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10. on the first day of school, mr. thibodeaux gave his third - grade s…

Question

  1. on the first day of school, mr. thibodeaux gave his third - grade students 9 new spelling words to learn. on each day of school after that, he gave the students 6 new spelling words. how many new spelling words had he given the students by the end of the 30th day of school?

f. 174
g. 180
h. 183
j. 189
k. 195

  1. what is the solution to the equation \\(\frac{3x - 6}{2}+5 = 18\\)?

a. \\(\frac{16}{3}\\)
b. \\(\frac{32}{3}\\)
c. \\(\frac{37}{3}\\)
d. \\(\frac{38}{3}\\)
e. \\(\frac{52}{3}\\)

  1. while her mother drives their car along the highway, mia is noticing some of the mile marker signs. she sees mile marker 117 at noon, and exactly 20 minutes later she sees mile marker 97. what is the average speed, in miles per hour, of their car over these 20 minutes?

f. 23.4
g. 39
h. 60
j. 70
k. 100

Explanation:

Step1: Calculate words after first day

There are \(30 - 1=29\) days after the first - day. Each of these 29 days has 6 new words. So the number of words from these 29 days is \(29\times6 = 174\).

Step2: Add first - day words

On the first day, there are 9 new words. So the total number of new words by the end of the 30th day is \(174 + 9=183\).

Step1: Isolate the fraction

For the equation \(\frac{3x - 6}{2}+5 = 18\), first subtract 5 from both sides: \(\frac{3x - 6}{2}=18 - 5=13\).

Step2: Eliminate the denominator

Multiply both sides of the equation \(\frac{3x - 6}{2}=13\) by 2 to get \(3x-6 = 26\).

Step3: Solve for x

Add 6 to both sides: \(3x=26 + 6=32\). Then divide both sides by 3, so \(x=\frac{32}{3}\).

Step1: Calculate distance traveled

The distance traveled is the difference in mile - markers. So \(d=117 - 97 = 20\) miles.

Step2: Convert time to hours

The time \(t = 20\) minutes. Since 1 hour = 60 minutes, \(t=\frac{20}{60}=\frac{1}{3}\) hours.

Step3: Calculate speed

Using the formula \(v=\frac{d}{t}\), substitute \(d = 20\) miles and \(t=\frac{1}{3}\) hours. So \(v=\frac{20}{\frac{1}{3}}=60\) miles per hour.

Answer:

H. 183