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KIM [24]
3 years ago
7

5 4. y = 1 D 1 1 1 I B

Mathematics
1 answer:
stellarik [79]3 years ago
4 0

Answer:

not sure what you need but I would be happy to help

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I will mark as brainliest please i am newbie in 12​
Vaselesa [24]

Answer:

I can't see the pic

Step-by-step explanation:

8 0
3 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.

∵ B² - 4AC = 0 , if a conic exists, it will be a parabola.

∵ B² - 4AC > 0 , if a conic exists, it will be a hyperbola.

* 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'

6 0
3 years ago
Read 2 more answers
Describe a sequence of transformations that can be performed to polygon ABCDE to prove that it is similar to polygon FGHIJ
maxonik [38]

Answer:

answer below

Step-by-step explanation:

ABCDE go through dilation over center (6 , -2) with factor of 1/2 to FGHIJ

AB // FG slope: -2 , √20:√5 = 2: 1

BC // GH // X axis 8:4 = 2:1

CD // HI, slope= 1 , √8:√2 = 2:1

DE // IJ // x axis, 4:2 = 2:1

EA // JF // y axis, 2:1

8 0
3 years ago
Lol I’m dumb please help
NeX [460]

The answer is 50, just divide 10 and 2, you get 5 and multiply the 5 with the 10

5 0
3 years ago
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These are two lines that do not intersect and have the same slope. what are they called?​
Len [333]

Parallel lines is the answer

4 0
3 years ago
Read 2 more answers
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