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Fynjy0 [20]
3 years ago
5

What is the range of the function on the graph?

Mathematics
2 answers:
Gre4nikov [31]3 years ago
8 0

Answer:

All the real numbers greater than or equal to –3

Step-by-step explanation:

Because the rounded part at the bottom is exactly at -3 (y), which means that anything greater than it can also be a y

This is the right answer pls believe

vagabundo [1.1K]3 years ago
7 0

Answer:

All the real number greater than or equal to 2

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5^x + 5^x+1= 150<br> Find x<br> Help me pls i need soon<br> Tksss
raketka [301]

Answer:

x=2

Step-by-step explanation:

So we have the equation:

5^x+5^{x+1}=150

First, separate the second term:

5^x+(5\cdot5^x)=150

Let u equal 5ˣ. So:

u+5u=150

Combine like terms:

6u=150

Divide both sides by 6:

u=25

Substitute back u:

5^x=25

Take the base 5 logarithm of both sides:

\log_5(5^x)=\log_5(25)

The left cancels:

x=\log_5(25)

Evaluate the right:

x=2

So, our answer is 2.

And we're done!

5 0
3 years ago
Camille collects stickers. Her sticker book holds 5 stickers in each row. When a page is full, it holds 65 stickers. How many ro
bezimeni [28]
There are 13 rows because 65 divided by 5 is equal to 13.
5 0
3 years ago
Read 2 more answers
Part 3 - Discussion/Explanation Question
SpyIntel [72]

Step-by-step explanation:

Vertical asymptote can be Identites if there is a factor only in the denominator. This means that the function will be infinitely discounted at that point.

For example,

\frac{1}{x - 5}

Set the expression in the denominator equal to 0, because you can't divide by 0.

x - 5 = 0

x = 5

So the vertical asymptote is x=5.

Disclaimer if you see something like this

\frac{(x - 5)(x + 3)}{(x - 5)}

x=5 won't be a vertical asymptote, it will be a hole because it in the numerator and denominator.

Horizontal:

If we have a function like this

\frac{1}{x}

We can determine what happens to the y values as x gets bigger, as x gets bigger, we will get smaller answers for y values. The y values will get closer to 0 but never reach it.

Remember a constant can be represent by

a \times  {x}^{0}

For example,

1 = 1 \times  {x}^{0}

2 =  2 \times {x}^{0}

And so on,

and

x =  {x}^{1}

So our equation is basically

\frac{1 \times  {x}^{0} }{ {x}^{1} }

Look at the degrees, since the numerator has a smaller degree than the denominator, the denominator will grow larger than the numerator as x gets larger, so since the larger number is the denominator, our y values will approach 0.

So anytime, the degree of the numerator < denominator, the horizontal asymptote is x=0.

Consider the function

\frac{3 {x}^{2} }{ {x}^{2}  + 1}

As x get larger, the only thing that will matter will be the leading coefficient of the leading degree term. So as x approach infinity and negative infinity, the horizontal asymptote will the numerator of the leading coefficient/ the leading coefficient of the denominator

So in this case,

x =  \frac{3}{1}

Finally, if the numerator has a greater degree than denominator, the value of horizontal asymptote will be larger and larger such there would be no horizontal asymptote instead of a oblique asymptote.

8 0
2 years ago
an employee adds 160 fluid ounces of chemical to a feature that holds 120,000 gallons of water. did the employee add yhe correct
Hitman42 [59]
We are given a volume of 160 fluid ounces of chemical which is added to a container that holds 120,000 gallons of water. Assuming that the chemical has the same density as water, we just need to convert 120,000 gallons to ounces.

A conversion factor is taken from literature, 1 gallon is equivalent to 128 fluid ounces. So 160 fluid ounces is only 1.25 gallons, thus occupying minimal space in the container. The employee could add more of the chemical in the container. He can actually add 15360000 fluid ounces in total. 
5 0
3 years ago
Who is the person in the end of intergalctic?
andrew11 [14]
Its lil turbo and mr beef cKE

6 0
3 years ago
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