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Burka [1]
2 years ago
14

Consider the statement “x is a number such that x < 2

Mathematics
1 answer:
polet [3.4K]2 years ago
8 0

Anything less than 2, can even be 1.999999. What ever way the open part is facing is the side that has more than the other.


Step-by-step explanation:


You might be interested in
Which expression is equivalent to 24x2 - 22x + 5?​
enot [183]

Answer:

C

Step-by-step explanation:

Given

24x² - 22x + 5

Consider the factors of the product of the x² term and the constant term which sum to give the coefficient of the x- term.

product = 24 × 5 = 120 and sum = - 22

The factors are - 12 and - 10

Use these factors to split the x- term

24x² - 12x - 10x + 5 ( factor the first/second and third/fourth terms )

= 12x(2x - 1) + 5(2x - 1)  ← factor out (2x - 1) from each term

= (12x - 5)(2x - 1) ← in factored form → C

4 0
3 years ago
Given that P = xy.<br> Find P when:<br> x = -5 and y = -9
boyakko [2]

Answer:

45

Step-by-step explanation:

8 0
2 years ago
Read 2 more answers
Help me please thank you!!
Alexxandr [17]

The answer to your problem is A because you must add them altogether but i might be wrong im just trying to help you out

5 0
3 years ago
Find the vertices and foci of the hyperbola with equation quantity x plus one squared divided by sixteen minus the quantity of y
katrin2010 [14]

Answer:

The vertices are (3 , -5) , (-5 , -5)

The foci are (4 , -5) , (-6 , -5)

Step-by-step explanation:

* Lets study the equation of the hyperbola

- The standard form of the equation of a hyperbola with  

  center (h , k) and transverse axis parallel to the x-axis is

  (x - h)²/a² - (y - k)²/b² = 1

- The length of the transverse axis is 2 a

- The coordinates of the vertices are  (h  ±  a  ,  k)

- The coordinates of the foci are (h ± c , k), where c² = a² + b²

- The distance between the foci is  2c

* Now lets solve the problem

- The equation of the hyperbola is (x + 1)²/16 - (y + 5)²/9 = 1

* From the equation

# a² = 16 ⇒ a = ± 4

# b² = 9 ⇒ b = ± 3

# h = -1

# k = -5

∵ The vertices are (h + a , k) , (h - a , k)

∴ The vertices are (-1 + 4 , -5) , (-1 - 4 , -5)

* The vertices are (3 , -5) , (-5 , -5)

∵ c² = a² + b²

∴ c² = 16 + 9 = 25

∴ c = ± 5

∵ The foci are (h ± c , k)

∴ The foci are (-1 + 5 , -5) , (-1 - 5 , -5)

* The foci are (4 , -5) , (-6 , -5)

4 0
3 years ago
Read 2 more answers
Solve the equation graphically: √ x = - x.
Nikolay [14]
Sqrt[x] = -x   
x= x ^ 2  
x - x^2 = 0
-x(x-1)=0
x(x-1)=0
x = 1 or x = 0
sqrt 1 isn not equal to 0

x= 0

6 0
3 years ago
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