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MrRa [10]
2 years ago
11

Elijah’s home is 1.1km from the library

Mathematics
1 answer:
julsineya [31]2 years ago
5 0

Elijah was travelling before his stop than afterwards with speed of 0.5 km/h.

<h3>What is formula to calculate speed or velocity?</h3>

To solve for distance use the formula for distance d = vt, or distance equals speed times time. Rate and speed are similar since they both represent some distance per unit time like miles per hour or kilometers per hour.

Before his stop, the velocity =\frac{0.4km}{0.1 h}

                                              = 4 km/h

After his stop, the velocity =\frac{2.7km}{2.2 h}

                                               = 3.5 km/h

So, 4-3.5 km/h= 0.5 km/h

Elijah was travelling before his stop than afterwards with speed of 0.5 km/h

Learn more about speed and velocity here:

brainly.com/question/13271165

#SPJ1

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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
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