Question

I need help with Graphing Square Root Functions. Please anyone help.

Answer 1

1. What is the domain of the function \[y=3\sqrt{6x+42}\]

Answer 2

do you know that you can not take the square root of a number and get are real answer unless the number is positive or zero?

Answer 3

The choices are

A. x ≥ 0
B. x ≤ 7
C. x ≥ –6
D. x ≥ –7

Answer 4

Yes I know that.

Answer 5

I think that answer is D.

Answer 6

But I was confused because I thought you had to flip the equal to or greater than sign when you divide.

Answer 7

so set up an inequality that mean the terms inside the square root are equal to or greater than zero

Answer 8

you only have to flip the inequality sign when you are dividing by a negative

Answer 9

\[6x+42\ge0\]

Answer 10

thats it ...

Answer 11

\[6x \ge-42\]Subtract 42 get

Answer 12

Divide by 6 and get\[x \le-6\]

Answer 13

Oops I mean\[x \le -7\]

Answer 14

\[42=6\times7\]

Answer 15

Because when you divide you flip the sign, right?

Answer 16

Yes but the 42 is negative.

Answer 17

you dont have to flip the sign in the question, because you were dividing by positive 6

Answer 18

So is it D. and your not supposed to flip the sign?

Answer 19

in this*

Answer 20

Oh....Ok

Answer 21

So it is D.

Answer 22

I thought so.

Answer 23

Not to waste more of you time but that one I had already knew the answer I just wanted to figure out why I kept flipping the sign. So could you help me out with 2 more and then I'm done?

Answer 24

we can check

\[y=3\sqrt{6(-7)+42}=3\sqrt0, \]

Answer 25

Ok cool

Answer 26

Yes, more

Answer 27

2. What are the domain and range of the function \[y=2\sqrt{3x+4}-5\]? (1 point)


3. When will the dependent variable in the equation equal or exceed 4?
(1 point)x ≥ –0.17
x ≥ 45
x ≥ 48
x ≥ 53

Answer 28

so, can you set up the inequality like before?

Answer 29

No because I didn't know what to do with the 5...

Answer 30

Because it is out of the sqrt.

Answer 31

the only restricted terms are either in radicals, denominators or some other strange places,

Answer 32

so just look at the terms under the square root here

Answer 33

Ok so could you set it up fo r me so I learn?

Answer 34

\[3x+4-5\ge0\]

Answer 35

Is that it?

Answer 36

if the 5 isn't in the square root
\[y=2\sqrt{(3x+4)}-5\]
then it wont effect the domain

\[3x+4≥0\]

Answer 37

Ok

Answer 38

so 4 divided by 3?

Answer 39

Thats 1.3 repeating.

Answer 40

\[3x+4≥0\]take away four from both sides\[3x+4-4≥0-4\]divide by three

Answer 41

Oh you leave it at \[\frac{ 4 }{ 3 }\]

Answer 42

-4/3

Answer 43

yeah but it should be -4/3

Answer 44

yeah

Answer 45

so you have \(x≥-\tfrac43\)

Answer 46

so is it \[x \ge -\frac{ 4 }{ 3 };y \ge-5\]

Answer 47

Is that right?

Answer 48

\[\color{red}\checkmark\]

Answer 49

\[x \le -\frac{ 4 }{ 3 };y \le-5\]Or would it be

Answer 50

well when you look at
\[y=2\sqrt{(3x+4)}-5\]
there are two main term
the term with the square root can only work if it equals zero or greater

Answer 51

When will the dependent variable in the equation \[y=\sqrt{x+4}-3\] equal or exceed 4?


x ≥ –0.17
x ≥ 45
x ≥ 48
x ≥ 53

Answer 52

Ok last one.

Answer 53

do you remember if the dependent variable is x or y?

Answer 54

The last three choices, when I plug them in, all range from 4 to 4.56968 or somthing like that.

Answer 55

It would be y because that is the outcome.

Answer 56

so when \(y≥4\)
\[\quad\sqrt{x+4}-3≥4\]

Answer 57

So is it B.?

Answer 58

add three to both sides
then square both side,
the take away four from both sides

Answer 59

you add 3 and thats \[\sqrt{x+4}\ge\]

Answer 60

ok

Answer 61

what happened to RHS?

Answer 62

i forgot to put 3 on the other side

Answer 63

i forgot to put 3 on the other side

Answer 64

so it 3-4 squared?
so its A.

Answer 65

Right?

Answer 66

i got 1 when i did (3-4)^2

Answer 67

um wait

Answer 68

Ok so what did you get?

Answer 69

\[\quad\sqrt{x+4}-3≥4\]add three to both sides
\[\quad\sqrt{x+4}-3+3≥4+3\]then, square both sides,
\[\quad(\sqrt{x+4}-3+3)^2≥(4+3)^2\]
the take away four from both sides
\[\quad(\sqrt{x+4}-3+3)^2-4≥(4+3)^2-4\]

Answer 70

Ok now to simplify

Answer 71

ok its 45

Answer 72

\[x≥\]

Answer 73

I was right when I said b

Answer 74

yes

Answer 75

Thanks fo rall of you help.

Answer 76

1. What is the domain of the function ? (1 point)
(0 pts) x ≥ 0
(0 pts) x ≤ 7
(0 pts) x ≥ –6
(1 pt) x ≥ –7
1 /1 point
2. What are the domain and range of the function ? (1 point)
(1 pt)
(0 pts)
(0 pts)
(0 pts)
1 /1 point
3. When will the dependent variable in the equation equal or exceed 4?
(1 point)
(0 pts) x ≥ –0.17
(1 pt) x ≥ 45
(0 pts) x ≥ 48
(0 pts) x ≥ 53
1 /1 point
4. Which function is shown on the graph below?



(1 point)
(0 pts)
(0 pts)
(1 pt)
(0 pts)
1 /1 point
Note: The below question was entered in error. You will receive credit for ANY answer.
5. The function models the time (t) in seconds that an object has been falling after the object has fallen d feet. When will the time be more than 1 minute?
(1 point)
(1 pt) d ≥ 16 ft
(1 pt) d ≥ 225 ft
(1 pt) d ≥ 3,600 ft
(1 pt) d ≥ 57,600 ft
1 /1 point

The final score is 5/5 (100%).

Answer 77

Bye Im login out

Answer 78

\[5/5=100\%\quad\color{red}{\checkmark\checkmark\checkmark\checkmark\checkmark}\]

Answer 79

@UnkleRhaukus check you mail

Answer 80

your*