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galben [10]
2 years ago
6

Find the missing number 1:2 = 3:_

Mathematics
2 answers:
Andrei [34K]2 years ago
7 0

Answer:

the answer is 6

Step-by-step explanation:

Angelina_Jolie [31]2 years ago
4 0
The answer is 6, hope this helped!
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1/2x-2/3>3 enter your answer using interval notation
Nata [24]
So basically u keep the x on one side and the rest on the other. Get a common denominator. to get rid of the half u times both sides by 2.
hope this helps

7 0
2 years ago
The number of customers in a store on the first day is represented by: (6x - 3) The number of customers on the second day is rep
Scorpion4ik [409]

Answer:the expression is 5x - 2

Step-by-step explanation:

The number of customers in a store on the first day is represented by: (6x - 3).

The number of customers on the second day is represented by: (x - 1)

An expression to find how many more customers visited the store on the first day would be the difference between the number of customers that visited the store on the first day and the number of customers that visited the store on the second day. The expression becomes

6x - 3 - (x - 1)

= 6x - 3 - x + 1

= 6x - x - 3 + 1

= 5x - 2

7 0
3 years ago
HELP me plz someone!
Kamila [148]

Answer:

4,095 ft³

Step-by-step explanation:

V=LWH

V = 3 x 15 x 14

V= 4,095

4 0
2 years ago
Which equation represents a line which is perpendicular to the y-axis?
erma4kov [3.2K]

Answer:

y = 5

Step-by-step explanation:

The y-axis and any lines that are parallel to it are vertical. Therefore, a line that would be perpendicular to the y-axis would need to be horizontal. All horizontal lines have a slope equal to 0 and are represented by the equation y = a constant, or a single number with no variables.  

The only option that follows that format is y = 5, thus that is the answer.  

6 0
2 years ago
Can you help me with my work
andrezito [222]

Answer:

1. x = ±9

2. x=\pm \sqrt{13}

3. 12 and -12.

4. Antoine is incorrect. There exists two solutions x=5 and x= -5.

Step-by-step explanation:

According to the questions,

Problem 1. x^{2}-81=0 i.e. x^{2}=81 i.e. x = ±9.

Problem 2. 2x^{2}-26=0 i.e. x^{2}-13=0 i.e. x^{2}=13 i.e. x=\pm \sqrt{13}

Problem 3. [tex]f(x)=x^{2}-144[tex]

To find the roots, we take, [tex]x^{2}-144=0[tex] i.e. [tex]x^{2}=144[tex] i.e. x = ±12.

Thus, the options are 12 and -12.

Problem 4. We have [tex]f(x)=x^{2}+25[tex]

For the roots, we take, [tex]x^{2}+25=0[tex] i.e. [tex]x^{2}=25[tex] i.e. x = ±5.

Thus, Antoine is not correct and two solutions namely x=5 and x= -5 exists.

8 0
2 years ago
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