Ask question

Ask question

All categories

3 years ago

14

Which function is shown in the graph?

2 answers:

marta [7]3 years ago

7 0

Answer:

f(x)=log5x

x>0

all real number

Step-by-step explanation:

ladessa [460]3 years ago

5 0

Answer:

The next two parts are Domain: x>0 and Range: all real numbers

Step-by-step explanation:

Edge2020

You might be interested in

Sherry needs $120 by the end of the month. So far, she has only raised one-fourth of the amount that she needs. How much money h

nexus9112 [7]

The answer is C. Hope this helps

4 0

3 years ago

Read 2 more answers

Helppppp question 21,22

Anna35 [415]

21. B or 6

22. B or 2a/h

3 0

1 year ago

Select the correct word to describe the function. the equation y = startfraction 5 x cubed over 8 endfraction represents functio

balandron [24]

The functions and their descriptions are:

Polynomial: $y = \frac{5x^3}{8}$ and $y = 5x^3 + 8$
Logarithm: $y = 6 + 0.25\ln(x)$
Exponential $y = \frac{18}{1 + 6e^{-0.25x}}$

<h3>How to describe the functions?</h3>

The functions are given as:

$y = \frac{5x^3}{8}$

$y = 6 + 0.25\ln(x)$

$y = \frac{18}{1 + 6e^{-0.25x}}$

$y = 5x^3 + 8$

The functions would be described based on the type of function they represent.

Functions that use the "ln" keyword are logarithmic functions, while functions that use the "e" keyword are exponential functions.

Read more about functions at:

brainly.com/question/4025726

#SPJ1

7 0

2 years ago

Someone please help me I need help .

natta225 [31]

Answer:

It’s the last option; 10/24 and 20/24

3 0

3 years ago

Given the quadratic function f(x) = 4x^2 - 4x + 3, determine all possible solutions for f(x) = 0

solong [7]

Answer:

The solutions to the quadratic function are:

$x=i\sqrt{\frac{1}{2}}+\frac{1}{2},\:x=-i\sqrt{\frac{1}{2}}+\frac{1}{2}$

Step-by-step explanation:

Given the function

$f\left(x\right)\:=\:4x^2\:-\:4x\:+\:3$

Let us determine all possible solutions for f(x) = 0

$0=4x^2-4x+3$

switch both sides

$4x^2-4x+3=0$

subtract 3 from both sides

$4x^2-4x+3-3=0-3$

simplify

$4x^2-4x=-3$

Divide both sides by 4

$\frac{4x^2-4x}{4}=\frac{-3}{4}$

$x^2-x=-\frac{3}{4}$

Add (-1/2)² to both sides

$x^2-x+\left(-\frac{1}{2}\right)^2=-\frac{3}{4}+\left(-\frac{1}{2}\right)^2$

$x^2-x+\left(-\frac{1}{2}\right)^2=-\frac{1}{2}$

$\left(x-\frac{1}{2}\right)^2=-\frac{1}{2}$

$\mathrm{For\:}f^2\left(x\right)=a\mathrm{\:the\:solutions\:are\:}f\left(x\right)=\sqrt{a},\:-\sqrt{a}$

solving

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

$x-\frac{1}{2}=\sqrt{-1}\sqrt{\frac{1}{2}}$ ∵ $\sqrt{-\frac{1}{2}}=\sqrt{-1}\sqrt{\frac{1}{2}}$

as

$\sqrt{-1}=i$

so

$x-\frac{1}{2}=i\sqrt{\frac{1}{2}}$

Add 1/2 to both sides

$x-\frac{1}{2}+\frac{1}{2}=i\sqrt{\frac{1}{2}}+\frac{1}{2}$

$x=i\sqrt{\frac{1}{2}}+\frac{1}{2}$

also solving

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

$x-\frac{1}{2}=-i\sqrt{\frac{1}{2}}$

Add 1/2 to both sides

$x-\frac{1}{2}+\frac{1}{2}=-i\sqrt{\frac{1}{2}}+\frac{1}{2}$

$x=-i\sqrt{\frac{1}{2}}+\frac{1}{2}$

Therefore, the solutions to the quadratic function are:

$x=i\sqrt{\frac{1}{2}}+\frac{1}{2},\:x=-i\sqrt{\frac{1}{2}}+\frac{1}{2}$

4 0

2 years ago