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lana66690 [7]
3 years ago
14

Two cell phone companies have different rate plans. Runfast has monthly charges $10 plus $4 per gig of data. B A &D’s monthl

y charge is $6 plus $6 per gig of data. Your task is to determine under what circumstances each company has the better pricing. You will derive and solve a set of equations, graph those equations, and evaluate the meanings for these events. Don’t worry; I have a series of questions to guide you through the process. Good luck!
1. (1pt) Determine the equation for the monthly charges for Runfast and B A&D.

Runfast= ______________________

BA&D=________________________

2. (4 pts)Use your mathematical skill to solve this system of equations using the substitution method. (show your work here)
Mathematics
1 answer:
leva [86]3 years ago
4 0
Let x be the number of gb used.

Question (1):
Runfast = 10 + 4x
<span>BA&D= 6 + 6x

Question (2):
</span> 10 + 4x = 6 + 6x
2x = 4
x = 2

Both companies will cost the same if you use 2 Gb of data.
BA&D will be cheaper if you use less than 2 Gb of data.
Runfast will be cheaper if you use more than 2 Gb of data.

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<u>Answer: </u>

Required five terms of sequence are 19 , 12 , 5 , -2 and -9 .

<u> Solution: </u>

Need to find the five terms of the sequence.

Given recursive rule is f(x) = f(x-1) -7

Substituting x = 2 , f(2) = f(2-1)-7

= f(2) = f(1) – 7    ------(1)

Also given that f(2) = 12.  

On substituting the given value of f(2) in eq (1) we get

12 = f(1) – 7

f(1) = 12 + 7 = 19

Using given recursive rule and given value of f(2) calculating f(3)

Substituting x = 3 ,  

f(3) = f(3-1) – 7  

= f(2) – 7  

= 12 – 7  

= 5

Using given recursive rule and calculated  value of f(3) calculating f(4)

Substituting x = 4,  

f(4) = f(4-1) – 7  

= f(3) – 7  

= 5– 7  

= -2

Using given recursive rule and calculated  value of f(4) calculating f(5)

Substituting x = 5,  

f(5) = f(5-1) – 7  

= f(4) – 7  

= -2– 7  

= -9

Hence required five terms of sequence are 19 , 12 , 5 , -2 and -9 .

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20 feet of ribbon, cutting ribbon into 7 1/2 inches or 6 3/4 inch lengths to make bracelets. Write an algebraic expression that
IRINA_888 [86]

<u>ANSWER:  </u>

The algebraic expression for number of ribbons is \frac{20 \text { feet }-\left\{\frac{20 \text { feet }}{x}\right\}}{x} and the algebraic expression for length of ribbon is \frac{25 x}{3} \text { feet }

<u>SOLUTION: </u>

Given, 20 feet of ribbon, cutting ribbon into 7\frac{1}{2} inches or 6\frac{3}{4} inch lengths to make bracelets.

Let us convert the mixed fraction to improper fractions.

7 \frac{1}{2}=\frac{7 \times 2+1}{2}=\frac{15}{2} \text { and } 6 \frac{3}{4}=\frac{6 \times 4+3}{4}=\frac{27}{4}

Let, the length of bracelets can be made be “x”

First we have to remove the excess length of ribbon and then we have to divide the remaining with length of bracelet we want.

\text { number of ribbons }=\frac{\text { length of ribbon-excess ribbon.}}{\text {length of bracelet}}

\left.=\frac{20 \text { feet }-\left\{\frac{20 \text { feep }}{x}\right.}{x} \text { [we know that, }\{x\} \text { is fractional part of } x\right ]

So, the algebraic expression for number of ribbons is  \frac{20 \text { feet }-\left\{\frac{20 \text { feet }}{x}\right\}}{x}

Now, let us find the algebraic expression for feet of ribbon to make 100 ribbons.

We have 100 bracelets of length x, then total length = 100 × x

length of ribbon = 100x inches

\begin{array}{l}{=100 \mathrm{x} \times \frac{1}{12} \text { feet }} \\\\ {=\frac{25 x}{3} \text { feet }}\end{array}

So, the algebraic expression for length of ribbon is \frac{25 x}{3} \text { feet }

Hence, the algebraic expression for number of ribbons is  \frac{20 \text { feet }-\left\{\frac{20 \text { feet }}{x}\right\}}{x} and the algebraic expression for length of ribbon is \frac{25 x}{3} \text { feet }

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