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3 years ago

Based on the graph below, what is a reasonable estimate for the у value when x is 4?

Mathematics

1 answer:

san4es73 [151]3 years ago

6 0

Y=6
Simply double the y value
Since the line is keeping a constant slope.

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3 years ago

Solve the inequality 2x>30+5/4x

insens350 [35]

Answer:

Step-by-step explanation:

$2x > 30+\frac{5}{4x} \\2x-\frac{5}{4x} > 30\\\frac{8x^2-5}{4x} > 30\\case~1\\if~x > 0\\8x^2-5 > 120x\\8x^2-120x > 5\\x^2-15x > \frac{5}{8} \\adding~(-\frac{15}{2} )^2~to~both~sides\\(x-\frac{15}{2} )^2 > \frac{5}{8}+\frac{225}{4} \\(x-\frac{15}{2} )^2 > \frac{455}{8} \\x-\frac{15}{2} < -\sqrt{\frac{455}{8} } \\x < \frac{15}{2}-\sqrt{\frac{455}{8} } \\or~x < 0\\rejected~as~x > 0$

$x-\frac{15}{2} > \sqrt{\frac{455}{8} } \\x > \frac{15}{2} +\sqrt{\frac{455}{8} }$

case~2

$if~x < 0\\8x^2-5 < 120x\\8x^2-120x < 5\\x^2-15x < \frac{5}{8} \\adding~(-\frac{15}{2} )^2\\(x-\frac{15}{2} )^2 < \frac{5}{8} +(-\frac{15}{2} )^2\\|x-\frac{15}{2} | < \frac{5+450}{8} \\-\sqrt{\frac{455}{8} } < x-\frac{15}{2} < \sqrt{\frac{455}{8} } \\\frac{15}{2} -\sqrt{\frac{455}{8} } < x < \frac{15}{2} +\sqrt{\frac{455}{8} } \\but~x < 0\\7.5-\sqrt{\frac{455}{8} } < x < 0$

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2 years ago

What are the parallel lines and the perpendicular lines

djyliett [7]

Parallel is MN and QP and perpendicular is MQ and OP

4 0

3 years ago

(2x+3)²=5.find the roots of the equation

bija089 [108]

Answer:

-2.61803398875, -0.38196601125

(-3±√5)/2

Step-by-step explanation:

First, expand. (2x+3)² = 4x²+12x+9=5

4x²+12x+4 = 0

Quadratic equations take the form ax²+bx+c = 0

In this case,

a = 4

b = 12

c = 4

You can solve with the equation x=(-b±√b²-4ac)/2a

Plug in:

(-12±√144-64)/8

(-12±√80)/8

(-12±√2^4×5)/8

(-12±4√5)/8

(-3±√5)/2

Final answer x = (-3±√5)/2

If you want decimal form: -2.61803398875, -0.38196601125

4 0

2 years ago