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lisabon 2012 [21]
3 years ago
14

What is the value of x?12 units15 units20 units25 units

Mathematics
2 answers:
Sholpan [36]3 years ago
8 0

Answer:

The value of x equals 12 units

Step-by-step explanation:

simple maths :)

scZoUnD [109]3 years ago
4 0

Answer:

∴ The value is 12 units

Step-by-step explanation:

From Geometric Mean Theorem we know that,

In a right triangle, the altitude from the right angle to the hypotenuse divides the hypotenuse into two segments.

The length of the altitude is the geometric mean of the lengths of the two segments.

So, Above theorem apply in the diagram;

\frac{ST}{RT}=\frac{RT}{TQ}

Plug RT=x , ST=9 and TQ=16 from diagram;

\frac{9}{x}=\frac{x}{16}

x^{2} =9\times 16

x^{2} =144

x=12

∴ The value is 12 units

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Answer:

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Step by step:

20yards( 2 ft )+ 30 yards +20 yards + 30 yards=

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I hope it's help.

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Multiply or divide as indicated. x^4 • x^-2
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The answer is 0 unless I did the math wrong


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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
I don't get this problem
Dominik [7]
20/4.... 5cm is ur answer
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