For her ninth birthday, isabella’s best friend, jackson, gave her a full bag of m& m’s. isabella determined she had 200 m& m’s in the bag. since isabella loves candy, she ate half of the m& m’s the first day. her mom told her if she continues to eat the m& m’s at that rate, they would all be gone by tomorrow. isabells decided she would only eat half of the candy each day and this would make it last forever! how much candy does isabella’s have at the end of 7 days? will the candy really last forever? include the following information in your final 1. a data table of values 2. write an equation to represent the data provided in the problem 3. a hand graphed graph of the data table 4. explain your findings mathematically and compare and contrast them to what you just graphed. 5. use complete sentences and thoughts when answering the following questions. 1. do you think is

1) 25/8 or 3 1/8  M&Ms 2) No, it won't last forever 3)4) The quantity of chocolate decreases exponentially at a rate of 1/2, for a each period of time.

Step-by-step explanation:

This problem works like a Geometric Sequence, whose quotient is 1/2.

Supposing that Isabella has eaten 1/2 and has bee each day eating only half of what's been left.

Setting on a Table:

Day | Candy

1       |  200

2      | 100

3      |  50

4      |  25

5      |  12.5

6      | 6.25

7      |  3.125

2) The equation used is the Recursive Formula for the Geometric Sequence, adjusted to the data:

3) Check the graph below

4) The quantity of chocolate decreases exponentially at a rate of 1/2, for a each period of time. Each day the amount goes getting smaller and smaller till it gets to zero. Then you can insert more values in the Domain you'll get zero, this is the reason why the curve goes closer and closer until gently intercepts x-axis.

answer: ouch. thats difficult

step-by-step explanation:

option a, c,d are correct.

step-by-step explanation:

from the given figure, it is given that z is equidistant from the sides of the triangle rst, then from triangle tzb and triangle szb, we have

tz=sz(given)

bz=zb(common)

therefore, by rhs rule,δtzb ≅δszb

by cpctc, sz≅tz

also, from δctz and δasz,

tz=sz(given)

∠tcz=∠saz(90°)

by rhs rule, δctz ≅ δasz, therefore by cpctc, ∠ctz≅∠asz

also,from δasz and δzsb,

zs=sz(common)

∠zbs=∠saz=90°

by rhs rule, δasz ≅δzsb, therefore, by cpctc, ∠asz≅∠zsb

hence, option a, c,d are correct.