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4 years ago

F(x) = x2 + 3 Is it an even odd function, both or neither?

Mathematics

2 answers:

Virty [35]4 years ago

7 0

Answer: Both because it depends on the number you put in there

Step-by-step explanation:such as 1^2 +3=4 an even number 2^2+3=7 an odd number

k0ka [10]4 years ago

3 0

Seems to be an even function.

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Simply plz hurry thx

Lunna [17]

The answer to this question is D 17/24

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3 years ago

What expression is equivalent to 18c^8 d^9/9c^3 d^6

Annette [7]

First, since all of the terms are multiplied, you can cancel them out to simplify them:

18/9 = 2

When a term has exponents on the top and bottom of a fraction, you subtract the bottom from the top:

c^8/c^3 = c^5

d^9/d^6 = d^3

So now, we can get rid of the fraction bar and put all of these terms into one expression:

2(c^5)(d^3)

7 0

4 years ago

There\ were\ 750\ questions\ in\ a\ mathematics\ book.\ \ Sam\ answered\ 82\%\ of\ them\ correctly.\ \ how\ many\ questions\ did

aniked [119]

Answer: 135

Step-by-step explanation:

Multiply 750 times .82 and then substract the result from 750

6 0

3 years ago

Write 3 ½% as a fraction

jeka57 [31]

Answer:

7/200

Step-by-step explanation:

3.5% = 3.5/100 = (3.5/100)×(2/2) = 7/200

_____

The % symbol is a shorthand way to write "/100", so 3.5% = 3.5/100.

3 0

4 years ago

Please Help! will give brainlest!

zzz [600]

Answer:

$\theta = \frac{\pi}{3} , \pi, \frac{5\pi}{ 3}$

Step-by-step explanation:

we want to solve the following trigonometric equation:

$\displaystyle - 2 { \sin}^{2} (\theta )+ \cos( \theta) + 1 = 0, \: \theta \in [0,2\pi)$

The first step of solving trigonometric equation is to rewrite the equation in terms of one trigonometric function . With Pythagorean theorem, we know that sin²x=1-cos²x . It will be helpful to rewrite the equation in terms of one trig functions. Therefore, substitute 1-cos² $\theta$ in the place of sin² $\theta$ :

$\displaystyle - 2 (1 - \cos ^{2} ) + \cos( \theta) + 1 = 0, \: \theta \in [0,2\pi)$

simplify:

$\displaystyle\implies - 2 (1 - \cos ^{2} (\theta) + \cos( \theta) + 1 = 0, \: \theta \in [0,2\pi)\\\implies- 2 + \cos ^{2} (\theta)+ \cos( \theta) + 1 = 0, \: \theta \in [0,2\pi)\\\implies\cos ^{2} (\theta)+ \cos( \theta) -1 = 0$

Consider cos² $\theta\implies$ x. Thus,

$2 {x}^{2} + x - 1 = 0$

solving the quadratic equation yields:

${x}^{} = \frac{1}{2} \\ x = - 1$

back-substitute:

$\begin{cases} \cos( \theta) = \dfrac{1}{2} \\ \cos( \theta) = - 1 \end{cases} \theta \in[0,2\pi)$

take inverse trig in both sides

$\implies \begin{cases} \theta = \dfrac{\pi}{3} + 2n\pi\\\theta= \frac{5\pi}{3} + 2n\pi \\ \theta = \pi + 2n\pi\end{cases} \theta \in[0,2\pi) \\\\\implies\theta= \frac{\pi}{3}+\dfrac{2n\pi}{3},\theta\in [0,2\pi)\\\text{when n=0}\\ \implies \theta = \frac{\pi}{3} \\ \text{when n=1}\\ \theta= \pi\\ \text{when n=2}\\\theta=\frac{5\pi}{ 3}$

In conclusion,

$\displaystyle\theta = \frac{\pi}{3} , \pi, \frac{5\pi}{ 3}$

7 0

2 years ago