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3 years ago

What is the inverse of X^2-2x+3

1 answer:

4 0

So you have x, then you square it to get x^2, then you minus 2x, then +3
The inverse is going backwards so you -3 then +2x then square root x
So you get square root x + 2x -3
I'm pretty sure this is right, sorry if I'm wrong.

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So i figured out the first but I don't know the rest please help?

Pani-rosa [81]

For the first two, multiply 13 in place of the X. For example, 4(13)+71 = 123.

6 0

2 years ago

Find f(2) and f(3).<br> f(1) = 3<br> f(n) = 7f(n = 1)

Kay [80]

Answer:

do it got a picture

on the edge2020

Step-by-step explanation:

4 0

2 years ago

Using the given zero, find one other zero of f(x). Explain the process you used to find your solution.

IRISSAK [1]

Answer:

One other zero is 2+3i

Step-by-step explanation:

If 2-3i is a zero and all the coefficients of the polynomial function is real, then the conjugate of 2-3i is also a zero.

The conjugate of (a+b) is (a-b).

The conjugate of (a-b) is (a+b).

The conjugate of (2-3i) is (2+3i) so 2+3i is also a zero.

Ok so we have two zeros 2-3i and 2+3i.

This means that (x-(2-3i)) and (x-(2+3i)) are factors of the given polynomial.

I'm going to find the product of these factors (x-(2-3i)) and (x-(2+3i)).

(x-(2-3i))(x-(2+3i))

Foil!

First: x(x)=x^2

Outer: x*-(2+3i)=-x(2+3i)

Inner: -(2-3i)(x)=-x(2-3i)

Last: (2-3i)(2+3i)=4-9i^2 (You can just do first and last when multiplying conjugates)

---------------------------------Add together:

x^2 + -x(2+3i) + -x(2-3i) + (4-9i^2)

Simplifying:

x^2-2x-3ix-2x+3ix+4+9 (since i^2=-1)

x^2-4x+13 (since -3ix+3ix=0)

So x^2-4x+13 is a factor of the given polynomial.

I'm going to do long division to find another factor.

Hopefully we get a remainder of 0 because we are saying it is a factor of the given polynomial.

x^2+1

---------------------------------------

x^2-4x+13| x^4-4x^3+14x^2-4x+13

-( x^4-4x^3+ 13x^2)

------------------------------------------

x^2-4x+13

-(x^2-4x+13)

-----------------

So the other factor is x^2+1.

To find the zeros of x^2+1, you set x^2+1 to 0 and solve for x.

$x^2+1=0$

$x^2=-1$

$x=\pm \sqrt{-1}$

$x=\pm i$

So the zeros are i, -i , 2-3i , 2+3i

7 0

3 years ago

Write your answer as a whole number and remainder.<br> 31+4=R

ki77a [65]

Answer:

Step-by-step explanation:

Add 31 and 4 to get 35.

8 0

3 years ago

Use the fundamental identities and appropriate algebraic operations to simplify the following expression. (18 +tan x) (18-tan x)

andrezito [222]

Answer:

a) $\left(18+\tan \left(x\right)\right)\left(18-\tan \left(x\right)\right)+\sec ^2\left(x\right)=325$

b) The lowest point of $y=\cos \left(x\right)$ , $0\leq x\leq 2\pi$ is when x = $\pi$

Step-by-step explanation:

a) To simplify the expression $\left(18+\tan \left(x\right)\right)\left(18-\tan \left(x\right)\right)+\sec ^2\left(x\right)$ you must:

Apply Difference of Two Squares Formula: $\left(a+b\right)\left(a-b\right)=a^2-b^2$

$a=18,\:b=\tan \left(x\right)$

$\left(18+\tan \left(x\right)\right)\left(18-\tan \left(x\right)\right)=18^2-\tan ^2\left(x\right)=324-\tan ^2\left(x\right)$

$324-\tan ^2\left(x\right)+\sec ^2\left(x\right)$

Apply the Pythagorean Identity $1+\tan ^2\left(x\right)=\sec ^2\left(x\right)$

From the Pythagorean Identity, we know that $1=-\tan ^2\left(x\right)+\sec ^2\left(x\right)$

Therefore,

$324[-\tan ^2\left(x\right)+\sec ^2\left(x\right))]\\324[+1]\\325$

b) According with the below graph, the lowest point of $y=\cos \left(x\right)$ , $0\leq x\leq 2\pi$ is when x = $\pi$

3 0

3 years ago