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bagirrra123 [75]
3 years ago
14

Answer to this problem

Mathematics
1 answer:
Assoli18 [71]3 years ago
7 0
Y = 3^x 

To the x power.

y = 3^(1) = 3
y = 3^(2) = 9
y= 3^(3) = 27 and so on... 
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Which is an equation of the line that passes through the point
Allisa [31]

Answer:

y = \frac{1}{2} x+3

Step-by-step explanation:

4 0
2 years ago
Determine the derivative of y=x^2+xy
ratelena [41]
Hey I just started this test to that's funny lol
7 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
How to solve derivative of (sin3x)/x using first principle ​
Leona [35]

\dfrac{d}{dx}(\dfrac{\sin(3x)}{x})

First we must apply the Quotient rule that states,

(\dfrac{f}{g})'=\dfrac{f'g-g'f}{g^2}

This means that our derivative becomes,

\dfrac{\dfrac{d}{dx}(\sin(3x))x-\dfrac{d}{dx}(x)\sin(3x)}{x^2}

Now we need to calculate \dfrac{d}{dx}(\sin(3x)) and \dfrac{d}{dx}(x)

\dfrac{d}{dx}(\sin(3x))=\cos(3x)\cdot3

\dfrac{d}{dx}(x)=1

From here the new equation looks like,

\dfrac{3x\cos(3x)-\sin(3x)}{x^2}

And that is the final result.

Hope this helps.

r3t40

7 0
2 years ago
Read 2 more answers
Plsss helpppp 6x^2+4x-42
melisa1 [442]

Answer:

Step-by-step explanation:

6x^2+4x-42

2(3x^2+2x-21)=0

3x^2+2x-21=0

3x^2+(9-7)x-21=0

3x^2+9x-7x-21=0

3x(X+3)-7(X+3)=0

(3X-7)(X+3)=0

So Answer is (3x-7)(X+3)

6 0
2 years ago
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