All categories

2 years ago

What is the first step in solving the equation Two-thirds x + one-third x + 2 = 5?

Mathematics

1 answer:

andreev551 [17]2 years ago

7 0

Answer:

Step-by-step explanation:

Subtract One-third from each side of the equation.

Add 2 to each side of the equation.

Combine like terms.

Multiply each side of the equation by 5.

You might be interested in

Write an equation in slope intercept form for the line with slope three and Y intercept -1

gladu [14]

The equation in slope intercept form for the line with slope 3 and Y intercept -1 is y = 3x -1 .

3 0

3 years ago

Find the length of the arch

posledela

Use the formula....central angle/360= arc length/2* pie* R
150/360=x/2*3.14*7.5
150/360=x/47.1
360x=7065
x=7065/360
x=19.625 in

8 0

2 years ago

What is (-33x-78)+(5x8x9x10)divided by9.75

kherson [118]

Answer:

6,174

Step-by-step explanation:

(-33 * - 78) + (5 * 8 * 9 * 10)

Remove the brackets:

-33 * - 78 + 5 * 8 * 9 * 10

Solve like so:

-33 * - 78 = 2,574

5 * 8 * 9 * 10 = 3,600

(2,574) + (3,600)

2,574 + 3,600 = 6,174

PLEASE DO MARK ME AS BRAINLIEST UWU

7 0

3 years ago

Which expression is equivalent to 54.5-6?

pentagon [3]

Answer: 5-6•54

Hope this helps (:

3 0

3 years ago

The coordinates of the endpoints of AB and CD are A(2,

Ratling [72]

Answer:

Option 1: CD is a perpendicular bisector of AB

Step-by-step explanation:

Let us find out the slopes of various line segments and the Distances and then we will draw the conclusions accordingly.

Formula to find slope

$m= \frac{y_2-y_1}{x_2-x_1}$

Formula to Find Distance between two points

$D=\sqrt{(y_2-y_1)^2+(x_2-x_1)^2}$

mAB ( represents , Slope of AB )

1. $mAC= \frac{3-2}{2-5}=\frac{1}{-3}=-\frac{1}{3}$

2. $mBC=\frac{2-1}{5-8}=\frac{1}{-3}=-\frac{1}{3}$

3. $mCD=\frac{5-2}{6-5}=\frac{3}{1}=3$

4. $AC=\sqrt{(3-2)^2+(2-5)^2} =\sqrt{(1)^2+(-3)^2}=\sqrt{1+9}=\sqrt{10}$

5. $BC=\sqrt{(2-1)^2+(5-8)^2} =\sqrt{(1)^2+(-3)^2}=\sqrt{1+9}=\sqrt{10}$

mAC = mBC , and C is common point , hence these three are collinear points making a straight line whole slope is $-\frac{1}{3}$

$mAB=-\frac{1}{3}$

$mCD=3$

$mAB \times mCD = -\frac{1}{3} \times 3 = -1$

Hence CD ⊥ AB

Also

From Point 4 and point 5 above , we see that

AC = CB

Hence CD bisect AB at C, also CD ⊥ AB

There fore

CD is a perpendicular bisector of AB

Therefor option 1 is true

4 0

3 years ago