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mezya [45]
3 years ago
10

Find the solution to the following system using substitution or elimination

Mathematics
1 answer:
Bas_tet [7]3 years ago
5 0

Answer:

Step-by-step explanation:

3x - 8 = -2x + 7

5x - 8 = 7

5x = 15

x = 3

y = -2(3) + 7 = -6 + 7 = 1

(3, 1)

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Write and solve an equation to find the value of x and the missing angle measures.
Oksana_A [137]

Answer:

Step-by-step explanation:

1). Since, both the angles are vertically opposite angles,

  Measure of both the angles will be same.

  6x = 30

   x = 5

2). Since, both the angles are the linear pair of angles,

   (4 + 5x)° + (x + 2)° = 180°

    6x + 6 = 180

    6x = 180 - 6

    x = \frac{174}{6}

    x = 29

    Therefore, (4 + 5x)° = 4 + 5(29)

                                     = 149°

    And (x + 2)° = (29 + 2)

                        = 31°

3). Since, both the angles are linear pair of angles,

    5x° + (3x + 12)° = 180°

    8x + 12 = 180

    8x = 180 - 12

    x = \frac{168}{8}

    x = 21

    Therefore, 5x° = 5(21)

                             = 105°

    (3x + 12)° = 3(21) + 12

                    = 75°

8 0
3 years ago
two angles are a linear pair. the measure of the first angle minus 39 is equal to twice the measure of the second angle. What ar
baherus [9]

Step-by-step explanation:

Linear pairs = 180°!!!!!

  1. 1st angle is 2[x]-39 2nd angle is just x
  2. soooo add the x. 3x-39=180
  3. 3x=141 x=47

X is just the second angle!!!!!

lol. i toonk geo last year and i cant believe i still remember all this hopefully i got it right!

8 0
3 years ago
I need help plizzzzz
Nataly [62]
Try this dude i hope it is right
5 0
3 years ago
Anyone know how to solve this?? Number 5!!
iragen [17]
P=6+1+2
6+1+2=9 x 3= 27
8 0
3 years ago
Find the area of triangle ABC with vertices A(2,1), B (12,2), C (12,8). Hence, or otherwise find the perpendicular distance from
Lapatulllka [165]
<h2>The area of a triangle is =54 square units</h2><h2>The perpendicular distance from B to AC is = \frac{108}{\sqrt{149} } units</h2>

Step-by-step explanation:

Given a triangle ABC with vertices A(2,1),B(12,2) and C(12,8)

x_1=2,y_1=1,x_2=12,y_2=2,x_3=12 and      y_3=8

The area of a triangle is= \frac{1}{2} [x_1(y_2-y_3) +x_2 (y_3- y_1)+x_3(y_1-y_2)]

=|\frac{1}{2} [2(2-8+12(8-1)+12(1-2)]|

=|-54| = 54 square units

The length of AC = \sqrt{(x_1-x_3)^{2} +(y_1-y_3)^2}

                          = \sqrt{(2-12)^{2} +(1-8)^2}

                         =\sqrt{149} units

Let the perpendicular distance from B to AC be = x

According To Problem

\frac{1}{2} \times  x \times \sqrt{149} = 54

⇔x =\frac{108}{\sqrt{149} } units

Therefore the perpendicular distance from B to AC is = \frac{108}{\sqrt{149} } units

5 0
3 years ago
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