body{margin:0;} noscript{z-index: 10000; position: fixed; display: block; height: 100%; width: 100%; background: #fff;} noscript .fa{font-size: 40px; display:block; color: #daa520;} noscript div{margin-top: 6%; padding-left: 20px; padding-right: 20px; text-align: center;} ! You must have JavaScript enabled to use this site.

Big Ideas Math Integrated I, 2016

BI

Big Ideas Math Integrated I, 2016 View details

Chapter Test

Finish the assignment

Continue to next subchapter

Exercise 8 Page 211

Practice makes perfect

a We are told that the first row has 42 seats and every row after has 3 more than the previous one. We can model this situation using a linear function in slope-intercept form, where the number of seats depends on the row number.

s(r)=mr+b The slope of our function will be equal to the increase of seats per row 3. s(r)= 3r+b In order to find b we will substitute the only point we know, ( 1, 42), into the function.

s(r)=3r+b

r= 1, s(r)= 42

42=3( 1)+b

▼

Solve for b

Identity Property of Multiplication

42=3+b

LHS-3=RHS-3

39=b

Rearrange equation

b=39

The function is s(r)=3r+39. Now we can calculate the number of seats in the 25th row.

s(r)=3r+39

r= 25

s( 25)=3( 25)+39

Multiply

s(25)=75+39

Add terms

s(25)=114

There are 114 seats in the Row 25.

b We can find the row with 90 seats using the function from Part A.

s(r)=3r+39 Let's substitue 90 for s(r) and calculate r.

s(r)=3r+39

s(r)= 90

90=3r+39

▼

Solve for r

LHS-39=RHS-39

51=3r

.LHS /3.=.RHS /3.

17=r

Rearrange equation

r=17

Therefore, there are 90 seats in the 17th row.

Subchapter links

Exercises p.211