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2 years ago

Solve the inequality 7(x + 4) > 0. x < -4 x > -4 x < 4 x > 4

2 answers:

7 0

7(x+4)>0

Distribute 7. (multiply by x and 4)

7x+28>0

Subtract 28 on both sides.

7x>-28

Divide by 7 into both sides.

x>-4

x>-4 is the answer.

I hope this helps!
~kaikers

MaRussiya [10]2 years ago

4 0

Answer:

x>-4

Step-by-step explanation:

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2 years ago

Read 2 more answers

What are the possible values of x in 8x2 + 4x = -1?

Westkost [7]

Answer:

The possible values of x are:

x= $\frac{-1+\sqrt{3}}{4} \,\,or\,\, x= 0.18$

and

x= $\frac{-1-\sqrt{3}}{4} \,\, or \,\, x= -0.68$

Step-by-step explanation:

$8x^2+4x+1=0$

Using Quadratic formula to solve this equation:

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\a = 8 \,\, b = 4\,\, c = -1\\Putting \,\, values \,\, in \,\, the\,\, equation\\x= \frac{-4\pm\sqrt{(4)^2-4(8)(-1)}}{2(8)}\\x= \frac{-4\pm\sqrt{16+32}}{16}\\x= \frac{-4\pm\sqrt{48}}{16}\\x= \frac{-4+ \sqrt{48}}{16} \,\, and \,\, x= \frac{-4- \sqrt{48}}{16}\\x= \frac{-1+ \sqrt{3}}{4} \,\, and \,\, x= \frac{-1- \sqrt{3}}{4}$

The possible values of x are:

x= $\frac{-1+\sqrt{3}}{4} \,\,or\,\, x= 0.18$

and

x= $\frac{-1-\sqrt{3}}{4} \,\, or \,\, x= -0.68$

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kondor19780726 [428]

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Step-by-step explanation:

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malfutka [58]

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