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2 years ago

Which function is a quadratic function?

Mathematics

2 answers:

MariettaO [177]2 years ago

8 0

Answer:

h(x) = 3x(x - 2) - 4

Step by-step explanation

A quadratic function has a degree of 2.

Firstly, we expand h(x):

h(x) = 3x (x - 2) - 4

OR h(x) = 3x squared - 6x - 4
Therefore, it is h(x)

Musya8 [376]2 years ago

4 0

Answer:

h(x) = 3x(x - 2) - 4

Step-by-step explanation:

A quadratic function is a polynomial having a degree of 2.

==============================================================

On expanding h(x) :

⇒ h(x) = 3x (x - 2) - 4

⇒ h(x) = 3x² - 6x - 4

=============================================================

Hence, h(x) is a quadratic function.

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3 years ago

Is 16% of 40 the same as 40% of 16

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Answer:

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Step-by-step explanation:

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3 years ago

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Assuming that the equations define x and y implicitly as differentiable functions x=f(t),y=g(t) find the slope of the curve x=f(

Mumz [18]

The given equations are
$x(t+1)-4t \sqrt{x} =9$ (1)
$2y+4y^{3/2}=t^{3}+t$ (2)

When t=0, obtain
$x=9 \\ 2y+4y^{3/2}=0 \,\,=\ \textgreater \ \, y(1+2 \sqrt{y} )=0 \,=\ \textgreater \ \,y=0$

Obtain derivatives of (1) and find x'(0).
x' (t+1) + x - 4√x - 4t*[(1/2)*1/√x = 0
x' (t+1) + x - 4√x -27/√x = 0
When t=0, obtain
x'(0) + x(0) - 4√x(0) = 0
x'(0) + 9 - 4*3 = 0
x'(0) = 3
Here, x' means $\frac{dx}{dt}$ .

Obtain the derivative of (2) and find y'(0).
2y' + 4*(3/2)*(√y)*(y') = 3t² + 1
When t=0, obtain
2y'(0) +6√y(0) * y'(0) = 1
2y'(0) = 1
y'(0) = 1/2.
Here, y' means $\frac{dy}{dt}$ .

Because $\frac{dy}{dx} = \frac{dy}{dt} / \frac{dx}{dt}$ , obtain
$\frac{dy}{dx} |_{t=0}\, = \frac{1/2}{3}= \frac{1}{6}$

Answer:
The slope of the curve at t=0 is 1/6.

3 0

3 years ago