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attashe74 [19]
3 years ago
9

What is the solution for -10 -1 x > -19 omg

Mathematics
2 answers:
Digiron [165]3 years ago
6 0

Answer:

x>9

Step-by-step explanation:

-10-1x>-19

+10      +10

-x>-9

/-   /-

x>9

amid [387]3 years ago
5 0

Answer:

n

Step-by-step explanation:

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The measure of each angles
Aleks [24]
Supplementary angles is when 2 angles add up to 180 degrees.

We will name the smaller angle a and the bigger angle b.

therefore:

b - 172.6 = a

a + b = 180

Now we substitute:

a + b = 180
b - 172.6 + b = 180
2b - 172.6 = 180
2b = 352.6
b = 176.3

To find a:

a + b = 180
a + 176.3 = 180
a = 3.7

Therefore,

angle a = 3.7 degrees
angle b = 176.3 degrees.

Hope it helps and have a great day ahead
4 0
4 years ago
The side of a square are 3 cm long. One vertex of the square is at (2,0) on a square coordinate grid marked in centimeters units
yuradex [85]
(2,0) (5.0) (5,3) (2,3)
(2,0) (-1,0) (2,3) (-1,3)
3 0
3 years ago
What is the circumference of a circle with a radius of 8 in.? Use 3.14 for t.​
Ket [755]

Answer:

201

Step-by-step explanation:

3.14*8*8=201

4 0
3 years ago
Which of the following is the number of sides for a regular polygon that will not form a regular tessellation?
trasher [3.6K]

Answer with explanation:

A regular polygon is simple closed geometrical shape, only made up of line segments.

And Regular Tessellation, is geometrical shape, made up of regular polygons having, 3 ,or 4 or 6 sides.

It is combination of, only single kind of polygon,either, polygon having 3 sides, or ,4 sides or 6 sides.

→Regular polygon having 3 sides is Called Equilateral Triangle.

→Regular polygon having 4 sides is Called Square.

→Regular polygon having 6 sides is Called Regular Hexagon.

⇒Regular polygon having , 8 sides will not form a regular tessellation.

Option D

6 0
3 years ago
Consider the following proability density function: f(x) = { kx. 0<= x< 2, k(4-x). 2<= x <=4, 0. otherwise. find the
Gnesinka [82]

The value of K for which f(x) is a valid probability density function is 1/4.

<h3>How to solve for the value of K</h3>

\int\limits^4_0 {fx(x)} \, dx =1

\int\limits^2_0 {Kx} \, dx +\int\limits^4_2 {K(4-x)} \, dx =1

K[\frac{2^2}{2} -0]+[K[4(4-2)-(\frac{4^2}{2} -\frac{2^2}{2} )]=1

open the equation

K\frac{4}{2}+K[8 - (\frac{16}{2}  -\frac{4}{2} )] = 1\\

2K+K[\frac{4}{2} ]=1

2K + 2K = 1

4K = 1

divide through by 4

K = 1/4

Read more on probability density function here

brainly.com/question/15714810

#SPJ4

4 0
1 year ago
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