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Ede4ka [16]
3 years ago
9

Which algebraic expression is equivalent to the expression below?

Mathematics
2 answers:
Andreas93 [3]3 years ago
6 0
The answer is D hope this helps
kirza4 [7]3 years ago
5 0

Answer:

D

Step-by-step explanation:

(16x +1) -5x

you basically subtract the -5x from 16x

11x +1

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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
A ,b, c, d , e, help please
Lelu [443]

Answer:

a= 131 21/30

b= 132 21/30

c= 132

Step-by-step explanation:

7 0
3 years ago
Please help me out with this question please!!!
bixtya [17]

Answer:

3 days

Step-by-step explanation:

company A would have the equation 50x+40 =y  and company B would be 30x + 100 =y. we're tying to find the days that these cost the same so you would do 50x+40 = 30+100. When you isolate x you get 3

7 0
3 years ago
Please help with 4, part a and b
Troyanec [42]
Z<=16+2
z-2<=16
Zina spent at most 18 dollars
4 0
4 years ago
in a standard casino dice game the roller wins on the first roll if he rolls a sum of 7 or 11. what is the probability of winnin
Alborosie

<u>Answer-</u>

<em>The probability of winning on the first roll is </em><em>0.22</em>

<u>Solution-</u>

As in the game of casino, two dice are rolled simultaneously.

So the sample space would be,

|S|=6^2=36

Let E be the event such that the sum of two numbers are 7, so

E = {(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)}

|E|=56

\therefore P(E)=\dfrac{|E|}{|S|}=\dfrac{6}{36}

Let F be the event such that the sum of two numbers are 11, so

F = {(6,5), (5,6)}

|F|=2

\therefore P(F)=\dfrac{|F|}{|S|}=\dfrac{2}{36}

Now,

P(\text{sum is 7 or 11)}=P(E\ \cup\ F)=P(E)+P(F)=\dfrac{6}{36}+\dfrac{2}{36}=\dfrac{8}{36}=\dfrac{2}{9}=0.22

8 0
4 years ago
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