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2 years ago

Convert 0.9 into a fraction using 100 for denominator

Mathematics

2 answers:

Arisa [49]2 years ago

7 0

Remove the decimal sign and divide by 100 to get the answer

In this case: 0.9 = 90 / 100

Nuetrik [128]2 years ago

6 0

$0.9 = \frac{90}{100} = \boxed{ \frac{9}{10} \ Answer}$

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Consider the equation and the relation “(x, y) R (0, 2)”, where R is read as “has distance 1 of”. For example, “(0, 3) R (0, 2)”

Leviafan [203]

Answer:

The equation determine a relation between x and y

x = ± $\sqrt{1-(y-2)^{2}}$

y = ± $\sqrt{1-x^{2}}+2$

The domain is 1 ≤ y ≤ 3

The domain is -1 ≤ x ≤ 1

The graphs of these two function are half circle with center (0 , 2)

All of the points on the circle that have distance 1 from point (0 , 2)

Step-by-step explanation:

* Lets explain how to solve the problem

- The equation x² + (y - 2)² and the relation "(x , y) R (0, 2)", where

R is read as "has distance 1 of"

- This relation can also be read as “the point (x, y) is on the circle

of radius 1 with center (0, 2)”

- “(x, y) satisfies this equation , if and only if, (x, y) R (0, 2)”

* Lets solve the problem

- The equation of a circle of center (h , k) and radius r is

(x - h)² + (y - k)² = r²

∵ The center of the circle is (0 , 2)

∴ h = 0 and k = 2

∵ The radius is 1

∴ r = 1

∴ The equation is ⇒ (x - 0)² + (y - 2)² = 1²

∴ The equation is ⇒ x² + (y - 2)² = 1

∵ A circle represents the graph of a relation

∴ The equation determine a relation between x and y

* Lets prove that x=g(y)

- To do that find x in terms of y by separate x in side and all other

in the other side

∵ x² + (y - 2)² = 1

- Subtract (y - 2)² from both sides

∴ x² = 1 - (y - 2)²

- Take square root for both sides

∴ x = ± $\sqrt{1-(y-2)^{2}}$

∴ x = g(y)

* Lets prove that y=h(x)

- To do that find y in terms of x by separate y in side and all other

in the other side

∵ x² + (y - 2)² = 1

- Subtract x² from both sides

∴ (y - 2)² = 1 - x²

- Take square root for both sides

∴ y - 2 = ± $\sqrt{1-x^{2}}$

- Add 2 for both sides

∴ y = ± $\sqrt{1-x^{2}}+2$

∴ y = h(x)

- In the function x = ± $\sqrt{1-(y-2)^{2}}$

∵ $\sqrt{1-(y-2)^{2}}$ ≥ 0

∴ 1 - (y - 2)² ≥ 0

- Add (y - 2)² to both sides

∴ 1 ≥ (y - 2)²

- Take √ for both sides

∴ 1 ≥ y - 2 ≥ -1

- Add 2 for both sides

∴ 3 ≥ y ≥ 1

∴ The domain is 1 ≤ y ≤ 3

- In the function y = ± $\sqrt{1-x^{2}}+2$

∵ $\sqrt{1-x^{2}}$ ≥ 0

∴ 1 - x² ≥ 0

- Add x² for both sides

∴ 1 ≥ x²

- Take √ for both sides

∴ 1 ≥ x ≥ -1

∴ The domain is -1 ≤ x ≤ 1

* The graphs of these two function are half circle with center (0 , 2)

* All of the points on the circle that have distance 1 from point (0 , 2)

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2 years ago