Rewrite standard form equation in graphing form and then sketch the graph

Which statement best describes how to determine wether f(x)=x4-x3 is even a function

egoroff_w [7]

X4-x3=1 -4+3
×4 +3
×=1

6 0

3 years ago

For the relation S:{(2,3),(−1,4), (1,6)}. a Find the domain. B. Find range

Rudik [331]

Answer:

domain: {2, -1, 1}

range: {3, 4 6}

Step-by-step explanation:

In the relation of the form (x,y)

Set of all possible value of X is called Domain

Set of all possible value of Y is called Range

To get a list of domain and range it is required to separate value of x and y

_________________________________

Relation given in the problem

S:{(2,3),(−1,4), (1,6)}

domain: {2, -1, 1}

range: {3, 4 6}

8 0

3 years ago

How to solve this <br><br> 4x-3(x-2)=21

Natasha_Volkova [10]

We can start by getting rid of parenthesis:

4x-3x+6=21

Then we can combine like terms:

x=21-6

x=15

So our end product is x=15.

Hope I helped soz if I'm wrong ouo.

~Potato.

Copyright Potato 2019.

4 0

3 years ago

Which is the inverse of the function a(d)=5d-3? And use the definition of inverse functions to prove a(d) and a-1(d) are inverse

Drupady [299]

Answer:

$a'(d) = \frac{d}{5} + \frac{3}{5}$

$a(a'(d)) = a'(a(d)) = d$

Step-by-step explanation:

Given

$a(d) = 5d - 3$

Solving (a): Write as inverse function

$a(d) = 5d - 3$

Represent a(d) as y

$y = 5d - 3$

Swap positions of d and y

$d = 5y - 3$

Make y the subject

$5y = d + 3$

$y = \frac{d}{5} + \frac{3}{5}$

Replace y with a'(d)

$a'(d) = \frac{d}{5} + \frac{3}{5}$

Prove that a(d) and a'(d) are inverse functions

$a'(d) = \frac{d}{5} + \frac{3}{5}$ and $a(d) = 5d - 3$

To do this, we prove that:

$a(a'(d)) = a'(a(d)) = d$

Solving for $a(a'(d))$

$a(a'(d)) = a(\frac{d}{5} + \frac{3}{5})$

Substitute $\frac{d}{5} + \frac{3}{5}$ for d in $a(d) = 5d - 3$

$a(a'(d)) = 5(\frac{d}{5} + \frac{3}{5}) - 3$

$a(a'(d)) = \frac{5d}{5} + \frac{15}{5} - 3$

$a(a'(d)) = d + 3 - 3$

$a(a'(d)) = d$

Solving for: $a'(a(d))$

$a'(a(d)) = a'(5d - 3)$

Substitute 5d - 3 for d in $a'(d) = \frac{d}{5} + \frac{3}{5}$

$a'(a(d)) = \frac{5d - 3}{5} + \frac{3}{5}$

Add fractions

$a'(a(d)) = \frac{5d - 3+3}{5}$

$a'(a(d)) = \frac{5d}{5}$

$a'(a(d)) = d$

Hence:

$a(a'(d)) = a'(a(d)) = d$

7 0

2 years ago

8x^2+18x-5=0 find the discriminant

Shalnov [3]

The formula for the discriminant is the square root of b^2 - 4ac

8x^2+18x-5= 0

a = 8, b=18, c= -5

Square root of 324 +160

Square root of 484 = 22 The discriminant is 22

7 0

3 years ago