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Alenkinab [10]
2 years ago
5

F(x)=(x^3-3)÷(x^5) determine if odd or even

Mathematics
1 answer:
VladimirAG [237]2 years ago
8 0
I am pretty sure that it is even
You might be interested in
How do u factor this
PilotLPTM [1.2K]

You can factor a parabola by finding its roots: if

p(x)=x^2+bx+c

has roots x_1,\ x_2, then you have the following factorization:

p(x) = (x-x_1)(x-x_2)

In order to find the roots, you can use the usual formula

x_{1,2} = \dfrac{-b\pm\sqrt{b^2-4ac}}{2a}

In the first example, this formula leads to

x_{1,2} = \dfrac{-2\pm\sqrt{4+4}}{2} = \dfrac{-2\pm\sqrt{8}}{2} = \dfrac{-2\pm2\sqrt{2}}{2} = 1 \pm \sqrt{2}

So, you can factor

x^2-2x-1 = (x-1-\sqrt{2})(x-1+\sqrt{2})

The same goes for the second parabola.

As for the third exercise, simply plug the values asking

f(1.5)=-5.25

you get

f(-1.5) = 1.5c-3 = -5.25

Add 3 to both sides:

1.5c = -2.25

Divide both sides by 1.5:

c = 1.5

7 0
3 years ago
Subtract these polynomials
SCORPION-xisa [38]
So I couldn’t do square roots so I had to take a photo

4 0
2 years ago
Read 2 more answers
#1. WILL REWARD BRAINLIEST
Gemiola [76]

c is the answer for 1'


6 0
3 years ago
If K = (AB)/(A+B) , then B = ?
marysya [2.9K]

Lets do

\\ \sf\longmapsto K=\dfrac{AB}{A+B}

\\ \sf\longmapsto \dfrac{1}{K}=\dfrac{A+B}{AB}

\\ \sf\longmapsto \dfrac{1}{K}=\dfrac{1}{A}+\dfrac{1}{B}

\\ \sf\longmapsto \dfrac{1}{B}=\dfrac{1}{K}-\dfrac{1}{A}

\\ \sf\longmapsto \dfrac{1}{B}=\dfrac{K-A}{AK}

\\ \sf\longmapsto B=\dfrac{AK}{K-A}

6 0
2 years ago
III
irakobra [83]

Answer:

101in

Step-by-step explanation:

9*5=45

9*4=36

5*4=20

45+36+20=101

3 0
2 years ago
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