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shepuryov [24]
3 years ago
12

Which function is a quadratic function?

Mathematics
2 answers:
BigorU [14]3 years ago
8 0
It's the first oneeeeeeeeeeeeeee

inessss [21]3 years ago
6 0

Answer:

u(x) is the only quadratic function.

Step-by-step explanation:

We are given four algebraic functions in x and we have to find which one is a quadratic function.

Let us arrange those polynomials in descending powers of x.

i) u(x) = 3x^2-x-8

The highest degree is 2.  Hence this is quadratic function.

ii) v(x) = 8x^3+2x^2+9x

The highest degree is 3 and hence this is cubic

iii) y(x) = 3x^5+x^2+4

The highest degree is 5 and hence this cannot be quadratic

iv) z(x) = 2x^3+7x^2-3

The highest degree is 3 and hence this is cubic

Hence u(x) is the only quadratic function.

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Answer:

add a peireiod

Step-by-step explanation:

7 0
3 years ago
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It takes 48 minutes for Mickey to walk 6 laps around the park. How many minutes does it take Mickey to walk 1 lap?
Feliz [49]

Since it takes 48 minutes to walk 6 laps, you can divide 48 by 6 to get how long it takes for him to walk one lap. Mickey will take 8 minutes to walk 1 lap.

48÷6=8

6 0
3 years ago
Which is the correct input-output table for the function f(x) = 3x2 - 1+4?
krok68 [10]

Answer:

the question is incorrect. please write your points as coordinates or scan the diagram.

4 0
2 years ago
Use the formula d = ( r − c ) t to find c if d = 6 , r = 4 , and t = 2 . c=
Minchanka [31]

First, we are going to want to plug in the values we are given. In this case, we will end up with the equation:

6 = (4 - c) 2


From here, we can solve the equation to find c:

6 = 2(4 - c)

  • Apply the commutative property to rearrange the terms on the right-side of the equation to make the distributive property more apparent

6 = 8 -2c

  • Apply the distributive property

-2 = -2c

  • Subtract 8 from both sides of the equation

1 = c

  • Divide both sides of the equation by -2

We have found that c = 1.

7 0
3 years ago
How do you simplify <img src="https://tex.z-dn.net/?f=%5Cfrac%7B%5Csqrt%7B2%7D%20%2B%5Csqrt%7B6%7D%20%7D%7B%5Csqrt%7B8%7D%20%2B%
blondinia [14]

The trick is to exploit the difference of squares formula,

a^2-b^2=(a-b)(a+b)

Set a = √8 and b = √6, so that a + b is the expression in the denominator. Multiply by its conjugate a - b:

(\sqrt8+\sqrt6)(\sqrt8-\sqrt6)=(\sqrt8)^2-(\sqrt6)^2=8-6=2

Whatever you do to the denominator, you have to do to the numerator too. So

\dfrac{\sqrt2+\sqrt6}{\sqrt8+\sqrt6}=\dfrac{(\sqrt2+\sqrt6)(\sqrt8-\sqrt6)}{(\sqrt8+\sqrt6)(\sqrt8-\sqrt6)}=\dfrac{(\sqrt2+\sqrt6)(\sqrt8-\sqrt6)}2

Expand the numerator:

(\sqrt2+\sqrt6)(\sqrt8-\sqrt6)=\sqrt{2\cdot8}+\sqrt{6\cdot8}-\sqrt{2\cdot6}-(\sqrt6)^2

(\sqrt2+\sqrt6)(\sqrt8-\sqrt6)=\sqrt{16}+\sqrt{48}-\sqrt{12}-6

(\sqrt2+\sqrt6)(\sqrt8-\sqrt6)=4+\sqrt{48}-\sqrt{12}-6

(\sqrt2+\sqrt6)(\sqrt8-\sqrt6)=-2+\sqrt{12}(\sqrt4-1)

(\sqrt2+\sqrt6)(\sqrt8-\sqrt6)=-2+\sqrt{12}(2-1)

(\sqrt2+\sqrt6)(\sqrt8-\sqrt6}=-2-\sqrt{12}

So we have

\dfrac{\sqrt2+\sqrt6}{\sqrt8+\sqrt6}=-\dfrac{2+\sqrt{12}}2

But √12 = √(3•4) = 2√3, so

\dfrac{\sqrt2+\sqrt6}{\sqrt8+\sqrt6}=-\dfrac{2+2\sqrt3}2=\boxed{-1-\sqrt3}

7 0
3 years ago
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