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ASHA 777 [7]
3 years ago
14

PLEASE HELP ME ITS IXL ​

Mathematics
2 answers:
Natasha_Volkova [10]3 years ago
5 0

Answer:

Step-by-step explanation:

Blababa [14]3 years ago
3 0

Answer:

upper diagram

12

Step-by-step explanation:

Diagram:

The upper diagram shows 4 penalty kicks made and 1 missed.

That is a ratio of 4 to 1 of penalty kicks made to penalty kicks missed.

Answer: upper diagram

Number of penalty goals made and missed:

We can guess the numbers until the numbers work.

For each number of missed kicks we guess, we guess 4 times as many kicks he made.

For example, we guess 1 missed kick and 4 kicks he made.

First guess:

Maybe he made 4 and missed 1.

4 - 1 = 3, not 9, so this is not the answer.

Second guess:

Maybe he made 8 and missed 2.

8 - 2 = 6, not 9, so this is not the answer.

Third guess:

Maybe he made 12 and missed 3.

12 - 3 = 9, which is what we need, so this is the answer.

He made 12 and missed 3

Answer: 12

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40

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2 years ago
Write and equation of the translated or rotated graph in general form (picture below)
WINSTONCH [101]

Answer:

The answer is hyperbola; (x')² - (y')² - 16 = 0 ⇒ answer (a)

Step-by-step explanation:

* At first lets talk about the general form of the conic equation

- Ax² + Bxy + Cy²  + Dx + Ey + F = 0

∵ B² - 4AC < 0 , if a conic exists, it will be either a circle or an ellipse.

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* Now we will study our equation:

 xy = -8

∵ A = 0 , B = 1 , C = 0

∴ B² - 4 AC = (1)² - 4(0)(0) = 1 > 0

∴ B² - 4AC > 0

∴ The graph is hyperbola

* The equation xy = -8

∵ We have term xy that means we rotated the graph about

  the origin by angle Ф

∵ Ф = π/4

∴ We rotated the x-axis and the y-axis by angle π/4

* That means the point (x' , y') it was point (x , y)

- Where x' = xcosФ - ysinФ and y' = xsinФ + ycosФ

∴ x' = xcos(π/4) - ysin(π/4) , y' = xsin(π/4) + ycos(π/4)

∴ x' = x/√2 - y/√2 = (x - y)/√2

∴ y' = x/√2 + y/√2 = (x + y)/√2

* Lets substitute x' and y' in the 1st answer

∵ (x')² - (y')² - 16 = 0

∴ (\frac{x-y}{\sqrt{2}})^{2}-(\frac{x+y}{\sqrt{2}})^{2}=

 ( \frac{x^{2}-2xy+y^{2}}{2})-(\frac{x^{2}+2xy+y^{2}}{2})-16=0

* Lets open the bracket

∴ \frac{x^{2}-2xy+y^{2}-x^{2}-2xy-y^{2}}{2}-16=0

* Lets add the like terms

∴ \frac{-4xy}{2}-16=0

* Simplify the fraction

∴ -2xy - 16 = 0

* Divide the equation by -2

∴ xy + 8 = 0

∴ xy = -8 ⇒ our equation

∴ Answer (a) is our answer

∴ The answer is hyperbola; (x')² - (y')² - 16 = 0

* Look at the graph:

- The black is the equation (x')² - (y')² - 16 = 0

- The purple is the equation xy = -8

- The red line is x'

- The blue line is y'

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