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Novay_Z [31]
3 years ago
11

The quotient of 50 and n

Mathematics
1 answer:
omeli [17]3 years ago
3 0

Answer:

D is the answers for the question

Step-by-step explanation:

please mark me as brainlest

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7x 1 1/5 enter answer as a mixed number
Leno4ka [110]

The answer is

8 \frac{2}{5}

6 0
2 years ago
Help please !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
SpyIntel [72]
Hang in there just a sec
4 0
3 years ago
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)”

* <em>Lets solve the problem</em>

- 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)

8 0
3 years ago
You flip a coin.
SVEN [57.7K]
50 percent is the probability
7 0
3 years ago
Read 2 more answers
Part A: Sydney made $18.50 selling lemonade, by the cup, at her yard sale. She sold each cup for $0.50 and received a $3 tip fro
prisoha [69]

Answer:

Part A: x0.50 + 3 = 18.50

Part B: x0.75 + 3 - x0.10 = 21

Part C: The equations from Part A and Part B differ because of the cost of plastic cup.

Step-by-step explanation:

Let x represent the number of cup of lemonade sold. Therefore, we have:

Part A:

This situation can be represented by the following equation:

x0.50 + 3 = 18.50

Part B:

This situation can be represented by the following equation:

x0.75 + 3 - x0.10 = 21

Part C:

The equations from Part A and Part B differ because of the cost of plastic cup.

For equation from Part A, revenue is the same as profit as Sydney does incur any cost to buy plastic cup before selling her lemonade.

For equation from Part B, revenue is different from profit because Daria has to incur the cost of plastic cup which $0.10 per cup of lemonade before selling her lemonade.

8 0
3 years ago
Read 2 more answers
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