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Kay [80]
4 years ago
5

How many real solutions does the equation 0.2x^5-2x^3+1.8x+k=0 have when k=1?

Mathematics
1 answer:
Tems11 [23]4 years ago
5 0

A graphing calculator shows it has 3 real solutions, near x ∈ {-3.033, 1.241, 2.964}.

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Solve for x.<br><br> −4/5x+3/10&lt;8/10
Genrish500 [490]

Answer:

for (−4/5)x + 3/10 < 8/10:  solution is x > -5/8

for −4/(5x) + 3/10 < 8/10: solution is x > -8/5

Step-by-step explanation:

I'm not exactly sure if you mean -4 divided by 5x or (-4/5)x, so I've done both! :)

-\frac{4}{5}x+ \frac{3}{10} < \frac{8}{10}

subtract 3/10 from both sides:  -\frac{4}{5}x < \frac{5}{10}

Divide both sides by 4/5:  -x < \frac{5}{8}

Divide both sides by -1 (reverse sign):  x > -\frac{5}{8}

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subtract 3/10 from both sides:  -\frac{4}{5x} < \frac{5}{10}

Multiply both sides by 5x:  -4 < \frac{5}{2}x

Divide both sides by 5/2:  x>- \frac{8}{5}

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