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Flura [38]
4 years ago
12

Which lines are perpendicular to the line y – 1 = (x+2)? Check all that apply. y + 2 = –3(x – 4) y − 5 = 3(x + 11) y = -3x – y =

x – 2 3x + y = 7
Mathematics
1 answer:
Harlamova29_29 [7]4 years ago
4 0
The answers:

a, c, & e


your welcome
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2. What is the value of x if the angles<br> within a triangle are 2x +1, 7x-2 and 3x<br> +5?
noname [10]

Answer:

 x=14.67\\°

Step-by-step explanation:

The sum of all the three angles of any triangle is 180°

Putting sum of all the angles given equals to 180°

                      (2x+1)+(7x-2)+(3x+5)=180°

                              2x+7x+3x+1-2+5=180°

                                                         12x+4=180°

                                                                   x=\frac{180-4}{12}

                                                                   x=14.67\\°

The value of the x is found on solving the given angles of the triangles.

8 0
3 years ago
In ΔBCD, \overline{BD} BD is extended through point D to point E, m∠CDE = (6x-13)^{\circ}(6x−13) ∘ , m∠BCD = (2x+10)^{\circ}(2x+
AveGali [126]

Answer:

  m∠CDE = 35°

Step-by-step explanation:

The exterior angle CDE is equal to the sum of the opposite interior angles, BCD and DBC. So, you have ...

  6x -13 = (2x+10) +(x +1)

  3x -24 = 0 . . . . . . . . . . . . subtract the right side, simplify

  x -8 = 0 . . . . . . . . . . . . . . divide by 3

  x = 8 . . . . . . . . . . . . . . . . . add 8

  6x -13 = 6·8 -13 = 35 . . . . substitute for x in the expression for CDE

The measure of angle CDE is 35°.

7 0
3 years ago
L1 : y = 2x , (2) find the equation of the line L2 perpendicular to L1 passing through the point P = (1, 2).
just olya [345]

Answer:

<h2>2y+x = 5</h2>

Step-by-step explanation:

Given the line L1 as y = 2x perpendicular to an unknown line L2 passing through the point P = (1, 2), we are to find the equation of line L2. to find the equation of the line L2, we will use the point-slope equation of a line expressed as y-y₀ = m(x-x₀)

m is the slope of the unknown line

(x₀, y₀) is the given point.

First is to get the slope of the known line:

comparing the line L1: y = 2x with the standard equation of the line y = mx+c, it can be seen that m = 2

Then we will calculate the slope of the required line.

Since L1 is perpendicular to L2, the product of their slope will be -1 i.e

mm₁ = -1 where m₁ is the slope of the required line L2.

Given m =2

m₁ = -1/m

m₁ = -1/2

Finally we will calculate the equation of line L2 by substituting the slope of line L2 and the point in the point slope equation above;

y-y₀ = m(x-x₀)

Given (x₀, y₀) = (1,2) and m₁ = -1/2

y-2 = -1/2(x-1)

open the parenthesis

y-2 = -x/2+1/2

multiply through by 2:

2y-4 = -x+1

2y+x = 1+4

2y+x = 5

<em>Hence the equation of the line L2 is 2y+x = 5</em>

<em></em>

8 0
3 years ago
Please help me if possible
Andrew [12]

Answer:

14√2

Step-by-step explanation:

Using 45 45 90 triangle properties,

b = 28/√2 = 14√2

8 0
3 years ago
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Ratling [72]
314....1020.50/3.25 the decimal moves two places to the right in both numbers
7 0
3 years ago
Read 2 more answers
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