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LenaWriter [7]
2 years ago
13

How do you solve 7x-5=30?

Mathematics
2 answers:
Nadya [2.5K]2 years ago
8 0
7x - 5 = 30

* add 5 to both sides

7x = 30 + 5

7x = 35

* divided both sides by 7

(7/7)x = 35/7

<u><em>x = 5</em></u>
kenny6666 [7]2 years ago
7 0
7x-5=30\ \ |+5\ to\ both\ sides\\\\7x=30+5\\\\7x=35\ \ |:7\\\\x=\frac{35}{7}\\\\x=5\\\\Answer\ is\ x=5.
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At a sale, dresses were sold for $90 each. This price was 75% of a
gulaghasi [49]

Answer:

A dress originally costs $120

Step-by-step explanation:

To find the original cost of the dress, we will follow the steps below;

let x represent the original cost of the dress

75% of x  = $90

\frac{75}{100}  ×    x   = $90

\frac{75X}{100}  =   $90

MULTIPLY both-side of the equation by 100

\frac{75X}{100} × 100 =   $90×100

At the right-hand side of the equation, 100 will cancel out 100, leaving us with just 75x

75x = $9000

DIVIDE both-side of the equation by 75

75x/75 = $9000/75

At the left-hand side of the equation 75 will cancel-out 75 leaving us with just x

x=  $9000/75

x = $120

A dress originally costs $120

3 0
3 years ago
What solution is 2x+6=6+2x?
Reptile [31]

Answer:

<h2>x = 2</h2>

Step-by-step explanation:

2x + 6 = 6 + 2x

4 + 6 = 6 + 4

10 = 10

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2 years ago
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timama [110]

Answer:

Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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Questions 16-17 | Math 1 - 0 points Solve the graph Help needed !!
iris [78.8K]

Answer:

16) The area of the circle is 25.1 units²

17) JKLM is a parallelogram but not a rectangle

Step-by-step explanation:

16) Lets talk about the area of the circle

- To find the area of the circle you must find the length of the radius

- In the problem you have the center of the circle and a point on

 the circle, so you can find the length of the radius by using the

 distance rule

* Lets solve the problem

∵ The center of the circle is (1 , 3)

∵ The point on the circle is (3 , 5)

- Using the rule of the distance between two points

* Lets revise it

- The distance between the two points (x1 , y1) and (x2 , y2) is:

 Distance = √[(x2 - x1)² + (y2 - y1)²]

∴ r = √[(3 - 1)² + (5 - 3)²] = √[4 + 4] = √8 = 2√2 units

∵ The area of the circle = πr²

∴ The area of the circle = π (2√2)² = 8π = 25.1 units²

* The area of the circle is 25.1 units²

17) To prove a quadrilateral is a parallelogram, prove that every

     to sides are parallel or equal or the two diagonal bisect

     each other

* The parallelogram can be rectangle if two adjacent sides are

  perpendicular to each other (measure of angle between them is 90°)

 or its diagonals are equal in length

- The parallel lines have equal slopes, then to prove the

  quadrilateral is a parallelogram, we will find the slopes of

  each opposite sides

* Lets find from the graph the vertices of the quadrilateral

∵ J = (0 ,2) , K (2 , 5) , L (5 , 0) , M (3 , -3)

- The opposite sides are JK , ML and JM , KL

- The slope of any line passing through point (x1 , y1) and (x2 , y2) is

 m = (y2 - y1)/(x2 - x1)

∵ The slop of JK = (5 - 2)/(2 - 0) = 3/2 ⇒ (1)

∵ The slope of LM = (-3 - 0)/(3 - 5)= -3/-2 = 3/2 ⇒ (2)

- From (1) and (2)

∴ JK // LM

∵ The slope of KL = (0 - 5)/(5 - 2) = -5/3 ⇒ (3)

∵ The slope of JM = (-3 - 2)/(3 - 0)= -5/3 ⇒ (4)

- From (3) and (4)

∴ KL // JM

∵ Each two opposite sides are parallel in the quadrilateral JKLM

∴ It is a parallelogram

- The product of the slopes of the perpendicular line is -1

* lets check the slopes of two adjacent sides in the JKLM

∵ The slope of JK = 3/2 and the slope of KL = -5/3

∵ 3/2 × -5/3 = -5/2 ≠ -1

∴ JKLM is a parallelogram but not a rectangle

4 0
3 years ago
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