are perpendicular and satisfies the given condition.

c). Given terminal point is $\vec a$ is (-1, 1, 2, -2)

We know that,

Position vector = terminal point - initial point

Initial point = terminal point - position point

= (-1, 1, 2, -2) - (1, 2, 3, 4)

= (-2, -1, -1, -6)

d). $\vec b = (4,3,2,1)$

Let us say a vector $\vec d = (d_1, d_2,d_3,d_4)$ is perpendicular to $\vec b.$

Then, $\vec b.\vec d = 0$

$4d_1+3d_2+2d_3+d_4=0$

$d_4=-4d_1-3d_2-2d_3$

There are infinitely many vectors which satisfies this condition.

Let us choose arbitrary $d_1=1, d_2=1, d_3=2$

Therefore, $d_4=-4(-1)-3(1)-2(2)$

= -3

The vector is (-1, 1, 2, -3) perpendicular to given $\vec b.$

6 0

3 years ago

Are 3/4 and 12/9 equivalent explain your answer?

julia-pushkina [17]

yes

12/9=4/3

we used reciprocal here

7 0

3 years ago

Anyone help!!!!<br> Please I grant brainliest.

lakkis [162]

Answer:

hope this helps you.........

7 0

3 years ago

Read 2 more answers