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2 years ago

Classify 3

2 answers:

7 0

A because 3 is rational, and it’s a natural and whole number because it’s about 0. It’s an integer because integers include negative and positive numbers.

Hunter-Best [27]2 years ago

5 0

It c I had this question I know it's c

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What is the solution of

kobusy [5.1K]

Answer:

Third option: x=0 and x=16

Step-by-step explanation:

$\sqrt{2x+4}-\sqrt{x}=2$

Isolating √(2x+4): Addind √x both sides of the equation:

$\sqrt{2x+4}-\sqrt{x}+\sqrt{x}=2+\sqrt{x}\\ \sqrt{2x+4}=2+\sqrt{x}$

Squaring both sides of the equation:

$(\sqrt{2x+4})^{2}=(2+\sqrt{x})^{2}$

Simplifying on the left side, and applying on the right side the formula:

$(a+b)^{2}=a^{2}+2ab+b^{2}; a=2, b=\sqrt{x}$

$2x+4=(2)^{2}+2(2)(\sqrt{x})+(\sqrt{x})^{2}\\ 2x+4=4+4\sqrt{x}+x$

Isolating the term with √x on the right side of the equation: Subtracting 4 and x from both sides of the equation:

$2x+4-4-x=4+4\sqrt{x}+x-4-x\\ x=4\sqrt{x}$

Squaring both sides of the equation:

$(x)^{2}=(4\sqrt{x})^{2}\\ x^{2}=(4)^{2}(\sqrt{x})^{2}\\ x^{2}=16 x$

This is a quadratic equation. Equaling to zero: Subtract 16x from both sides of the equation:

$x^{2}-16x=16x-16x\\ x^{2}-16x=0$

Factoring: Common factor x:

x (x-16)=0

Two solutions:

1) x=0

2) x-16=0

Solving for x: Adding 16 both sides of the equation:

x-16+16=0+16

x=16

Let's prove the solutions in the orignal equation:

1) x=0:

$\sqrt{2x+4}-\sqrt{x}=2\\ \sqrt{2(0)+4}-\sqrt{0}=2\\ \sqrt{0+4}-0=2\\ \sqrt{4}=2\\ 2=2$

x=0 is a solution

2) x=16

$\sqrt{2x+4}-\sqrt{x}=2\\ \sqrt{2(16)+4}-\sqrt{16}=2\\ \sqrt{32+4}-4=2\\ \sqrt{36}-4=2\\ 6-4=2\\ 2=2$

x=16 is a solution

Then the solutions are x=0 and x=16

5 0

3 years ago

1. Find the domain and range of the function<br> f(x)=√1-4x².

slamgirl [31]

$f(x) = \sqrt{1 - 4{ x}^{2} }$

$1 - 4x {}^{2} \geqslant 0$

$- \infty \: \: \: \: \: \: \: \: - 0.5 \: \: \: \: \: \: \: \: \: 0.5 \: \: \: \: \: \: \: \: \infty \\ - \: \: \: \: \: \: \: \: \: \: \: \: \: \:0 \: \: \: \: \: + \: \: \: \:0 \: \: \: \: \: -$