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qwelly [4]
3 years ago
14

Does changing the compound inequality x > −3 and x < 3 from “and” to “or” change the solution set? explain.

Mathematics
2 answers:
topjm [15]3 years ago
6 0
Sure yes.

1) x > - 3 or x < 3 means that the solution set is all the numbers greater than - 3 and all the numbers smaller than 3. That is, this union of intervals: (-3,∞) U (-∞,3),  which is all the real numbers.

2) x > -3 and x <3 means that the solution set is the numbers that are at the same time greater than -3 and smaller than 3. That is the intersection of the intervals: (-3,∞) ∩ (-∞,3), which is the interval (-3, 3).
8_murik_8 [283]3 years ago
6 0
Yes. In interval notation: 

<span>x > -3 and x < 3 = (-3, 3) </span>

<span>x > -3 or x < 3 = (-inf, inf) </span>

<span>x > -3 xor x < 3 = (-inf, -3] U [3, inf) where xor is the "exclusive or", either-or but not both.</span>
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Which function has zero at x=10 and x=2
Marizza181 [45]

Answer:

Step-by-step explanation: A function y=(x-10)(x-2) will have its zeros at x=10 and x=2. y=x²-12x+20.

Step-by-step explanation:

3 0
3 years ago
HELPPPPPPPPPPPPPPPPP!!
goldfiish [28.3K]

Answer:

option 3

Step-by-step explanation:

V= πr²h

to make h the subject

h= V÷πr²

just take πr² to the other side

as it in multiplication on the left hand side, it will in division in right hand side.

hope it helps!

5 0
2 years ago
Which value of n makes the equation true?<br><br> 2/3n = -12
vivado [14]

Answer:

-1/18

Step-by-step explanation:

Let's multiply left and right by 3n:

2 = -12 * 3n =>

-36n = 2

now divide by -36:

n = 2/-36 = -1/18

5 0
2 years ago
Read 2 more answers
People are entering a building at a rate modeled by f (t) people per hour and exiting the building at a rate modeled by g (t) pe
erastova [34]

Answer:

The correct option is D) f'(t)-g'(t) > 0

Step-by-step explanation:

Consider the provided information.

People are entering a building at a rate modeled by f (t) people per hour and exiting the building at a rate modeled by g (t) people per hour,

The change of number of people in building is:

h(x)=f(t)-g(t)

Where f(t) is people entering in building and g(t) is exiting from the building.

It is given that "The functions f and g are non negative and differentiable for all times t."

We need to find the the rate of change of the number of people in the building.

Differentiate the above function with respect to time:

h'(x)=\frac{d}{dt}[f(t)-g(t)]

h'(x)=f'(t)-g'(t)

It is given that the rate of change of the number of people in the building is increasing at time t.

That means h'(x)>0

Therefore, f'(t)-g'(t)>0

Hence, the correct option is D) f'(t)-g'(t) > 0

4 0
2 years ago
5÷x-4 - 2÷x+2 = 4÷x-4
Alborosie
I'd suggest you begin by subtracting 4 / (x-4) from both sides.  Doing that would leave you with    1 / (x-4)   -     2 / (x+2) =  0.

LCD is (x-4)(x+2).  Mult. all three terms by (x-4)(x+2).  The resulting equation is 

x+2 - 2(x-4) = 0.     Then x+2 = 2x - 8  =>  x = 10

Subst. 10 for x in the original equation to verify that 10 is indeed a solution.


8 0
3 years ago
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