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vovangra [49]
3 years ago
13

S f(-x)= x^2 -1 odd, even or neither

Mathematics
1 answer:
natka813 [3]3 years ago
5 0

Answer:

  f(x) = f(-x) = x^2 -1 is an <u>even</u> function

Step-by-step explanation:

When f(x) = f(-x), the function is symmetrical about the y-axis. That is the definition of an <em>even</em> function.

___

An odd function is symmetrical about the origin: f(x) = -f(-x).

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PLEASE HELP ANYONE
LUCKY_DIMON [66]
For this problem all you have to do is plug in the value that they is in parenthesis for x. If it says g (x) = x and then it asks g(5) = ?, it is saying what happens if i put 5 in for every x. in this case it would be g (5) = 5. I just replaced x with 5.

So g (-2) we sub -2 for x
g (-2) = -2 (-2)^2 + 3 (-2) - 5
= -2 (4) - 6 - 5
= -19

g (0) = -2 (0) + 3 (0) - 5
= 0 + 0 - t
= -5

g (3) = -2 (3)^2 + 3 (3) - 5
= -18 + 9 - 5
= -14
6 0
4 years ago
A bank is advertising the following savings account.
Vlad1618 [11]

Answer:

6.66%

Step-by-step explanation:

((1 + r/m)^n) - 1

Interest rate , r = 6.5%

m = number of compounding times per period = 4 (quarterly interest)

((1 + 0.065/4)^4) - 1

((1 + 0.01625) ^4) - 1

((1.01625)^4) - 1

1.06660 - 1

= 0.0666

0.0666 * 100%

= 6.66%

8 0
3 years ago
Bella is buying balloons for a back to school celebration. She sees there are 4
loris [4]

Answer:

The first deal is better because it has more balloons for not too much of a price (individually)

Step-by-step explanation:

15*4= 60

60/18= 3.33

Deal 1= 3.33 for each balloon

10*5= 50

50/16=3.125

3 0
3 years ago
Read 2 more answers
Prove that in each triangle the total angles are 180 degrees
dmitriy555 [2]

This is a proof that the angles in a triangle equal 180°:

The top line (that touches the top of the triangle) is
running parallel to the base of the triangle.

So:

<span> <span>angles A are the same </span> <span>angles B are the same </span> </span>

And you can easily see that A + C + B does a complete rotation from one side of the straight line to the other, or <span>180°</span>

8 0
3 years ago
Read 2 more answers
Find the absolute extrema of f(x) = e^{x^2+2x}f ( x ) = e x 2 + 2 x on the interval [-2,2][ − 2 , 2 ] first and then use the com
fredd [130]

f(x)=e^{x^2+2x}\implies f'(x)=2(x+1)e^{x^2+2x}

f has critical points where the derivative is 0:

2(x+1)e^{x^2+2x}=0\implies x+1=0\implies x=-1

The second derivative is

f''(x)=2e^{x^2+2x}+4(x+1)^2e^{x^2+2x}=2(2x^2+4x+3)e^{x^2+2x}

and f''(-1)=\frac2e>0, which indicates a local minimum at x=-1 with a value of f(-1)=\frac1e.

At the endpoints of [-2, 2], we have f(-2)=1 and f(2)=e^8, so that f has an absolute minimum of \frac1e and an absolute maximum of e^8 on [-2, 2].

So we have

\dfrac1e\le f(x)\le e^8

\implies\displaystyle\int_{-2}^2\frac{\mathrm dx}e\le\int_{-2}^2f(x)\,\mathrm dx\le\int_{-2}^2e^8\,\mathrm dx

\implies\boxed{\displaystyle\frac4e\le\int_{-2}^2f(x)\,\mathrm dx\le4e^8}

5 0
3 years ago
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