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ss7ja [257]
3 years ago
8

6x - 2(x + 2) > 2 - 3(x+3)Solve the inequality plz help ​

Mathematics
3 answers:
harina [27]3 years ago
8 0

Answer:

x>-3/7

Step-by-step explanation:

6x-2x-4>2-3x-9

4x-4>-3x-7

7x-4>-7

7x>-3

x>-3/7

Mumz [18]3 years ago
7 0

Answer:

x > -3/7

Step-by-step explanation:

6x - 2(x + 2) > 2 - 3(x + 3)

Distribute the two and three inside the parenthesis.

6x - 2x - 4 > 2 - 3x - 9

Combine like terms.

4x - 4 > -3x - 7

Add 4 to both sides.

4x > -3x - 3

Add 3x to both sides.

7x > -3

Divide both sides by 7.

x > -3/7

panboo2 years ago
0 0

x>-3/7 I think

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What is 3 1/3 divided by 1/3
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Convert 3 1/3 into a improper fraction:
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4 years ago
Read 2 more answers
F(x) = <img src="https://tex.z-dn.net/?f=%5Csqrt%7Bx%2B7%7D%20-%5Csqrt%7Bx%5E2%2B2x-15%7D" id="TexFormula1" title="\sqrt{x+7} -\
elena-s [515]

Answer:

x >= -7  ................(1a)

x >= 3   ...............(2a1)

Step-by-step explanation:

f(x) =  \sqrt{x+7}-\sqrt{x^2+2x-15}  .............(0)

find the domain.

To find the (real) domain, we need to ensure that each term remains a real number.

which means the following conditions must be met

x+7 >= 0  .....................(1)

AND

x^2+2x-15 >= 0 ..........(2)

To satisfy (1),  x >= -7  .....................(1a) by transposition of (1)

To satisfy (2), we need first to find the roots of (2)

factor (2)

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This implis

(x+5) >= 0 AND (x-3) >= 0.....................(2a)

OR

(x+5) <= 0 AND (x-3) <= 0 ...................(2b)

(2a) is satisfied with x >= 3   ...............(2a1)

(2b) is satisfied with x <= -5 ................(2b1)

Combine the conditions (1a), (2a1), and (2b1),

x >= -7  ................(1a)

AND

(

x >= 3   ...............(2a1)

OR

x <= -5 ................(2b1)

)

which expands to

(1a) and (2a1)   OR  (1a) and (2b1)

( x >= -7 and x >= 3 )  OR ( x >= -7 and x <= -5 )

Simplifying, we have

x >= 3  OR ( -7 <= x <= -5 )

Referring to attached figure, the domain is indicated in dark (purple), the red-brown and white regions satisfiy only one of the two conditions.

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