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3 years ago

Please help questions 18-20 for 49 POINTS

Mathematics

2 answers:

yanalaym [24]3 years ago

8 0

Ok, so since they arre all the same length....

18. sides are all same length
so
5-2x=-4x+9
solve
add 4x both sides
5+2x=9
mins 5 both sides
2x=4
x=2

sub back
5-2x=5-2(2)=5-4=1
the length of each side is 1
3 sides so the perimiter is 3 units

19.
same side length so
3(x-1)=5x-6
distribute
3x-3=5x-6
minus 3x both sides
-3=2x-6
add 6 both sides
3=2x
divide 2
3/2=x
sub back
5x-6
5(3/2)-6
15/2-6
7.5-6
1.5

5 sides so
1.5 times 5=7.5 units is the perimiter
which is about 15/2 units

20.
we can't see what's after the x+? so this is a mystery quesiton

Sauron [17]3 years ago

3 0

<h2><em>I cant see Question 20 tho</em></h2><h2><em>I wan to help if I can see it</em></h2>

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LAST OF MY GEOMETRY PLZ HELP

makvit [3.9K]

Answer:

Step-by-step explanation:

14) 2x - 10 = 80 {alternate interior angles are equal}

2x = 80 + 10

2x = 90

x = 90/2

x = 45°

15) 3x + 120 = 180 { co interior angles are supplementary}

3x = 180 - 120

3x = 60

x = 60/3

x = 20°

16) 3x + 15 = 135 {corresponding angles are equal}

3x = 165 - 15

3x = 150

x = 150/3

x = 50°

4 0

2 years ago

(12x + y + z = 26

mel-nik [20]

Option D. D has the matrix of constants [[12], [11], [4]].

Step-by-step explanation:

Step 1:

With the given equations, we can form matrices to represent them.

The coefficients of x, y, and z form a matrix of order 3 ×3, the variables x, y, and z form a matrix of order 1 ×3 and the constants form a matrix of order 1 ×3.

Step 2:

The linear system A is represented as

$\left[\begin{array}{ccc}12&1&1\\1&-11&0\\1&-1&4\end{array}\right]$ $\left[\begin{array}{ccc}x\\y\\z\end{array}\right]$ $= \left[\begin{array}{ccc}26\\17\\23\end{array}\right]$ .

Step 3:

The linear system B is represented as

$\left[\begin{array}{ccc}4&1&1\\1&-11&0\\1&-1&12\end{array}\right]$ $\left[\begin{array}{ccc}x\\y\\z\end{array}\right]$ $= \left[\begin{array}{ccc}23\\17\\26\end{array}\right]$ .

Step 4:

The linear system C is represented as

$\left[\begin{array}{ccc}1&1&1\\1&-1&0\\1&-1&1\end{array}\right]$ $\left[\begin{array}{ccc}x\\y\\z\end{array}\right]$ $= \left[\begin{array}{ccc}4\\11\\12\end{array}\right]$ .

Step 5:

The linear system D is represented as

$\left[\begin{array}{ccc}1&1&1\\1&-1&0\\1&-1&1\end{array}\right]$ $\left[\begin{array}{ccc}x\\y\\z\end{array}\right]$ $= \left[\begin{array}{ccc}12\\11\\4\end{array}\right]$ .