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3 years ago

Solve for X 15= 5/7x

2 answers:

4 0

15 = 5/7x

To simplify, you can either divide both sides by 5/7, which is roughly 0.71, or you can multiply both sides by the inverse of 5/7, which is 7/5 (or 1.4).

15 * (7/5) = 5/7(7/5)x

21 = x

GenaCL600 [577]3 years ago

3 0

15 = 5/7x
7/5 7/5

21 = x

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A farmer has a field in the shape of a semicircle of diameter 50 m. First,

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Answer:

157.1 m

Step-by-step explanation:

Here, we want to calculate the length of the semi-circle

This simply refers to the circumference of the semi-circle

Mathematically, the circumference is half that of a full circle

The circumference of a full circle is;

C = pi * d

For half circle or semi-circle;

It is ;

C = (pi * d)/2

D = 50 m

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C = 50 * 3.142

C = 157.1 m

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3 years ago

Which equation represents a proportional relationship?

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4 years ago

Which expression is equivalent to 7 (x y)?

joja [24]

Answer:

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Step-by-step explanation:

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2 years ago

Evaluate the following: (½)(10) −(−12) −25

mr Goodwill [35]

Answer:

-8

Step-by-step explanation:

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3 years ago

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

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3 years ago