All categories

3 years ago

What is the value of x

Mathematics

1 answer:

xxTIMURxx [149]3 years ago

8 0

Answer:

x=-6

Step-by-step explanation:

18x+30=12x-6

-12x -12x

6x+30=-6

-30 -30

6x= -36

-36/6

x=-6

You might be interested in

Graph the line for y+1=−3/5(x−4) on the coordinate plane.

Naddika [18.5K]

Please enclose that -3/5 in parentheses: y+1= (−3/5)(x−4)

Let's use the intercept method to graph this line.
First, set x = 0 and find y; this y will be the y-coordinate of the vertical intercept:

y+1= (−3/5)(0−4) => y = -1 + 12/5 = -5/5 + 12/5 = 7/5. Plot (0, 7/5).

Next, set y = 0 and find x; this will give us the coordinate of the horiz. int.:

0+1= (−3/5)(x−4) => -5/3 = x - 4, or x = 4- 5/3, or x = 7/3. Plot (7/3, 0).

Now draw a straight line thru these two points.

6 0

3 years ago

Read 2 more answers

A) single step Equation 1) X + 7 = 9

pshichka [43]

X=2
subtract 7 from both sides

6 0

3 years ago

I need help with intercepts from a graph

Rina8888 [55]

Explanation: The y axis is the line that goes up and down, vertical. The x axis is the line that goes straight across, horizontal. So, when looking for intercepts we have to see what numbers your graphed line goes through.

Make sure when you are writing your points you write it in (x,y) form. If you don't it will be wrong.

So, for our x intercept it would be -7,0.

This is because your graphed line goes through x at -7 and it has no y, so that would be 0.

For your y intercept, it would be 0,2

This is because our line does not go through x, so we make that 0. It does go through y at 2.

Hope this helps :)

If you need elaboration on anything let me know. I know sometimes with math it is kind of hard to explain through typing.

8 0

3 years ago

Name any two points on a horizontal line that is 4 units above the x-axis

hichkok12 [17]

(1,4) (2,4) (3,4) (4,4)

6 0

3 years ago

Maggie graphed the image of a 90 counterclockwise rotation about vertex A of . Coordinates B and C of are (2, 6) and (4, 3) and

Lemur [1.5K]

Answer:

A(2,2)

Step-by-step explanation:

Let the vertex A has coordinates $(x_A,y_A)$

Vectors AB and AB' are perpendicular, then

$\overrightarrow {AB}=(2-x_A,6-y_A)\\ \\\overrightarrow {AB'}=(-2-x_A,2-y_A)\\ \\\overrightarrow {AB}\perp\overrightarrow {AB'}\Rightarrow \overrightarrow {AB}\cdot \overrightarrow {AB'}=0\Rightarrow (2-x_A)(-2-x_A)+(6-y_A)(2-y_A)=0$

Vectors AC and AC' are perpendicular, then

$\overrightarrow {AC}=(4-x_A,3-y_A)\\ \\\overrightarrow {AC'}=(1-x_A,4-y_A)\\ \\\overrightarrow {AC}\perp\overrightarrow {AC'}\Rightarrow \overrightarrow {AC}\cdot \overrightarrow {AC'}=0\Rightarrow (4-x_A)(1-x_A)+(3-y_A)(4-y_A)=0$

Now, solve the system of two equations:

$\left\{\begin{array}{l}(2-x_A)(-2-x_A)+(6-y_A)(2-y_A)=0\\ \\(4-x_A)(1-x_A)+(3-y_A)(4-y_A)=0\end{array}\right.\\ \\\left\{\begin{array}{l}-4-2x_A+2x_A+x_A^2+12-6y_A-2y_A+y^2_A=0\\ \\4-4x_A-x_A+x_A^2+12-3y_A-4y_A+y_A^2=0\end{array}\right.\\ \\\left\{\begin{array}{l}x_A^2+y_A^2-8y_A+8=0\\ \\x_A^2+y_A^2-5x_A-7y_A+16=0\end{array}\right.$

Subtract these two equations:

$5x_A-y_A-8=0\Rightarrow y_A=5x_A-8$

Substitute it into the first equation:

$x_A^2+(5x_A-8)^2-8(5x_A-8)+8=0\\ \\x_A^2+25x_A^2-80x_A+64-40x_A+64+8=0\\ \\26x_A^2-120x_A+136=0\\ \\13x_A^2-60x_A+68=0\\ \\D=(-60)^2-4\cdot 13\cdot 68=3600-3536=64\\ \\x_{A_{1,2}}=\dfrac{60\pm8}{2\cdot 13}=\dfrac{34}{13},2$

Then

$y_{A_{1,2}}=5\cdot \dfrac{34}{13}-8 \text{ or } 5\cdot 2-8\\ \\=\dfrac{66}{13}\text{ or } 2$

Rotation by 90° counterclockwise about A(2,2) gives image points B' and C' (see attached diagram)

8 0

3 years ago