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3 years ago

A) 4x + 1

Mathematics

1 answer:

tia_tia [17]3 years ago

6 0

Answer:

a) 4x + 1 expression

b) C = 15+ 3n “it might be a Formula”

c) 3x + 4 = 13 equation

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What is the solution to the inequality

Dvinal [7]

Answer:

p > 5 and p <-8

Step-by-step explanation:

To solve this, you first need to isolate p.

First add 6 on both sides of the equation:

$-6 + |2p+3| >7\\\\(+6) -6 + |2p+3| >7 +6\\\\2p + 3 > 13$

Then subtract 3 from both sides of the equation.

$2p+3-3>13-3\\\\2p > 10\\$

The divide both sides by 2.

$\dfrac{2p}{2}>\dfrac{10}{2}\\\\p>5$

Another solution is possible because of the absolute value.

If $|2p+3|>13$

Then $|2p+3|$

Thus we can solve the second solution:

$|2p+3|$

$2p+3$

Isolate P again by subtracting both sides by 3

$2p+3-3$

$2p$

Then divide both sides by 2:

$\dfrac{2p}{2}$

$p$

3 0

3 years ago

Read 2 more answers

Determine the zeroes of the polynomial

Anna71 [15]

Answer:

3,7/6

Step-by-step explanation:

$(\sqrt{x^2-4x+3} )+(\sqrt{x^{2} -9} )-(\sqrt{4x^2-14x+6} )\\=(\sqrt{x^2-x-3x+3} )+(\sqrt{(x^2-3^2})-(\sqrt{4x^2-2x-12x+6})\\ =(\sqrt{x(x-1)-3(x-1)} )+\sqrt{(x+3)(x-3)}-\sqrt{2x(2x-1)-6(2x-1)} \\=\sqrt{(x-1)(x-3)}+\sqrt{(x+3)(x-3)} -\sqrt{2(2x-1)(x-3)} \\=\sqrt{x-3} (\sqrt{x-1} +\sqrt{x+3} -\sqrt{2(2x-1)} )\\$

$\sqrt{x-3} =0~gives~x=3\\or~\sqrt{x-1} +\sqrt{x+3} -\sqrt{2(2x-1)} =0\\or~ \sqrt{x-1} +\sqrt{x+3} =\sqrt{2(2x-1)} \\squaring\\x-1+x+3+2\sqrt{x-1} \sqrt{x+3} =2(2x-1)\\2x+2+2\sqrt{(x-1)(x+3)} =4x-2\\2\sqrt{x^2-x+3x-3} =2x-4\\\sqrt{x^2+2x-3} =x-2\\again ~squaring\\x^2+2x-3=x^2-4x+4\\\\2x+4x=4+3\\6x=7\\x=\frac{7}{6}$