Help me ill give 100 pts

Small pizza is for 1 person, Large pizza is for 5 people

Skills needed: System of Equations

Step-by-step explanation:

1) First, let's make two variables:

--> = how many people a large pizza can feed

--> = how many people a small pizza can feed

2) We need to use a system of equations to solve this --> A system is 2 or more equations with specific solutions for the variables involved.

Example:

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We can solve with elimination:

(Rewriting the first equation to look like the second one)

---> Then we add both:

When we add both equations, we eliminate the variable , solving for easily. After this, we can solve for

--->

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This is the elimination method of a system (Note: we can also subtract equations if needed)

3) Let's make a system for this problem:

---> 3 small pizzas (multiplied by the number of people it can feed) + 2 large pizzas (multiplied by the number of people it can feed) is for 13 people.

---> 3 small pizzas (multiplied by the number of people it can feed) + 4 large pizzas (multiplied by the number of people it can feed) is for 23 people.

4) We have two equations:

We can eliminate by subtracting both equations.

--->

We subtracted from which is just 0 (so it does not have to be shown)

We subtracted from which is just

On the right side: subtracted from which is just

----> , divide by -2,

, which means a large pizza is for 5 people.

Substituting into the first equation (it does not matter which one you substitute into) will get

Small pizza is for 1 person.

geometric series converge the common ratio r has magnitude less than one.

which is equivalent to

for the example we need

|2-x| < 1

-1 < 2-x   and 2-x < 1

x < 3 and x > 1

we can write that

answer: 1 < x < 3

1a. common ratio r=250/25=10 > 1     diverges

1b.   r=50/100=1/2 converges

when these converge they converge to

where a is the first term.

answer: 200

1c. r=-1   so it's not the case that |r| < 1.     diverges

2a. x is the common ratio so we converge when |x|< 1

2b.   r=-10x / -2 = 5x.

we converge when

|5x| < 1

|x| < 1/5

2c.   r=3-2x

-1 < 3-2x and 3-2x < 1

-4 < -2x and 2 < 2x

x < 2 and 1 < x

1 < x < 2

2d. [/tex] r=-48x^3/12 = 4x^3[/tex]

|4x^3| < 1

|x^3| < \frac 1 4

|x| < 4^{-1/3}