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disa [49]
2 years ago
12

Which represents a quadratic function? f(x) = −8x3 − 16x2 − 4x f (x) = x 2 + 2x − 5 f(x) = + 1 f(x) = 0x2 − 9x + 7

Mathematics
2 answers:
Katena32 [7]2 years ago
8 0
\text{A quadratic function:}\ f(x)=ax^2+bx+c\\\\\text{where}\ a\neq0
f(x)+-8x^3-16x^2-4x-NOT\\\\f(x)=x^2+2x-5-YES\\\\f(x)=?+1-NOT\\\\f(x)=0x^2-9x+7-NOT

Papessa [141]2 years ago
4 0

Answer:

f(x)=x^{2} +2x-5

Step-by-step explanation:

We need to find the function f(x) representing a quadratic function.

Now, we know that the general form of a quadratic function is ax^{2}+bx+c , where a\neq 0

So, Let us consider each function one by one.

f(x)=-8x^{3}-16x^{2} -4x

Clearly, the above function has a cubic term in it so it is a cubic function NOT a quadratic function.

Now, f(x)=x^{2} +2x-5

Clearly, the above function is of the form ax^{2}+bx+c so, it is a quadratic function.

Now, f(x)=0x^{2} -9x+7

Here, the coefficient of x^{2} is 0. So, it is not of the form ax^{2}+bx+c , where a\neq 0 .

So, it is NOT a quadratic function.

Hence, only f(x)=x^{2} +2x-5  is a quadratic function among the all functions.

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5 0
3 years ago
Let θ
konstantin123 [22]

Answer:

θ is decreasing at the rate of \frac{21}{16} units/sec

or \frac{d}{dt}(θ) = \frac{-21}{16}

Step-by-step explanation:

Given :

Length of side opposite to angle θ is y

Length of side adjacent to angle θ is x

θ is part of a right angle triangle

At this instant,

x =  8 , \frac{dx}{dt} = 7

( \frac{dx}{dt} denotes the rate of change of x with respect to time)

y = 8 , \frac{dy}{dt} = -14

( The negative sign denotes the decreasing rate of change )

Here because it is a right angle triangle,

tanθ = \frac{y}{x}-------------------------------------------------------------------1

At this instant,

tanθ = \frac{8}{8} = 1

Therefore θ = π/4

We differentiate equation (1) with respect to time in order to obtain the rate of change of θ or \frac{d}{dt}(θ)

\frac{d}{dt} (tanθ) = \frac{d}{dt} (y/x)

( Applying chain rule of differentiation for R.H.S as y*1/x)

sec^{2}θ\frac{d}{dt}(θ) = \frac{1}{x}\frac{dy}{dt} - \frac{y}{x*x}\frac{dx}{dt}-----------------------2

Substituting the values of x , y , \frac{dx}{dt} , \frac{dy}{dt} , θ at that instant in equation (2)

2\frac{d}{dt}(θ) = \frac{1}{8}*(-14)- \frac{8}{8*8}*7

\frac{d}{dt}(θ) = \frac{-21}{16}

Therefore θ is decreasing at the rate of \frac{21}{16} units/sec

or  \frac{d}{dt}(θ) = \frac{-21}{16}

3 0
3 years ago
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