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3 years ago

Which of the following equations does not represent a linear relationship between x and y

Mathematics

1 answer:

Ksenya-84 [330]3 years ago

7 0

A linear relationship is an equation that has a constant slope

The equation that does not represent a linear relationship is (b) y = 2x^2 + 5x

<h3>How to determine the non-linear relationship</h3>

A linear equation can take any of the following forms:

y = mx + b

y - y1 = m(x - x1)

Ax + By = c

Any form different from the above forms is not a linear equation

Using the above format as a guide, the equation that does not represent a linear relationship is (b) y = 2x^2 + 5x

Read more about linear relationships at:

brainly.com/question/15602982

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Please look at picture attached! i will mark you brainlyest if right!

katrin2010 [14]

Amp is the number in front of sine so its 1/2 aka D :)

8 0

2 years ago

Can i get help please

Natasha2012 [34]

Answer:

Function 2 is linear

Step-by-step explanation:

It is linear because all of the numbers that in x roc is 1 and y is -4

3 0

3 years ago

Solve the differential. This was in the 2016 VCE Specialist Maths Paper 1 and i'm a bit stuck

Nimfa-mama [501]

$\sqrt{2 - x^{2}} \cdot \frac{dy}{dx} = \frac{1}{2 - y}$
$\frac{dy}{dx} = \frac{1}{(2 - y)\sqrt{2 - x^{2}}}$

Now, isolate the variables, so you can integrate.
$(2 - y)dy = \frac{dx}{\sqrt{2 - x^{2}}}$
$\int (2 - y)\,dy = \int\frac{dx}{\sqrt{2 - x^{2}}}$
$2y - \frac{y^{2}}{2} = sin^{-1}\frac{x}{\sqrt{2}} + \frac{1}{2}C$

$4y - y^{2} = 2sin^{-1}\frac{x}{\sqrt{2}} + C$
$y^{2} - 4y = -2sin^{-1}\frac{x}{\sqrt{2}} - C$
$(y - 2)^{2} - 4 = -2sin^{-1}\frac{x}{\sqrt{2}} - C$
$(y - 2)^{2} = 4 - 2sin^{-1}\frac{x}{\sqrt{2}} - C$

$y - 2 = \pm\sqrt{4 - 2sin^{-1}\frac{x}{\sqrt{2}} - C}$
$y = 2 \pm\sqrt{4 - 2sin^{-1}\frac{x}{\sqrt{2}} - C}$

At x = 1, y = 0.
$0 = 2 \pm\sqrt{4 - 2sin^{-1}\frac{1}{\sqrt{2}} - C}$
$-2 = \pm\sqrt{4 - 2sin^{-1}\frac{1}{\sqrt{2}} - C}$

$4 - 2sin^{-1}\frac{1}{\sqrt{2}} - C > 0$
$\therefore 2 = \sqrt{4 - 2sin^{-1}\frac{1}{\sqrt{2}} - C}$

$4 = 4 - 2sin^{-1}\frac{1}{\sqrt{2}} - C$
$0 = -2sin^{-1}\frac{1}{\sqrt{2}} - C$
$C = -2sin^{-1}\frac{1}{\sqrt{2}} = -2\frac{\pi}{4}$
$C = -\frac{\pi}{2}$

$\therefore y = 2 - \sqrt{4 + \frac{\pi}{2} - 2sin^{-1}\frac{x}{\sqrt{2}}}$

6 0

3 years ago

Help me plz:<br><br> 8. Multiply the following:<br><br> c. (3s^4)(-6s^5)

ratelena [41]

Answer:

-18 s9

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

What is the area of the shaded region?

USPshnik [31]

Answer:

Area of the whole rectangle = 11 x 12

= 132 cm^2

Area of the circle = πr^2

= 3.14 x 2 x 2

= 12.56 cm^2

Area of the shaded region = area of rectangle - area of circle

= 132 - 12.56

= 119.44 cm^2

Hope this helps!

7 0

3 years ago