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Julli [10]
2 years ago
15

Jim had $20 to spend at the video rental. He has a late fee of $4.00 and it will cost him $2.99 to rent each movie. Not includin

g tax, what inequality can Jim write to solve for the most number of movies,m, he can rent?
Mathematics
2 answers:
hammer [34]2 years ago
7 0

Answer:

he pays the late fee that gives him 16 dollers it cost 3 dollers for  1 video 16 divided by 3 equals 5 so 5 movies with 1 extra doller

Step-by-step explanation:

zubka84 [21]2 years ago
4 0

Answer:

20 > 4 + 2.99m

Step-by-step explanation:

Jim had to pay $4 adding $2.99 for each movie and that amount must be smaller than $20

You might be interested in
. Which numbers are all divisible by 5?
Lisa [10]
I think that the answer is #2. 
6 0
3 years ago
Read 2 more answers
Can someone please answer this please!
DaniilM [7]

Answer:

T = ±22

Step-by-step explanation:

Let's solve your equation step-by-step.

0=−16t2+7744

Step 1: Add 16t^2 to both sides.

0+16t2=−16t2+7744+16t2

16t2=7744

Step 2: Divide both sides by 16.

16t2

16

=

7744

16

t2=484

Step 3: Take square root.

t=±√484

t=22 or t=−22

6 0
3 years ago
Can you guys help me I need to turn this in at 11:00pm :( . <br> Don’t answer it if you don’t know .
Ostrovityanka [42]
Answers :

1.) Linear

2.) non-linear

3.) non-linear

A linear function is a line that creates a straight line. A non-linear function is a line that creates anything besides a straight line.
6 0
2 years ago
If f(x) = 2x² +5 and g(x)=x2-2, find (f-g)(x).
Rufina [12.5K]

Answer:D. x^2+7

Step-by-step explanation:

If f(x) = 2x² +5 and g(x)=x2-2, find (f-g)(x).A. x^2+3B. 3x^2+3C. 3x^2+7D. x^2+7

4 0
1 year ago
Does anyone know how to do this?? Help please!!!!
Doss [256]

Answer:

When we have a rational function like:

r(x) = \frac{x + 1}{x^2 + 3}

The domain will be the set of all real numbers, such that the denominator is different than zero.

So the first step is to find the values of x such that the denominator (x^2 + 3) is equal to zero.

Then we need to solve:

x^2 + 3 = 0

x^2 = -3

x = √(-3)

This is the square root of a negative number, then this is a complex number.

This means that there is no real number such that x^2 + 3 is equal to zero, then if x can only be a real number, we will never have the denominator equal to zero, so the domain will be the set of all real numbers.

D: x ∈ R.

b) we want to find two different numbers x such that:

r(x) = 1/4

Then we need to solve:

\frac{1}{4} = \frac{x + 1}{x^2 + 3}

We can multiply both sides by (x^2 + 3)

\frac{1}{4}*(x^2 + 3) = \frac{x + 1}{x^2 + 3}*(x^2 + 3)

\frac{x^2 + 3}{4} = x + 1

Now we can multiply both sides by 4:

\frac{x^2 + 3}{4}*4 = (x + 1)*4

x^2 + 3 = 4*x + 4

Now we only need to solve the quadratic equation:

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

x^2 - 4*x - 1 = 0

We can use the Bhaskara's formula to solve this, remember that for an equation like:

a*x^2 + b*x + c = 0

the solutions are:

x = \frac{-b +- \sqrt{b^2 - 4*a*c} }{2*a}

here we have:

a = 1

b = -4

c = -1

Then in this case the solutions are:

x = \frac{-(-4) +- \sqrt{(-4)^2 - 4*1*(-1)} }{2*(1)} = \frac{4 +- 4.47}{2}

x = (4 + 4.47)/2 = 4.235

x = (4 - 4.47)/2 = -0.235

5 0
2 years ago
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