Answer:
4600
Step-by-step explanation:
This is an arithmetic sequence with common difference d between consecutive terms.
d = - 2 - (- 6) = 2 - (- 2) = 6 - 2 = 4
The sum to n terms of an arithmetic sequence is
= 
[ 2a + (n - 1)d ]
where a is the first term
here a = - 6, d= 4 and n = 50, hence
= 
[ (2 × - 6) + (49 × 4) ]
= 25 ( - 12 + 196 )
= 25 × 184
= 4600
Fraction division is basically multiplication of a reciprocal.
For example (1/4)/(1/8) = (1/4)x(8/1)=2
Multiplication of the answer and the piece that divides (in this case 2 x 1/8), you should get the piece that is being divided (in this case 1/4). If that is correct, then your division is correct.
B. is the answer I believe since most are not functions
9 inches * 1 foot / 12 inches
= 3/4 feet deep
Area of the sidewalk:
= the sides of the pool * the width of the sidewalk + the corners of the sidewalk (width * width)
= 2 sides * 6 (feet / side) * 2 feet + 2 sides * 14 (feet / side) * 2 feet + 4 corners * (2 feet * 2 feet / corner)
= 12 feet * 2 feet + 28 feet * 2 feet + 4 * 2 feet * 2 feet
= 24 feet^2 + 56 feet^2 + 16 feet^2
= 96 feet^2
Volume:
= 96 feet^2 * 3/4 feet
= (96 * 3 / 4) feet^3
= (19 * 3) feet^3
= 57 feet^3
Answer:
Step-by-step explanation:
1. Write an expression to show how to calculate the number of buses needed. Make sure to include parenthesis to show what should be done first since order of operations is important.
(9*25)+4+(2*4)=237 (total number of people going to the museum
Each bus,excluding the driver,holds 44 people
237/44 = 5.3863636363.......
2. How many buses will be needed?
6 buses are needed
3. Why must the answer to the problem be a whole number?
Buses cannot be divided in fractions It should be or 5 buse or 6
But 5 buses are not enough
4. Why shouldn't you round the answer the usual way?
Even when rounded it is still a decimal and buses need to be counted by whole number
5. Can your answer be classified as a rational number? Explain why or why not.
6 is a rational number (whole number)
5.38636363.......is rational ( repeating decimal
