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nirvana33 [79]
4 years ago
9

PLEASE ANSWER I NEED THIS ASAP

Mathematics
2 answers:
Rudik [331]4 years ago
5 0
It would also be 42 because the triangle is divided in half.
PolarNik [594]4 years ago
4 0
42ft2 because it's 7x12 divided by 2
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Find an equation for the line tangent to the circle x^2 +y^2=25 at the point (3, -4)
Lostsunrise [7]
Hello : 
the center of this cercle is : O( 0, 0)
calculate the slope  m of the line : OA    .. A (3,-4)
m= (yA - yO) / (xA- xO) 
m = (-4-0)/(3-0)
m= -4/3
if : k the slope of the tangent : k×m = -1   (the tangent perpendiclar of : OA)
k× (-4/3) = -1
k = 3/4
an equation of this tangent is : y - (-4) = (3/4)(x-3)

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4 years ago
Pleaseeeeeee answer!!!!!!!!!!!!
Reil [10]

-154= -214 - -60

228= 14 - -214

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3 years ago
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Solve for n, please i would really appreciate it :)
Lelu [443]

Answer:

2

Step-by-step explanation:

{ {a}^{b} }^{c}  =  {a}^{bc } \\ in \: this \: case \\  { ({8}^{n}) }^{3}  =  {8}^{3n}  \\ \\  {8}^{3n}  =  {8}^{6 }  \\ 3n = 6 \\ n =  \frac{6}{3}  = 2

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3 years ago
Help! ASAP! <br> f(x)=x^2 . What is g(x)?
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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
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