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pashok25 [27]
3 years ago
8

A store sells wrapping paper for $3, greeting cards for $2, and ribbon for $1. The soccer

Mathematics
1 answer:
Angelina_Jolie [31]3 years ago
4 0

Answer:

A) (15 x $3) + (10 x $2) + (12 x $1) = X

B) $77      

Step-by-step explanation:

A) Each wrapping paper is 3 dollors, and they bought 15 (15 x 3)

Each greeting card is 2 dollors, and they bought 10 (10 x 2)

Each roll of ribbon is 1 dollors, and they bought 12 (12 x 1)

B) 15 times 3 is 45

10 times 2 is 20

12 x 1 is 12

45 + 20 + 12 = 77

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The total cost of tickets sold is $2016.

Step-by-step explanation:

What I did to get this answer was multiply 168 and 12

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Calculate, to four decimal places, the first ten terms of the sequence. an = 1 +(−4/9)^n
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The first ten terms of the sequence a_n=1 + (-\frac{4}{9} )^n is<em> 0.5556, 1.1975, 0.9122, 1.039, 0.9827, 1.0077, 0.9966, 1.0015, 0.9993 and 1.0003</em>

<h3>What is an equation?</h3>

An equation is an expression that shows the relationship between two or more variables and numbers.

Given that:

a_n=1 + (-\frac{4}{9} )^n\\\\When\ n=1:a_1=1 + (-\frac{4}{9} )^1=0.5556\\\\When\ n=2:a_2=1 + (-\frac{4}{9} )^2=1.1975\\\\When\ n=3:a_3=1 + (-\frac{4}{9} )^3=0.9122\\\\When\ n=4:a_4=1 + (-\frac{4}{9} )^4=1.039\\\\When\ n=5:a_5=1 + (-\frac{4}{9} )^5=0.9827\\\\When\ n=6:a_6=1 + (-\frac{4}{9} )^6=1.0077\\\\When\ n=7:a_7=1 + (-\frac{4}{9} )^7=0.9966\\\\When\ n=8:a_8=1 + (-\frac{4}{9} )^8=1.0015\\\\When\ n=9:a_9=1 + (-\frac{4}{9} )^9=0.9993\\\\

When\ n=10:a_{10}=1 + (-\frac{4}{9} )^{10}=1.0003\\

The first ten terms of the sequence a_n=1 + (-\frac{4}{9} )^n is<em> 0.5556, 1.1975, 0.9122, 1.039, 0.9827, 1.0077, 0.9966, 1.0015, 0.9993 and 1.0003</em>

Find out more on equation at: brainly.com/question/2972832

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