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Gemiola [76]
3 years ago
6

Can someone please help me thanks!!!!!

Mathematics
1 answer:
natta225 [31]3 years ago
3 0
Sorry what is the question I cannot see the question
You might be interested in
Consider U = {x|x is a real number}. A = {x|x ∈ U and x + 2 > 10} B = {x|x ∈ U and 2x > 10} Which statements are true? 5 ∉
-Dominant- [34]

x+2 > 10 solves to x > 8 after we subtract 2 from both sides

So set A is the set of real numbers that are larger than 8. The value 8 itself is not in set A. The same can be said about 5 as well.


Set B is the set of values that are larger than 5 since 2x > 10 turns into x > 5 after dividing both sides by 2. The value x = 5 is not in set B since x > 5 would turn into 5 > 5 which is false. The values x = 6, x = 8, and x = 9 are in set B.


----------------


Summarizing everything, we can say...

5 is not in set A. True

5 is in set B. False

6 is in set A. False

6 is not in set B. False

8 is not in set A. True

8 is in set B. True

9 is in set A. True

9 is not in set B. False


5 0
3 years ago
Read 2 more answers
Wright the expression 6a - 2 (a-1) in simplest form
vovikov84 [41]
6a-2(a-1)
6a-2a-2
4a-2
I think it is 4a-2
6 0
3 years ago
Kyle went to the store to purchase a hat. The hat’s normal price is $50, but he used a coupon which allowed him to save $20. Wha
tiny-mole [99]

Answer:

Kyle saves 40 percent

$50 is 100% of what he had

$20 would be 40%

50= 100

40= 80

30= 60

20= 40

10= 20

do what u will with that info I'm bad at wording things

4 0
3 years ago
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
4 years ago
I’m need this worked out step by step by tonight
Pie

9514 1404 393

Answer:

  -3 ≤ x ≤ 19/3

Step-by-step explanation:

This inequality can be resolved to a compound inequality:

  -7 ≤ (3x -5)/2 ≤ 7

Multiply all parts by 2.

  -14 ≤ 3x -5 ≤ 14

Add 5 to all parts.

  -9 ≤ 3x ≤ 19

Divide all parts by 3.

  -3 ≤ x ≤ 19/3

_____

<em>Additional comment</em>

If you subtract 7 from both sides of the given inequality, it becomes ...

  |(3x -5)/2| -7 ≤ 0

Then you're looking for the values of x that bound the region where the graph is below the x-axis. Those are shown in the attachment. For graphing purposes, I find this comparison to zero works well.

__

For an algebraic solution, I like the compound inequality method shown above. That only works well when the inequality is of the form ...

  |f(x)| < (some number) . . . . or ≤

If the inequality symbol points away from the absolute value expression, or if the (some number) expression involves the variable, then it is probably better to write the inequality in two parts with appropriate domain specifications:

  |f(x)| > g(x)   ⇒   f(x) > g(x) for f(x) > 0; or -f(x) > g(x) for f(x) < 0

Any solutions to these inequalities must respect their domains.

8 0
3 years ago
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