All categories

3 years ago

Solve the Equation. 6x−3(x+8)=9 x=?

2 answers:

7 0

First distribute -3 into the parenthesis. you will get 6x-3x-24=9. Then subtract 6x and 3x because they're like terms. you will get 3x -24=9. After that add 24 by both sides and get 3x=33. this step is easy just divide by both sides and you will get x.

Hunter-Best [27]3 years ago

3 0

Answer:

x = -5

Step-by-step explanation:

Simplifying

6x + -3(x + -8) = 9

Reorder the terms:

6x + -3(-8 + x) = 9

6x + (-8 * -3 + x * -3) = 9

6x + (24 + -3x) = 9

Reorder the terms:

24 + 6x + -3x = 9

Combine like terms: 6x + -3x = 3x

24 + 3x = 9

Solving

24 + 3x = 9

Solving for variable 'x'.

Move all terms containing x to the left, all other terms to the right.

Add '-24' to each side of the equation.

24 + -24 + 3x = 9 + -24

Combine like terms: 24 + -24 = 0

0 + 3x = 9 + -24

3x = 9 + -24

Combine like terms: 9 + -24 = -15

3x = -15

Divide each side by '3'.

x = -5

Hope this Helps

I hope this is correct

You might be interested in

What is 35÷110 equal

Thepotemich [5.8K]

0.32 is the answer you are looking for.

3 0

3 years ago

Read 2 more answers

When a certain number is multiplied by 14 and the product is then multiplied by 32, the result is 60. what is the number?

Stolb23 [73]

It might be 0.133928571.....? I am unsure though. Sorry if this doesn't help

7 0

3 years ago

Evaluate the triple integral ∭EzdV where E is the solid bounded by the cylinder y2+z2=81 and the planes x=0,y=9x and z=0 in the

dem82 [27]

Answer:

I = 91.125

Step-by-step explanation:

Given that:

$I = \int \int_E \int zdV$ where E is bounded by the cylinder $y^2 + z^2 = 81$ and the planes x = 0 , y = 9x and z = 0 in the first octant.

The initial activity to carry out is to determine the limits of the region

since curve z = 0 and $y^2 + z^2 = 81$

∴ $z^2 = 81 - y^2$

$z = \sqrt{81 - y^2}$

Thus, z lies between 0 to $\sqrt{81 - y^2}$

GIven curve x = 0 and y = 9x

$x =\dfrac{y}{9}$

As such,x lies between 0 to $\dfrac{y}{9}$

Given curve x = 0 , $x =\dfrac{y}{9}$ and z = 0, $y^2 + z^2 = 81$

y = 0 and

$y^2 = 81 \\ \\ y = \sqrt{81} \\ \\ y = 9$

∴ y lies between 0 and 9

Then $I = \int^9_{y=0} \int^{\dfrac{y}{9}}_{x=0} \int^{\sqrt{81-y^2}}_{z=0} \ zdzdxdy$

$I = \int^9_{y=0} \int^{\dfrac{y}{9}}_{x=0} \begin {bmatrix} \dfrac{z^2}{2} \end {bmatrix} ^ {\sqrt {{81-y^2}}}_{0} \ dxdy$

$I = \int^9_{y=0} \int^{\dfrac{y}{9}}_{x=0} \begin {bmatrix} \dfrac{(\sqrt{81 -y^2})^2 }{2}-0 \end {bmatrix} \ dxdy$

$I = \int^9_{y=0} \int^{\dfrac{y}{9}}_{x=0} \begin {bmatrix} \dfrac{{81 -y^2} }{2} \end {bmatrix} \ dxdy$

$I = \int^9_{y=0} \begin {bmatrix} \dfrac{{81x -xy^2} }{2} \end {bmatrix} ^{\dfrac{y}{9}}_{0} \ dy$

$I = \int^9_{y=0} \begin {bmatrix} \dfrac{{81(\dfrac{y}{9}) -(\dfrac{y}{9})y^2} }{2}-0 \end {bmatrix} \ dy$

$I = \int^9_{y=0} \begin {bmatrix} \dfrac{{81 \ y -y^3} }{18} \end {bmatrix} \ dy$

$I = \dfrac{1}{18} \int^9_{y=0} \begin {bmatrix} {81 \ y -y^3} \end {bmatrix} \ dy$

$I = \dfrac{1}{18} \begin {bmatrix} {81 \ \dfrac{y^2}{2} - \dfrac{y^4}{4}} \end {bmatrix}^9_0$

$I = \dfrac{1}{18} \begin {bmatrix} {40.5 \ (9^2) - \dfrac{9^4}{4}} \end {bmatrix}$

$I = \dfrac{1}{18} \begin {bmatrix} 3280.5 - 1640.25 \end {bmatrix}$

$I = \dfrac{1}{18} \begin {bmatrix} 1640.25 \end {bmatrix}$

I = 91.125

4 0

3 years ago

If w= -1, x =6 and y= -3, find the following<br><br> 3w² - 5w + 1

drek231 [11]

Answer:

Step-by-step explanation:

3(1)² - 5(1) + 1

3x3 = 9

9-5+1

9-6 = 3

8 0

3 years ago

What is the percent of change from 500,000 to 20,000?

photoshop1234 [79]

Answer:

96 decrease

Step-by-step explanation:

5000000-20000=

480000/500000 x 100

96% decreased

8 0

2 years ago

Read 2 more answers